States Of Matter Gases And Liquids - Class 11 Medical Chemistry - Extra Questions

Explain, why Dalton's law of partial pressure can be applied only to a mixture of non-reacting gases.

Dalton's law of partial pressure states that in a mixture of non reacting gases, the total pressure exerted is equal to the sum of the partial pressure of the individuals gases. and is related to the ideal gas law


Equal volumes of all gases contain equal number of molecules under similar conditions of temperature and pressure.
If true enter 1, else enter 0.

Avogadro's Law is the relation which states that at the same temperature and pressure, equal volumes of all gases contain equal number of molecules.
198474_98471_ans.gif


One of the assumptions of kinetic theory of gases is that there is no force of attraction between the
molecules of a gas.

Kinetic theory makes many assumptions in order to explain the reasons gases act the way they do. According to kinetic theory:
1.Gases consist of particles in constant, random motion. They continue in a straight line until they collide with something—usually each other or the walls of their container. 
2.Particles are point masses with no volume. The particles are so small compared to the space between them, that we do not consider their size in ideal gases.
3.No molecular forces are at work. This means that there is no attraction or repulsion between the particles.
4.Gas pressure is due to the molecules colliding with the walls of the container. All of these collisions are perfectly elastic, meaning that there is no change in energy of either the particles or the wall upon collision.  No energy is lost or gained from collisions.
5.The time it takes to collide is negligible compared with the time between collisions.
6.The kinetic energy of a gas is a measure of its Kelvin temperature. Individual gas molecules have different speeds, but the temperature and kinetic energy of the gas refer to the average of these speeds.
7.The average kinetic energy of a gas particle is directly proportional to the temperature.  An increase in temperature increases the speed in which the gas molecules move.
8.All gases at a given temperature have the same average kinetic energy.
9.Lighter gas molecules move faster than heavier molecules.


What are the assumptions of the kinetic molecular theory?

Kinetic molecular theory is based on the following postulates, or assumptions:
1. Gases are composed of a large number of particles that behave like hard, spherical objects in a state of constant, random motion.
2. These particles move in a straight line until they collide with another particle or the walls of the container.
3. These particles are much smaller than the distance between particles. Most of the volume of a gas is therefore empty space.
4. There is no force of attraction between gas particles or between the particles and the walls of the container.
5. Collisions between gas particles or collisions with the walls of the container are perfectly elastic. None of the energy of a gas particle is lost when it collides with another particle or with the walls of the container.
6. The average kinetic energy of a collection of gas particles depends on the temperature of the gas and nothing else.


The volume occupied by the molecules of an ideal gas is :

An ideal gas is defined to be a system in which there are no intermolecular/interatomic forces. Such a system can only exist as a gas. Any real system will approach ideal gas behaviour in the limit that the pressure is extremely low and the temperature is high enough to overcome attractive intermolecular forces. An ideal gas is a gas to go which the laws of Boyle and Charles are strictly applicable under all conditions of temperatures and pressures and volume occupied by the molecules of an ideal gas is zero.


Water vapours have more energy than water at same temperature.

Liquid water molecules have less freedom to move around. While molecules of water vapour have more freedom to move around as there is larger space between the molecules. Hence, the kinetic energy of water molecules is more as compared to the kinetic energy of liquid water.


Molarity of a substance ______ with increase in temperature.

Molarity is effected by the temperature molarity is the ratio between number of moles of solute to the volume of solution in lines and as on increasing temperature, the volume increases and hence with the increase in temperature, the molarity of solution decreases.

Molarity of a substance decreases with increase in temperature.




The units of molarity is ______.

Molarity (M): This unit of concentration relates the moles of solute per litre of solution that is mol lit$$^{ -1 } $$.
The units of molarity is mol lit$$^{ -1 } $$.



The $$P, \ V$$ measurements for $$100\ g$$ of phosgene gas at $$25^{\circ}C$$ are given below:
$$P/ atm$$$$V/ dm^{3}$$$$PV/ atm\ dm^{3}$$
$$0.1$$$$244.5$$$$24.45$$
$$0.4$$$$61.02$$$$24.42$$
$$2.0$$$$12.17$$$$24.34$$
$$8.0$$$$2.925$$$$23.40$$
$$9.8$$$$0.020$$$$0.20$$
$$20.0$$$$0.020$$$$0.40$$
By considering the above data and how inter-molecular forces affect the behaviour of gases, explain whether the sample of phosgene gas in behaving ideally.

Phosgene
Molecular wt. of phosgene(carbonyl chloride)
$$=99$$gm
$$\therefore$$ moles of phosgene(given)$$=\dfrac{100}{99}-1.0101$$
at temperature$$=25^o$$C(constant)
p/atm          v/$$dm^3$$          pv/atm $$dm^3$$
$$0.1$$              $$244.5$$            $$24.45$$
$$0.4$$              $$61.02$$            $$24.42$$
$$2.0$$              $$12.12$$            $$24.34$$
$$8.0$$              $$2.925$$            $$23.40$$
The above data shows at constant temperature the gas behave ideally(almost) by following
$$p_1v_1=p_2v_2=pv=$$constant
p/atm         $$v/dm^3$$             pv/atm $$dm^3$$
$$9.8$$              $$0.020$$                $$0.20$$
$$20.0$$            $$0.020$$                $$0.40$$
As volume become constant for above two data so as $$p_1v_1=p_2v_2$$ is not so followed but separately as $$P_1=9.8$$ and $$P_2=20.0$$
and then $$p_1v_1=\dfrac{p_2v_2}{2}$$, thus gas do not behave ideally.

1115026_826951_ans_18dd21e9441d44be9be33e39e8e125ab.jpg


The vapour pressure of water at $$80^{\circ}C$$ is 355 torr. A 100 ml vessel contained water - saturated oxygen at $$80^{\circ}C$$, the total gas pressure being 730 torr. The contents of the vessel were pumped into 50.0 ml, vessel at the same temperature. What were partial pressures of oxygen and of vapour and the total pressure in the final equilibrium state ? Neglect the volume of any water might condense.

Vapour pressure of water remains constant in $$100ml$$ & $$50ml$$ as temperature remains constant.
$$P_T=P_{O_2}+P_{V.P. ofH_2O}$$
$$\Rightarrow 730= P_{O_2}+355$$
$$\Rightarrow P_{O_2}= 730-355=375mm$$
Now, $$P_1V_1=P_2V_2$$
$$\Rightarrow 375 \times 100= P_2 \times 50$$
$$\Rightarrow P_2= \cfrac {375 \times 100}{50}= 750 mm$$= Partial pressure of oxygen
Now, $$P_T$$ in $$50ml$$ container
   = $$P_2+$$ Vapour pressure of water
   =$$750+355$$
   =$$1105mm$$.


Write ideal gas equation for '$$n$$' moles of the gas.

Ideal gas equation- The volume (V) occupied by the $$n$$ moles of any gas has pressure(P) and temperature (T) Kelvin,
the relationship for these variables $$PV = nRT$$ where $$R$$ gas constant is called the ideal gas law.


Write any three postulates of Kinetic theory of gases.

Postulates of the kinetic theory of gas-
  • The molecules in a gas are small and very far apart. Most of the volume which a gas occupies is empty space,
  • Gas molecules are in constant random motion. Just as many molecules are moving in one direction as in any other,
  • Molecules can collide with each other and with the walls of the container. Collisions with the walls account for the pressure of the gas,
  • When collisions occur, the molecules lose no kinetic energy; that is, the collisions are said to be perfectly elastic.


What mass of hydrogen peroxide will be present in 2 litres of a 5 molar solution? Calculate the mass of oxygen which will be liberated by decomposition of 200 ml of this solution. 

$$Molarity = \dfrac{Wt}{Mwt \times Volume}$$
$$5   =\dfrac{Wt}{34\times 2}$$
$$Wt = 340gram$$

$$H_2O_2 \longrightarrow H_2O + \dfrac{1}{2}O_2$$
$$1$$ mole of $$H_2O_2$$ will give $$\dfrac{1}{2}$$ mole of $$O_2$$.
So the gain volume of $$O_2$$ is $$100ml$$,
mass of $$O_2$$ is-
$$\dfrac{Wt}{Mwt} = \dfrac{Volume}{22.4}$$

$$ Wt   = \dfrac{32 \times 100}{22400} =\dfrac{1}{7}gram$$


A closed bulb of capacity $$200$$ ml containing $$CH_4, H_2$$ and He at $$300$$K. The ratio of partial pressures of $$CH_4, H_2$$ and He, respectively is $$2 : 3 :5$$. Calculate the ratio of their weights present in the container.

For mixture of gases under identical condition of Temperature and Volume
$$p_i = \chi_i. P$$
$$p_i$$= partial pressure of the gas
and $$\chi_i$$=mole fraction of the gas
P=total pressure of the gas mixture
also $$\chi_i=\frac{n_i}{n_T}=\frac{m_i}{M_i \times n_T}$$
$$n_i=$$moles of ith gas
$$M_i=$$molar mass of ith gas 
$$n_T=$$total number of moles of gases in the mixture
$$\therefore\ p_i =P. \frac{m_i}{M_i \times n_T}$$
or $$p_i  \propto \frac{m_i}{M_i}$$ (for a given mixture)
Given that $$p_{CH_4}:p_{H_2}:p_{He}::2:3:5$$
thus $$\frac{m_{CH_4}}{M_{CH_4}}:\frac{m_{H_2}}{M_{H_2}}:\frac{m_{He}}{M_{He}}::2:3:5$$
or $$\frac{m_{CH_4}}{16}:\frac{m_{H_2}}{2}:\frac{m_{He}}{4}::2:3:5$$
$$m_{CH_4}:m_{H_2}:m_{He}::2 \times 16:3 \times 2:5 \times 4$$
$$m_{CH_4}:m_{H_2}:m_{He}::16:3:10$$



Equal masses of $${ H }_{ 2 }$$, He and $${C H }_{ 4 }$$ are mixed in empty container at 300 K, when total pressure is 2.6 atm. The partial pressure of $${ H }_{ 2 }$$ in the mixture is:-

choose answer:
1) 0.5 atm
2) 1.22 atm
3) 0.8 atm
4) 0.2 atm

Mass of each gas=m
moles of $$H_2$$=$$\frac{m}{2}$$
moles of He=$$\frac{m}{2}$$
moles of $$CH_4$$=$$\frac{m}{16}$$
Total moles=n=$$\frac{17m}{16}$$
Partial pressure of $$H_2$$=$$\frac{moles of H_2}{Total moles}\times Total pressure$$
$$\frac{\frac{\frac{m}{2}}{17m}}{16}\times 2.6$$
$$\frac{8}{17}\times 2.6$$
$$1.22$$ atom
Partial pressure of $$H_2=1.22$$ atm


A straight glass tube has two inlets $$X$$ and $$Y$$ at the two ends. The length of the tube is $$200$$ cm. $$HCl$$ gas through inlet $$X$$ and $$NH_3$$ gas through inlet $$Y$$ are allowed to enter the tube at the same time. White fumes first appear at a point $$P$$ inside the tube. The distance of $$P$$ (in cm) from $$X$$ is:

According to graham's law, 
$$\dfrac{r_{HCl}}{r_{NH_3}}=\sqrt{\dfrac{M_{NH_3}}{M_{HCl}}}$$
$$\dfrac{d}{200-d}=\sqrt{\dfrac{17}{36.5}}$$
$$d=81.12\;cm$$
The distance of $$P$$ from $$X=81.12\;cm$$


Calculate the final volume (in L) of one mole of an ideal gas initially at $$0^o$$C and $$1$$ atm pressure, if it absorbs $$2000$$ cal of heat during a reversible isothermal expansion.

Let us consider the problem:
AT stp; the gas is the pressure and standard temperature condition
$${v^1} = 22.4\,d{m^3}\,(given)$$
$${v_2} = ?\,\,\,\,\,(find\,out)$$
Let we know that 
$$\Delta U = q + w$$
$$\Delta U = 0$$ (isotherm and resistible for given expression)
$$ \Rightarrow $$
$$q =  - w = 2000\,cal$$
$$ = 2000 \times 4.184 = 8368$$
woek done $$w{ = _n}RT{/_n}\left( {\frac{{{v_2}}}{{{v_1}}}} \right)$$
Therefore 
$$ - w = 8368J = Nrt/n\left( {\frac{{{v_2}}}{{{v_1}}}} \right)$$
$$\, \Rightarrow $$ $$(im01) \times (8.314J{k^{ - 1}}m{01^{ - 1}})(273k)/n\left( {\frac{{{v_2}}}{{02.5dm}}} \right)$$
hence,
$${v_2} = 242.50d{m^3}$$


Set the following solutions in increasing order of vapour pressure:
$$3m$$  $$Na_2SO_4$$, 
$$1m$$  $$Urea$$
 $$0.4m$$  $$AlCl_3$$
$$0.2m$$  $$NaCl$$

The correct way increasing order is:

$$3m$$  $$Na_2SO_4$$< $$1m$$  $$Urea$$< $$0.4m$$  $$AlCl_3$$<$$0.2m$$  $$NaCl$$


An iron ball of radius $$0.3\ cm$$ falls through a column of oil of density $$0.94\ g\ cm^{-3}$$. If it attains a terminal velocity of $$0.54\ m\ s^{-1}$$, what is the viscosity of oil? Density of iron is $$7.8\ g\ cm^{-3}$$.
A viscous force of $$1.018\times 10^{-7}\ N$$ acting on a raindrop makes it fall through air with a terminal velocity of $$1\ ms^{-1}$$. If the viscosity of air is $$0.018\times 10^{-3}\ PI$$, what is the radius of the raindrop ?

Given:
$$e=7.8\ gcm^{-2}$$
$$\sigma =0.94\ gcm^{-3}$$
$$v_e=\dfrac {2(e-\sigma )r^2 g}{9\nu}$$
$$\nu=\dfrac {2(e-\sigma )r^2}{9\ vt}g =\dfrac {2\times (7.8-0.94)\times (0.3)^2 \times 980}{9\times 0.5}$$
$$=268.9$$ Paise
radius of raindrop
$$r=\sqrt {\dfrac {9\nu \ vt}{2g(e-\sigma )}}=\sqrt {\dfrac {9\times 0.018\times 1}{2\times 980 \times (7.8-0.94)}}$$
$$=\sqrt {\dfrac {0.162}{3978}}=\sqrt {0.40\times 10^{-6}}$$
$$=0.2\times 10^{-3}\ cm$$


Aerosol cans carry a clear warning of the heating of the can. Why?

Aerosol cans carry a clear warning of heating of the can because on heating the kinetic energy of particles increases and after collision with wall they create high pressure that can leads to explosion.


One mole of an ideal gas expands reversibly and adiabatically from a temperature of $${ 27 }^{ }C$$. If the work done during the process is 3 kJ, then final temperature of the gas is:$$\left( { C }_{ V }=20\quad J/K \right) $$ 
(a) 100 K               (b) 150 K
(c) 195 K               (d) 255 K

The correct option is b
in short Explanation;
$$W=nC_v(T_1-T_2)$$ 0r $$T_1-T_2=\frac{w}{nC_v}$$
$$T_2=T_1-\frac{W}{nC_v}=300-\frac{3000}{1\times 20}=15o K$$


Certain amount of a gas occupies a volume of $$0.4 litre$$ at $$17^oC$$. To what temperature should it be heated so that its volume gets (a) doubled (b) reduced to half, pressure remaining constant.

(a)$$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$$ by Charles law
$$\dfrac{0.4}{273+17}$$ Kelvin=$$\dfrac{0.8}{T_2}$$
$$T_2=300\times2$$
$$T_2=600$$ Kelvin
$$T_2=600-273$$
$$T_2=327^0C$$

(b)$$\dfrac{0.4}{300}=\dfrac{0.2}{T_2}$$
$$T_2=\dfrac{300}{2}$$
$$T_2=150$$ Kelvin
$$T_2=150-273$$
$$T_2=-123^0C$$


The pressure in a bulb dropped from $$2000$$ to $$1500$$ mm of mercury in $$47$$ min when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight $$79$$ in the molar ratio of $$1:1$$ at a total pressure of $$4000$$ mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of $$74$$ min.

The change of pressure of oxygen after $$74$$ min= $$\cfrac {500}{47}\times 74=787.2 mm$$
The $$1:1$$ molar mixture of oxygen and another gas, each of them has an equal pressure of $$2000mm$$ because the total pressure is given to be $$4000mm$$ $$Hg$$. 
The pressure of oxygen left after $$74$$ min
$$=2000-787.2=1212.8 mm$$ $$ Hg$$
Graham's law of diffusion:
$$\cfrac {r_{gas}}{r_{O_2}}=\sqrt {\cfrac {M_{O_2}}{M_{gas}}}$$
$$\Rightarrow \cfrac {V_{gas}(diffused)}{V_{O_2}(diffused)}\times \cfrac {t_{O_2}}{t_{gas}}=\sqrt {\cfrac {M_{O_2}}{M_{gas}}}$$
$$\Rightarrow \cfrac {P_{gas}(diffused)}{P_{O_2}(diffused)}\times \cfrac {t_{O_2}}{t_{gas}}=\sqrt {\cfrac {M_{O_2}}{M_{gas}}}$$
Both diffuse for the same time, so
$$\Rightarrow \cfrac {P_{gas}(diffused)}{787.2}=\sqrt {\cfrac {32}{79}}$$
$$\Rightarrow P_{gas}(diffused)=500.8 mm$$ $$Hg$$
The pressure of gas left after $$74$$ min= $$2000-500.8=1499.2 mm$$ $$Hg$$
Molar ratio of the gas and oxygen left= $$\cfrac {1499.2}{1212.8}=1.236$$ .


State and explain Gay Lussac's law of combining volumes.

Gay Lussac's Law of Combining Volumes
It states that the chemical reaction in which the gaseous reactants combine together to form one or more gaseous products, the ratio of the volumes of reactants and products will be in the whole number ratio.
Example:
$${ H }_{ 2 }+{ Cl }_{ 2 }\longrightarrow 2HCl$$
$$1$$Vol.  $$1$$Vol.       $$2$$Vol.  


The volume occupied by $$3.5$$g of $$O_2$$ gas at $$27^o$$C & $$740$$ mm of Hg pressure$$[O=16]$$

Solution:-
Mass of $${O}_{2} = 3.5 \; g$$
Molar mass of $${O}_{2} = 32 \; g$$
$$\therefore$$ No. of moles of $${O}_{2}, \left( n \right) = \cfrac{3.5}{32} = 0.109 \text{ mol}$$
Given that:-
Pressure, $$\left( P \right) = 740 \text{ mm of Hg} = 0.97 \; atm$$
Temperature, $$\left( T \right) = 27 ℃ = \left( 27 + 273 \right) = 300 K$$
$$R = 0.0821 \text{ L atm } {K}^{-1} {mol}^{-1}$$
Now from ideal gas law,
$$PV = nRT$$
$$\Rightarrow V = \cfrac{nRT}{P}$$
$$\Rightarrow V = \cfrac{0.109 \times 0.0821 \times 300}{0.97} = 2.76 \; L$$
Hence the required answer is $$2.76 \; L$$.


What are the main postulates of kinetic molecular theory of gases? 

The kinetic-molecular theory of gases can be stated as four postulates:
  • A gas consists of molecules in constant random motion.
  • Gas molecules influence each other only by collision; they exert no other forces on each other.
  • All collisions between gas molecules are perfectly elastic; all kinetic energy is conserved.


400 mL of oxygen at $$27^oC$$ were cooled to - $$15^oC$$ without the change in pressure. Calculate the contraction in volume.

Given,
$$V_1=400mL$$
$$T_1=27^0C=300K$$
$$T_2=-5^0C=258K$$
Now using Charle's Law,
$$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$$
$$\implies \dfrac{400}{300}=\dfrac{V_2}{258}$$
$$\implies V_2=344mL$$


Calculate the volume occupied by $$8.8g$$ of $$ { CO }_{ 2 }$$ at $$31.{ 1 }^{ o }C$$ and 1 bar pressure. ($$R=0.083$$ bar $${ dm }^{ 3 }\quad { K }^{ -1 }{ mol }^{ -1 })$$

Molecular weight of $$C{O}_{2} = 44 g$$
Weight of $$C{O}_{2} = 8.8 g$$
$$\therefore$$ no. of moles of $$C{O}_{2}, \left( n \right) = \cfrac{wt.}{\text{mol. wt.}} = \cfrac{8.8}{44} = 0.2$$
Given that:-
Pressure, $$\left( P \right) = 1 bar$$
Volume, $$\left( V \right) = ?$$
Temperature, $$\left( T \right) = 31.1 ℃= \left( 273 + 31.1 \right) K = 304.1 K$$
$$R = 0.083 \; bar \; {dm}^{3} {K}^{-1} {mol}^{-1}$$
From ideal gas equation,
$$PV = nRT$$
$$\Rightarrow 1 \times V = 0.2 \times 0.083 \times 304.1$$
$$\Rightarrow \; V = 0.0166 \; {dm}^{3} $$


 A mixture of gases have $$O_2$$ and $$N_2$$ at the ratio of 1 : 4, then what will be the ratio of no. of molecular in mixture of gases?

We know,
No. of moles $$=\dfrac{\text {Given Mass}}{\text {Molar Mass}}$$
So,
$$\dfrac{N_{O_2}}{N_{N_2}}=\dfrac{\dfrac{1}{32}}{\dfrac{4}{28}}=7:32$$


A certain mass of a gas occupies a volume of $$0.2 dm^3$$ at $$273K$$. Calculate the volume (in $$dm^3$$)of the gas if its absolute temperature is doubled at the same pressure. (Round off to 1 decimal places)

$$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$$

$$\dfrac{0.2}{273} = \dfrac{V_2}{546} $$

$$\therefore V_2 = 0.4dm^3$$



Solids have infinite free surfaces while liquids have only one upper free surface.

Surface which is free or exposed to air with balance of force to maintain it so. If you go by this simple definition, then both liquid and solid have only one free surface. Therefore, we need to expand this definition in terms of force.
Solids have infinite no of surfaces because it has infinite no of shapes and molecules are closely packed to each other and are not able to flow like fluids .for example a brick is made up of mud and this mud is made up of small particles like dry dirt and some amount of water . When masonry makes brick ,by which we can initiate the building design .This small brick will convert into big building with infinite no of surfaces.
Liquids only flow by nature and can change their shape according to the surface . that is why we can conclude it with one surface because our naked eyes will never see the actual behavior of molecules of liquids


A spherical balloon of 21 cm diameter is to be filled with $$H_2$$ gas at STP. From a cylinder, containing the gas at 20 atm and $$17^oC$$. If the vol. of cylinder is $$2.82$$ L, what will be the no. of balloon that can be filled up by this pressure? 

Pressure in balloon $$=1$$ atm
Temperature $$={ 25 }^{ \circ  }$$
Vol$$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }=\cfrac { 4 }{ 3 } \times 3.14\times 9.26\times { 10 }^{ -3 }{ m }^{ 3 }\\ =38.76l$$
$$n=\cfrac { PV }{ RT } =\cfrac { (1)(38.76) }{ (0.0821)(298) } =1.5843$$
n in cylinder $$=\cfrac { (20)(2.82) }{ (0.0821)(290) } =2.368$$
$$\therefore$$ Only one balloon can be filled by this cylinder.


State and explain Gay Lussac's law.

Gay-Lussac's law states that the pressure of a gas is directly proportional to the absolute temperature provided mass & volume of gas is constant.

At constant volume, the pressure of the gas increases with increase in temperature and decreases with a decrease in temperature which is expressed as:

    $$P \quad \alpha \quad T$$ -------- (at constant volume & mass)


At constant volume of $$0.0821$$ litres for an ideal gas if $$\dfrac {d}{dT}(PT)=300$$ at $$150K$$. The number of moles of that gas is___

Ideal gas equation= $$PV=nRT$$
$$\left(\cfrac {dP}{dT}\right)_V= \cfrac {nR}{V} \rightarrow (1)$$
$$\Rightarrow \cfrac {d}{dT}(PT)= T \cfrac {dP}{dT}+P\cfrac {dT}{dT}$$
$$\Rightarrow 300= T \cfrac {dP}{dT}+P$$
$$\Rightarrow 300= T\left(\cfrac {nR}{V}\right)+\cfrac {nRT}{V}$$    [as $$PV=nRT$$]
$$\Rightarrow 300= \cfrac {2nRT}{V}$$
$$\Rightarrow 300= \cfrac {2\times n \times 0.0821 \times 150K}{0.0821L}$$
$$\Rightarrow 300= 300 \times n$$
$$\Rightarrow n= 1$$
Thus number of moles of the gas is $$1$$ .


A bulb of tree litre capacity filled with air is headed from $$27^{o}C$$ to $$t^{o}C$$. The air thus expelled measured $$1.45$$ litre at $$17^{o}C$$ considering the pressure to be $$1$$ atm, throughout the experiment and ignoring the expansion of bulb, calculate $$t$$.

$$\dfrac { { P }_{ 300 }\quad { V }_{ 300 } }{ { n }_{ 300 }\quad 300 } =\dfrac { { P }_{ t }{ V }_{ t } }{ { n }_{ t }t } $$

$${ P }_{ 300 }={ P }_{ t }=1\quad atm$$

$${ V }_{ 300 }=3L$$

$${ n }_{ 300 }={ n }_{ t }$$

$$So,\quad \dfrac { 3 }{ 300 } =\dfrac { 3+x }{ t } $$

$$Rearrange\quad them$$

$$t=\dfrac { 300\times (3+x) }{ 3 } ------(1)$$

$$Given:\\ $$

$$\dfrac { { V }_{ 290 } }{ { T }_{ 290 } } =\dfrac { { V }_{ t } }{ t } $$

$$\dfrac { 1.45 }{ 290 } =\dfrac { x }{ t } $$


$$x=\dfrac { 1.45\times t }{ 290 } ----------(2)$$
Put the value of x from equation (2) to euation(1) and solve for t

$$t=\dfrac { 300\times \left( 3+\dfrac { 1.45t }{ 290 }  \right)  }{ 3 } \\$$

$$ =\dfrac { 300\times \left( 3+(5\times 1{ 0 }^{ -3 }\times t \right)  }{ 3 } \\$$

$$ =\dfrac { 900+1.5t }{ 3 } =\dfrac { 900 }{ 3 } +\dfrac { 1.5t }{ 3 } \\$$

$$ t-\dfrac { 1.5t }{ 3 } =\dfrac { 900 }{ 3 } \\ \dfrac { 3t-1.5t }{ 3 } =300\\$$

$$ \dfrac { 1.5 }{ 3 } \times t=300\\$$

$$ 0.5\times t=300\\$$

$$ t=\dfrac { 300 }{ 0.5 } =600K$$

$$\dfrac { { P }_{ 300 }\quad { V }_{ 300 } }{ { n }_{ 300 }\quad 300 } =\dfrac { { P }_{ t }{ V }_{ t } }{ { n }_{ t }t } $$

$${ P }_{ 300 }={ P }_{ t }=1\quad atm$$


Two flask $$A$$ and $$B$$ of equal volumes maintained at temperature $$300K$$ and $$700 K$$ contain equal mass of $$He(g)$$ and $${ N }_{ 3 }(g)$$ respectively. What is the ratio of total translational kinetic energy of gas in flask $$A$$ to the flask $$B$$?

Total translational kinetic energy= $$\cfrac {3}{2} K_BT$$
$$\therefore KE \propto T$$
$$\therefore$$ Ratio of kinetic energy of gas in A to that in B= $$\cfrac {T_A}{T_B}=\cfrac {3}{7}$$


A gas occupies $$100.0 mL$$ at $${ 50 }^{ o }C$$ and 1 atm pressure. The gas is cooled at constant pressure so that volume is reduced to $$50.0 mL$$. What is the final temperature of the gas?

Given that:-
Initial volume $$\left( {V}_{1} \right) = 100 \; mL$$
Final volume $$\left( {V}_{2} \right) = 50 \; mL$$
Initial temperature $$\left( {T}_{1} \right) = 50 ℃ = \left( 273 + 50 \right) K = 323 K$$
Final temperature $$\left( {T}_{2} \right) = ?$$
From ideal gas law-
$$PV = nRT$$
Given that the pressure is constant, i.e.,
$$V \propto T$$
$$\Rightarrow \cfrac{{V}_{1}}{{V}_{2}} = \cfrac{{T}_{1}}{{T}_{2}}$$
$$\Rightarrow \cfrac{100}{50} = \cfrac{323}{{T}_{2}}$$
$$\Rightarrow {T}_{2} = \cfrac{323 \times 50}{100} = 161.5 K$$
Hence the requird answer is $$161.5 K$$.


$$250$$ml of nitrogen gas maintained at $$650$$mm pressure and $$380 $$ mL of oxygen gas maintained at $$650$$ mm pressure are out together in 1 L flask.If temperature is kept constant, what will be the final pressure of the mixture?

$$\cfrac {PV}{T}$$= constant or $$PV$$= constant as $$T$$ is constant.
$$\Rightarrow 250\times 10^{-3}\times 650+380\times 10^{-3}\times 650=P\times 1$$
$$\therefore P= 409.5 mm$$ of $$Hg$$


The volume expansivity of a gas under constant pressure is $$0.0037$$ or $$ \left( \dfrac { 1 }{ 273 }  \right) $$. Calculate its volume at $$-100^oC$$ if its volume at $$100^oC$$ is $$685 cm^3$$ .

$$ \text{Given :  Volume expansitivity of gas =$\cfrac{1}{273}$} \\ \text{At 100$^{\circ}$C volume = }685 cm^{3} \\ \text{At -100$^{\circ}$C volume = V} \\ \text{Volume at 100$^{\circ}$C = V+V(100 - (-100)) = } \cfrac{1}{273} = V( 1+ \cfrac{200}{273}) \\ V = \cfrac{685 \times 273}{473} = 395 cm^{3} \\ \text{Volume at -100$^{\circ}$C = 395 cm}^{3}$$ 


Write down all the postulates of kinetic molecular theory of gases.

The microscopic theory of gas behavior based on molecular motion is called the kinetic theory of gases. Its basic postulates are-

  1. The molecules in a gas are small and very far apart. Most of the volume which a gas occupies is empty space.
  2. Gas molecules are in constant random motion. Just as many molecules are moving in one direction as in any other.
  3. Molecules can collide with each other and with the walls of the container. Collisions with the walls account for the pressure of the gas.
  4. When collisions occur, the molecules lose no kinetic energy; that is, the collisions are said to be perfectly elastic. The total kinetic energy of all the molecules remains constant unless there is some outside interference with the
  5. The molecules exert no attractive or repulsive forces on one another except during the process of collision. Between collisions, they move in straight lines.



If $$200mL$$ of $$N_2$$ at $$25^oC$$ and a pressure of $$250mm$$ are mixed with $$350mL$$ of $$O_2$$ at $$25^oC$$ and a pressure of $$300mm$$ so that, the volume of resulting mixture is $$300mL$$, what would be the final pressure of the mixture at $$25^oC$$?

$$ \text{ P$_{1}$ = 250      P$_{2}$ = 300      V$_{f}$ = 300    V$_{1}$ = 200   V$_{2}$ = 350    P$_{f}$ = ?}   \\  P_{1} V_{1} + P_{2} V_{2} = P_{f} V_{f} \\ \Rightarrow (250)(200) + (300)(350) = P_{f} (300) \\ \Rightarrow P_{f} = \cfrac{350(300)}{300} + \cfrac{250 \times 200}{300} = 350 + \cfrac{500}{3} = 516.66 mm \\ \Rightarrow \text{Final Pressure of the mixture= 516.66 mm}$$


At a constant temperature, 250 mL of argon at 760 mm pressure and 600 mL of nitrogen at 500 mm pressure are put together in a one litre flask. Calculate the final pressure.

$$ P_{1} V_{1} + P_{2} V_{2} = P_{mix} V_{mix} \text{                (At const. temperature)} \\ \Rightarrow (760)(250) + (500)(600) = (1000) P_{mix}  \\ \Rightarrow P_{mix} = 300 + 190 = 490 mm = \text{Final Pressure}$$


Some faculty members of XYZ classes were trying to simplify a problem for students by mentioning data explicity about a gaseous mixture. Information given by them is as follow.
Faculty - 1 : Mixture consist of three gases A, B and C.
Faculty - 2 : Molar mass of B is twice of A & Molar mass of A is $$4$$ times of C.
Faculty 3 : The gas that is neither heaviest nor lightest is $$16\%$$ by mass.
Faculty - 4 : Molar ratio of heaviest to lightest gas is $$2 : 5$$
Based on the above information, answer the question that follows.
What would be the average molar mass of mixture.

Given that
$$A=4C$$
$$B=2A$$
$$\therefore B=8C$$
Given that $$\cfrac {B}{C}=\cfrac {2}{5}$$
If $$B=2, A=1$$
$$\therefore$$ The molar ratio of $$A:B:C=1:2:5$$
Average molar mass of mixture= $$\cfrac {1+2+5}{3}=\cfrac {8}{3}$$ .


A gas occipies $$700ml$$ at  S.T.P. Find the volume occupied by the gas when its pressure is $$400mm$$ of Hg and its temperature $$15^oC$$

$$V_1=700\text{ }ml\\P_1=1\text{ }atm=760\text{ }mm\text{ }Hg\\T_1=298\text{ }K$$            $$V_2=?\\P_2=400\text{ }mm\text{ }Hg\\T_1=288\text{ }K$$
Here number of moles is constant.
$$\therefore \cfrac{P_1V_1}{T_1}=\cfrac{P_2V_2}{T_2}\\V_2=\cfrac{P_1\times V_1\times T_2}{T_1\times P_2}=\cfrac{760\times 700\times 288}{298\times 400}=1285.37\text{ }ml$$
$$\therefore$$ Volume occupied by gas is $$1285.37\text{ }ml.$$


A balloon is filled with hydrogen at room temperature, it will burst if pressure exceeds $$0.2$$ bar. If at $$1$$ bar pressure the gas occupies $$2.27$$ L volume, upto what volume can the balloon be expanded?

From the given data,
$$V_1= 2.27L$$ and $$P_1= 1 bar$$ and $$P_2= 0.2 bar$$
$$V_2=?$$

According to Boyle's law
$$P_1V_1=P_2V_2$$

$$\Rightarrow V_2= \cfrac {1 bar \times 2.27L}{0.2 bar}=11.35L$$

Since balloon bursts at $$0.2 bar$$ pressure, the volume of the balloon should be less than $$11.35L$$ .


A gas occupies 100mL at 50degree C at 1atm pressure.The gas is cooled at constant pressure so that volume is reduced to 50mL.What is the final temperature of the gas 

Given $$V_2=50, V_1=100,T_1=323,T_2=?$$
$$\cfrac {V_1}{T_1}=\cfrac {V_2}{T_2}$$
$$\Rightarrow T_2=\cfrac {V_2\times T_1}{V_1}=\cfrac {50 \times 323}{100}=161.5K$$
$$\Rightarrow T_2= -111.5C$$ .


Explain ion-dipole interaction with suitable example.

      • An ion-dipole force is an attractive force that results from the electrostatic attraction between an ion and a neutral molecule that has a dipole.
      • Ion-dipole forces are generated between polar water molecules and a sodium ion. The oxygen atom in the water molecule has a slight negative charge and is attracted to the positive sodium ion. These inter molecular ion-dipole forces are much weaker than covalent or ionic bonds.


      Write note on Laws of multiple proportions.
      Write note on Dalton's atomic theory.
      The density of 3 M solution of NaCl is 1.25 g\ml. Calculate the molality of the solution.
      A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass
      is 98.96 g. What is its empirical and molecular formula?
      Calculate number of atoms present in 500 ml 0.05 M aqueous solution of H$$_2$$SO$$_4$$?

      $$\textbf{Part 1:}$$

      The Law of multiple proportions states that when two elements combine to form more than one compound, the weights of one element combined with a fixed weight of the other are in a ratio of small whole numbers.

      $$\textbf{Example:}$$The masses of oxygen combined with a fixed mass of carbon in $$CO_{2}$$ and $$CO$$ are 32 and 16, respectively. These masses of oxygen show a simple ratio of 32: 16 that is 2: 1 to each other.


      $$\textbf{Part 2:}$$

      The postulates of Dalton's atomic theory are:
      • All matter is made up of small indivisible particles called atoms.
      • All atoms of a element are identical in mass, size, and other properties.
      • Atoms can neither be created nor destroyed. Furthermore, atoms cannot be divided into smaller particles.
      • Atoms of different elements can combine with each other in fixed whole-number ratios in order to form compounds.
      • $$\textbf{Limitations of Dalton's atomic theory:}$$
      • It does not account for subatomic particles like proton, electron and neutron.
      • It does not account for isotopes: As per Dalton’s atomic theory, all atoms of an element have identical masses and densities. However, different isotopes of elements have different atomic masses.

      $$\textbf{Part 3:}$$
      $$\textbf{Step 1:}$$ Analyzing the given data
      3M solution of NaCl means 3 moles of NaCl is present in one litre of water.
      Molecular weight of NaCl= 58.44g
      $$\text{Density}=\dfrac{ Mass}{Volume}$$
      Putting all the values in above formula we get
      $$Mass= 1.25gm/mL\times{1000mL}=1250gm$$
      mass of water= mass of solution - mass of 3 mole of NaCl
                             = $$1250gm - 3\times{58.44}gm$$
                             = $$1074.68 gm$$

      $$\textbf{Step 2:}$$ Calculation for molality
      $$Molality=\dfrac{moles}{mass\ of\ solvent (g)}\times{1000}$$
      $$Molality=\dfrac{3}{1074.88}\times{1000}$$
      = 2.7915 molal
      So, the molality of the solution is 2.7915.


      $$\textbf{Part 4:}$$

      Hydrogen = 4.07%, Carbon=24.27%, Chlorine=71.65%

      Molar Mass = 98.96g

      To Calculate Empirical Formulae, we must use following steps:

      $$\textbf{Step 1:}$$ Calculation of number of moles of each atom.

      Hydrogen = $$\dfrac{4.07}{1}=4.07$$

      Carbon= $$\dfrac{24.27}{12}=2.02$$

      Chlorine=$$\dfrac{71.65}{35.5}=2.01$$

      $$\textbf{Step 2:}$$ Finding empirical formula of given compound.

      Now divide all the values with the lowest obtained value in step 1.

      Hydrogen = $$\dfrac{4.07}{2.01}=2$$

      Carbon= $$\dfrac{2.01}{2.01}=1$$

      Chlorine=$$\dfrac{2.01}{2.01}=1$$

      Therefore, the empirical formula is $$CH_{2}Cl$$

      $$\textbf{Step 3:}$$ Finding molecular formula of the given compound.

      To Calculate Molecular Formula, we need to find the weight of the empirical formula

      $$CH_{2}Cl_{2}$$$$\ $$=$$\ $$12 + 2(1) + 35.5 = 49.5

      The molecular weight- 98.96 is double of empirical weight. Therefore molecular formula is $$C_{2}H_{4}Cl_{2}$$


      $$\textbf{Part 5:}$$
       
      $$\textbf{Step 1:}$$ Calculation of the number of moles
      Molarity is defined as the number of moles of solute present in 1 litre or 1000mL of solution.

      $$Molarity=\dfrac{\text{Moles}}{\text{volume of solution(mL})}\times{1000}$$
      $$0.05=\dfrac{\text{Moles}}{500}\times{1000}$$ 
      $$moles= 0.025$$
      $$\textbf{Step 2:}$$ Calculation for number of atoms
      $$ 1 mole = 6.022\times10^{23} atoms$$ 
      0.025 mole = $$0.025\times{6.022}\times{10}^{23}$$= $$1.50\times10^{23}$$
      So, the number of atoms present in 500 ml 0.05 M aqueous solution of $$H_2SO_4$$ = $$1.50\times10^{23}$$


      Explain ion-dipole interaction with the help of suitable example.


      1345406_1131733_ans_b43ccaf0e0ee4701b0f53417fff30d89.jpg


      The density of a gas is 3.80 g $$L^{-1}$$  at S.T.P calculate is density at $$27^{0}C$$ and 700 torr pressure.

      $$\dfrac{{D}_{1}{T}_{1}}{{P}_{1}} = \dfrac{{D}_{2}{T}_{2}}{{P}_{2}}$$

      $$\dfrac{3.80 \times 273}{760} = \dfrac{{D}_{2} \times 300}{700}$$

      $${D}_{2} = 3.185\space g\space {L}^{-1}$$


      $${O}_{2}$$ is present in $$1$$ litre flask at a pressure of $$1.52$$ x $${10}^{3}$$ $$mm$$ of Hg. Calculate mass of $${O}_{2}$$ at $$0C$$.


      1351535_1141245_ans_a8cfd1dbceca401694dfde3a059e56fd.jpg


      What percent of a sample of nitrogen must be allowed to escape if its temperature, pressure and volume are to be changed from $$220C$$, $$3 atm$$ and $$1.65 litre$$ to $$110C$$, $$0.7 atm$$ and $$1.00 litre$$ respectively?


      1348553_1143368_ans_db2f994f6b7e4959a2eabe3fab3eba04.jpg


      What will be the volume of a given mass of a gas at a pressure 50 cm of Hg. If it occupies 250 ml at a pressure of 98 cm of Hg, keeping the temperature constant. 

      Use Boyle's law
      P1 * V1 = P2 * V2
      Put the values from the queations
      98* 260 = 50 * V2
      V2 = 98*260/50
      V2 = 509.6 ml



      Two closed bulbs of equal volume (v) containing an ideal gas initially at pressure $${ P }_{ 4 }$$ and temperature $${ T }_{ 1 }$$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $${ T }_{ 2 }$$. What is the final pressure $${ P }_{ f }$$?
      1151556_4aa3e60d3dd944beb2b755f5fc43fe7b.png

      Let us consider the problem:

      $${n_1} + {n_2} = n_1' + n_2'$$

      Therefore,
      $$\cfrac{{{p_i}V}}{{R{T_1}}} + \cfrac{{{p_i}V}}{{R{T_1}}} = \cfrac{{{p_f}V}}{{R{T_1}}} + \cfrac{{{p_f}V}}{{R{T_2}}}$$

      =>$$\cfrac{{2{p_i}V}}{{R{T_1}}} = \cfrac{{{p_f}V}}{R}\left( {\cfrac{{{T_1} + {T_2}}}{{{T_1}{T_2}}}} \right)$$

      =>$${P_f} = \cfrac{{2{p_i}{T_2}}}{{{T_1} + {T_2}}}$$


      A $$5\ lit$$ flask containing $$N_{2}$$ at $$1$$ bar and $$25^{o}C$$ is connected to a $$4 lit$$ flask containing $$N_{2}$$ at $$2$$ bar and $$0^{o}C$$. After the gases are allowed to mix keeping both flasks at their original temperature, what will be the pressure and amount of $$N_{2}$$ in the $$5\ lit$$ flask assuming ideal gas behaviour.

      After mixing, Total volume= $$5L+4L=9L$$
      We know that $$PV$$=constant
      So, for flask $$1$$,$$1 \times 5=P_2 \times 9 \Rightarrow P_2= 0.54$$atm
                  flask $$2$$, $$P_2=0.84$$ atm
      Hence, total pressure $$P= P_2+P_3=1.42$$ atm
      Amount= $$n= \cfrac {PV}{RT}= \cfrac {1.42 \times 5}{0.82 \times 298}=0.29$$ moles .


      Find out the numerical value of gas constant (R).

      Using the conditions at Standard Temperature and Pressure(STP),

      $$P=1\ atm$$
      $$T=273\ K$$
      $$V=22.4\ L$$
      $$n=1$$

      Using the relation, $$PV=nRT$$

      $$\Rightarrow R=\dfrac{PV}{nT}=\dfrac{1\times22.4}{1\times273}=0.082\ L\ atm\ mol^{-1}\ K^{-1}$$


      An open bulb containing air at $$19^0C$$ was cooled to a certain temperature at which the no. of moles of gaseous molecules increased by $$25$$%. What is the final temperature?

      By the ideal gas low:
      $$PV=nRT$$
      and acc. to question $$n_{2}=1.25n_{1}$$
      $$PV/R =n_{1} T_{1}$$ and $$PV/R=n_{2}T_{2}$$
      $$n_{1} T_{1}=n_{2} T_{2}$$
      $$n_{1} T_{1}=1.25n_{1}T_{2}$$
      $$T_{2}=\dfrac {T_{1}}{1.25}=\dfrac {292}{1.25}=233.6k=40^{o}C$$





      Calculate the percentage of free volume available in $$1$$ mole gaseous water at $$1$$ atm pressure and $$373\ K$$.


      1351721_1259631_ans_e7bb3a4d870247e2b45ba817133806ef.jpg


      Calculate the volume occupied by $$7.0\ g$$ of nitrogen gas at $$27^{o}C$$ and $$750\ mm$$ $$Hg$$ pressure,


      1352723_1287695_ans_f1250c6dda99488ca78c30aa67de5b15.jpeg


      According to Henrys Law, in gases, an increase in pressure increase______.

      solubility


      5.40 gm of an unknown gas at $${ 27 }^{ \circ  }C$$ Occupies the same volume as 0.14 gm hydrogen at $${ 17 }^{ \circ  }C$$ and same pressure. The molecular weight of unknown gas is 


      1359665_1472960_ans_413afda4fc684732bf565618d2e78c18.jpeg


      Write the postulates of "kinetic theory of gases' .

      1. Gases are made up of large number of the minute particles.
      2. Pressure is exerted by a gas 
      3. There is no loss of kinetic energy. 
      4. Molecules of gas attract on one another. 
      5. Kinetic energy of the molecule in directly proportional to absolute temperature. 
      6. Actual volume of the gaseous molecule very small. 
      7. Gaseous molecules are at always in motion. 
      8. There is more influence of gravity on movement of gaseous molecule.


      Kinetic energy of oxygen molecule at$$ { 0 }^{ \circ  }C$$ is $$5.64\times{10  }^{ -21 }J$$. Calculate the value of Avogadro's number. Given$$R=8.31J$$ mole$$^{-1}K^{-1}$$.


      1359748_1476915_ans_c3863b805fa14dfba6322c6ae3d3503a.jpeg


      Write ideal gas equation for one mole of a gas.

      $$PV=RT$$


      At $$525$$K, the equilibrium constant of the reaction, $$PCl_5\rightleftharpoons PCl_3+Cl_2$$ is $$1.78$$ atm $$(K_p)$$. At what pressure should an equimolar mixture of $$Cl_2$$ and $$PCl_3$$ be taken for the pressure of $$PCl_5$$ to be $$5\times 10^4$$ Pa at equilibrium, volume remaining constant?


      1607344_1739856_ans_3724ec67cec2482fb2f4071962efe792.jpg


      Give two characteristic properties of a liquid.

      The particles of liquid have the ability to flow. 

      The particles are loosely packed and can move from one place to another. This property is called fluidity.

      Liquids have the ability to acquire the shape of the container in which it is kept in. This is possible because of the fluidity of liquids.


      Match the graphs between the following variable with their names:

      (i) The graph at $$\text{constant molar volume}$$ is called $$Isochore$$.
      (ii) The graph at $$\text{constant temperature}$$ is called $$Isotherm$$.
      (iii) The graph at $$\text{constant pressure}$$ is called $$Isobar$$.


      "One of the assumptions of the kinetic theory of gases is that there is no force of attraction between the molecules of a gas". State and explain the evidence that shows that the assumptions are not applicable for real gases.

      Real gases can be liquified by cooling and applying high pressure on the gas. This proves that $$\text{forces of attraction exist between molecules}$$.


      Why do liquids flow ?

      Liquids flow because the particles in a liquid are not very tightly bound to each other and they have high intermolecular spaces between them, which allows the particles to be displaced or move causing the liquids to flow.


      Define the three states of Matter

      Solid State: The molecules are very close to each other hence intermodular spaces are small and intermodular force is strong. (See figure (a)).
      Hence solids have a definite volume, rigid, retain definite shape, and are incompressible.

      Liquids: The molecules are less closely packed have more intermolecular spaces than solid, less strong forces than solids. (See figure (b)).
      Hence liquids have a definite volume but no definite shape. They take the shape of the container in which they are put. 

      Gases: The molecules in the gases are far apart with the weakest force of attraction. (See figure (c)).
      Hence gases have neither a definite volume nor definite shape but easily compressible.

      1668681_1797863_ans_85e0a1d267324426aa46118c2d266091.png


      Define matter

      Anything that has mass and occupies space is called matter.


      For each of the following statements, say whether it describes a solid, a liquid, or a gas. 
      (a) Particles move about very quickly.
      (b) Particles are quite close together.
      (c) Particles are far apart and move in all directions.

      (a) Particles move about very quickly in Liquid 
      (b) Particles are quite close together in Solid 
      (c) Particles are far apart and move in all directions in Gas 


      Name the three states of matter.

      The three states of matter are :
      1. Solid State 
      2. Liquids 
      3. Gases 



      1668679_1797856_ans_f1f27e520ffd4585a21b50f0417c7343.png


      Give the occurrence of water in the three different states i.e. solid, liquid and gaseous.

      The occurrence of water in three different states are: 
      1. Solid state - As snow and frost.
      2. Liquid state - In sea water, river water and lake water. 
      3. Gaseous state - As water vapour in air the amount depending on climatic conditions.



      State three main characteristics of the particles of matter.

      Three main characteristics of the particles of matter are:
      1. Particles of matter have space between them. This space is called intermolecular space
      2. Particles of matter are always in random motion.
      3. Particles of matter are tiny in size
      4. Particles of matter attract each other.


      .......... deals with the study of matter and the changes it undergoes.

      Chemistry deals with the study of matter and the changes it undergoes.


      For the following statement, say whether it describes a solid, a liquid or a gas.
      Particles are quite close together.

      Ans: Solid
      In a solid, the particles are very attracted to each other so they are close together. The particles can move in their fixed positions but cannot slide past one another.


      The different forms of a substance are called ............... of matter.


      1990879_1819743_ans_3dc5ccbea0ad4bfd84de61944816b755.jpg


      Give reason for : Tyres of automobiles are inflated to lesser pressure in summer than in winter.

      In summer, due to higher temperature, the average kinetic energy of the air molecules inside the tyre increases, ie., molecules start moving faster. Hence, the pressure on the walls of the tube increases. If its pressure inside is not kept low at the time of inflation, at the higher temperature, the pressure may become so high that the tyre may burst.


      Write a short note on Gay Lussac's Law.

      Gay Lussac's Law: 
      This law states that the volume of gases taking part in a  chemical reaction shows a simple whole-number ratio to one another when those volumes are measured at the same temperature and pressure.


      Give two examples of substances that are rigid and not compressible. 

      Substances that are rigid and not compressible are Solids.
      e.g. : Glass, stone, pen. 


      State Dalton’s law of partial pressure.

      Dalton's law of partial pressure states that whenever two or more gases, which do not react chemically, are enclosed in vessel, the total  pressure is equal to sum of partial pressure of each gas.



      What is Dalton's law of partial pressure?

      The law was formulated by John Dalton in $$1801$$. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases, i.e, the pressure which these gases would exert if they were enclosed separately in the same volume and under the same condition of temperature. In a mixture of gases, the pressure exerted by the individual gas is called partial pressure.
      Mathematically,
      $$P_{Total}=P_{1}+P_{2}+P_{3}$$     (at constant $$T,V$$)
      Where, $$P_{Total}$$ is the total pressure exerted by the mixture of gases and $$P_{1},P_{2},P_{3}$$etc, are partial pressure of gases.
      Or,
      According to Dalton's law of partial pressure, "Total exerted pressure by mixture of all non-reactive gases is equal to the sum of partial pressure of all individual gases".
      At constant temperature $$T$$ and volume $$V$$,
      $$P_{Total}=P_{1}+P_{2}+P_{3}$$
      Where, $$P_{Total}=$$ total exerted pressure of mixture of all gases.
      $$P_{1},P_{2},P_{3}$$ is pressure exerted by individual gases known as partial pressure.


      What assumption can be made regarding the possibility of collision between gas molecules?

      assumption can be made regarding the possibility of collision between gas molecules are:-
      $$\text{The molecules collide each other.}$$


      The drops of liquid take spherical shape. Why?

      The inward forces on the surface molecules of a liquid drop tend to cause the surface to volume ratio as small as possible. Since surface-to-volume ratio is minimum for the spherical shape so a liquid drop is spherical.


      The boiling point of liquid changes on increasing the pressure. Why?

      The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal; to the atmospheric pressure. Also the vapour pressure of a liquid is a function of the temperature. As the temperature increases, the vapour pressure increases, the vapour pressure increases. Now, if we increase the external pressure, a higher temperature would be required so that the vapour pressure equal to that increased external pressure and hence, the boiling point of the liquid increases.


      Mixture of $$NH_{3}$$ and $$HCl$$ gases do not follow Dalton's law of partial pressure. Why?

      Mixture of $$NH_{3}$$ and $$HCl$$ gases do not follow Dalton's law of partial pressure because the gases are reactive. They react to form ammonium chloride which is solid and Dalton's law of partial pressure is not applicable to solids and reactive gases.
      $$NH_{3}+HCl\rightarrow NH_{4}Cl$$.


      Complete the table
      Energy of gas moleculesVery high
      Distance between the molecules.........
      Freedom of moment of molecules..........
      Attractive force between molecules...........

       The energy of gas molecules Very high
       Distance between the molecules Away from each other
       Distance between the molecules Very high
       The attractive force between molecules Very low






      State Ideal gas equation and law.

      Ideal gas law is a law relating the pressure, temperature and volume of an ideal gas. Ideal gas equation is combination of three laws (Boyle's law, Charles law and Avogadro law) and gives a single equation $$(PV=nRT)$$ called as Ideal gas equation.
      Mathematically,
      According to Boyle's law, at constant $$T$$ and $$n$$,
      $$V\propto 1/P............(1)$$
      According to Charle's law, at constant $$P$$ and $$n$$,
      $$V\propto T..............(2)$$
      According top Avogadro law, at constant $$T$$ and $$n$$,
      $$V\propto n..............(3)$$
      From equation $$1,2$$ and $$3$$, we get
      $$V\propto nT/P...........(4)$$
      or, $$V=RnT/P.............(5)$$
      or, $$PV=nRT..............(6)$$
      Then, $$R=PV/nT...........(7)$$
      Where, $$R$$ is a gas constant which is same for all gases and known as Universal gas constant and equation $$(6), PV=nRT$$ is known as ideal gas equation.


      Identify and picturise the arrangement of particles in different states of matter.

      Arrangements of particles in different states of matter:
      The particles of solids are very close to each other. The freedom of movement of particles is limited. The attractive force between particles is very high. The particles of liquid are relatively further apart and have more freedom for movement than in the solid state. The attractive force is less than that of solid state. The particles in gaseous state remains far away from one another and the attractive force is very low.
      1799968_1850114_ans_6fa548c0482244baab86630e2ff3be7a.png


      Explain the peculiarities of materials.

      Tiny particles of matter:
      The substances are made up of many tiny particles which cannot be seen by naked eye. Even though we cannot see the tiny particles of sugar in sugar solution, we can understand that tiny particles of sugar is dissolved in the solution while we taste it.
      All the substances are made up of tiny particles which cannot be seen by naked eye and these tiny particles bear all the properties of the substances.


      At $$25^oC$$ and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where temperatures is $$10^oC$$ and volume of the gas is 640 mL?

      Given, $$P_1=760 mm Hg$$, $$V_1=600 mL$$

                  $$T_1=273+25=298 K$$
                
                  $$P_2=?$$
       
                  $$V_2=640 mL$$ and $$T_2=273+10=283K$$

      According to combined gas law equation,

      $$\dfrac {P_1V_1}{T_1}=\dfrac {P_2V_2}{T_2}$$

      or $$P_2=\dfrac {P_1V_1T_2}{V_2T_1}$$

      $$\displaystyle =\frac {(760 mm Hg)\times (600 mL)\times (283 K)}{(640 mL)(298 K)}$$

      = $$676.6 mm Hg$$


      Match the following.
      State of matterAB
      1GasSoilMilk
      2SolidKeroseneOxygen
      3LiquidCarbon MonoxidePeas

      State of Matter                                A                             B
      1. Gas                            Carbon monoxide                  oxygen
      2.Solid                            Soil                                          peas
      3.Liquid                           Kerosene                               Milk


      For hydrogen gas $$C_p-C_v=a$$ and for oxygen gas $$C_p-C_v=b$$, 

      so the relation between $$a$$ and $$b$$ is $$a = nb$$. Value of $$n$$ is:

      The difference between specific heats always equals $$R$$ (Universal gas constant.)
      $$C_P-C_V=R$$

      For $$H_2, C_P-C_V=a\Longrightarrow a=R$$

      For $$O_2, C_P-C_V=b\Longrightarrow b=R$$ 

      $$a=nb.\\R=nR.\\n=1$$


      Match the items of List 1 with those in List 2.

      1. Boyle's law states that the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system.
      mathematically we can say that $$V\propto \frac { 1 }{ P } $$
      $$A\rightarrow 3$$
      2.      Charles law states that When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be directly related
      mathematically e can say $$V\propto T$$
      $$B\rightarrow 1$$
      3.      Gay- Lussac's law states that The pressure of a gas of fixed mass and fixed volume is directly proportional to the gas's absolute temperature.
      mathematically $$P\propto T$$
      $$C\rightarrow 4$$
      4.      Dalton's law states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases. That is $${ P }_{ t\quad  }=\sum _{ i=1 }^{ n }{ { P }_{ i } } $$ 
      so, $$D\rightarrow 2$$


      Match the following from $$Column-I$$ to $$Column-II$$.

      (A) Law of conservation of mass is illustrated by the following example. $$4.2 \:g \:MgCO_3$$ gives $$2.0$$ g residue on heating.
      (B) Law of multiple proportion is illustrated by the following example. $$S$$ and $$O_2$$ combine to form $$SO_2$$ and $$SO_3$$.
      (C) Law of definite proportion is illustrated by the following example. $$CH_4$$ has carbon and hydrogen in $$3 : 1$$ mass ratio.
      (D) Law of reciprocal proportion is illustrated by the following example. In $$H_2S$$ and $$SO_2$$ mass ratio of H and O w.r.t. sulfur is 1 :16, hence in $$H_2O$$, mass ratio of $$H$$ and $$O$$ is $$1 : 8$$.
      (E) Gay Lussac's Law is illustrated by the following example. $$10 \: mL \: N_2$$ combines with $$30 \:mL \:of \:H_2$$ to form $$20 \: mL \: of \: NH_3$$.


      300 litres of ammonia gas at $$20^0 C$$ and 20 atmosphere pressure are allowed to expand in a space of 600 litres capacity and to a pressure of one atmosphere .Calculate the drop in temperature.

      Here,

      $$\cfrac {PV}{T}=constant$$

      $$\therefore \cfrac {20\times 300}{293}=\cfrac {1\times 600}{T}$$

      $$\therefore T=29.3$$ $$K$$

      Drop in temperature $$=293-29.3=263.7$$ $$K$$ or $$263.7^oC$$



      Consider the of bulbs shown arrangement below.
      If the pressure of the system when all the stop cocks are opened is x (in atm) then find 100 x ? (760 mm = 1 atm)
      761396_4acc0162c9fd4c2c91bea63901b55baf.png

      If T is constant, then the PV is constant (Boyle's law). 
      $$\therefore P_1V_1 + P_2V_2 + P_3V_3 =P_4V_4$$ ---------(1)
      Where $$P_4$$ is a total pressure and $$V_4$$ is a total volume.
      Now, the total volume of all the container$$=V_4=V_1+V_2+V_3=1+1+0.5=2.5L$$
      Now, putting all the values in equation (1) we get,
      $$300\times 1+240\times1+440\times0.5=X\times 2.5$$
      $$\therefore X=\dfrac{760}{2.5}=304mm$$
      Now, 760mm is equivalent to 1atm
      $$\therefore$$ 304mm is equivalent to $$\dfrac{304}{760}=0.4atm$$
      Thus, x=0.4 atm
      $$\therefore 100x=40atm$$


      $$16gm$$ of $${O}_{2}$$ was filled in a container of capacity $$8.21$$ lit. at $$300K$$. Calculate
      (i) Pressure exerted by $${O}_{2}$$
      (ii) Partial pressure of $${O}_{2}$$ and $${O}_{3}$$ if $$50$$% of oxygen is converted into ozone at same temperature
      (iii) Total pressure exerted by gases if $$50$$% of oxygen is converted into ozone ($${O}_{3}$$) at temperature $$50K$$

      i)  $$PV=nRT$$
      $$\Rightarrow P=\dfrac { \left( 16/32 \right) \times 0.0821\times 300 }{ 8.21 } =1.5$$atm
      ii) $${ 3O }_{ 2 }\rightarrow 2{ O }_{ 3 }$$
      $$\therefore \quad 0.25$$ moles of $${ O }_{ 2 }$$ is converted to $$\dfrac { 0.25\times 2 }{ 3 } $$ moles i.e., $$0.167$$ moles of $${ O }_{ 3 }$$
      iii) At $$50K,\quad { p }_{ { O }_{ 2 } }=\dfrac { { n }_{ { O }_{ 2 } }RT }{ V } $$  and  $${ P }_{ { O }_{ 3 } }=\dfrac { { n }_{ { O }_{ 3 } }RT }{ V } $$
      $${ P }_{ total }={ p }_{ { O }_{ 2 } }+{ p }_{ { O }_{ 3 } }=\left( { n }_{ { O }_{ 2 } }+{ n }_{ { O }_{ 3 } } \right) RT/V$$
                $$=\dfrac { \left( 0.25+0.167 \right) \times 0.0821\times 50 }{ 8.21 } =0.2085$$ atm.


      Two flask of equal volume have been joined by a narrow tube of negligible volume. Initially both flasks are at $$300K$$ containing $$0.60$$ mole of $$O_2$$ gas at $$0.5atm$$ pressure. One of the flask is then placed in a thermostat at $$600K$$. Calculate final pressure and the number of $$O_2$$ gas in each flask.

      Ideal gas law-  $$PV= nRT$$
      $$ P_1V = n_1RT$$ [Volume is constant]
      $$ 0.5 V_1 =0.6R\times 300$$
      $$  V   = 360R$$
      in 2nd case the pressure will be same for both flask and sum of moles is .6
      Flask-1
      $$ P\times 360R = n_1R\times 300$$
      $$ n_1  =1.2P$$
      Flask-2
      $$ P\times 360R = n_2R\times 600$$
      $$ n_2 = 0.6P$$
      $$ n_1+n_2 = 0.6$$
      $$ 1.2P+0.6P =0.6$$
      $$ 1.8P    = 0.6$$
      Final pressure-
      $$  P      = \dfrac{1}{3}$$
      Number of moles of oxygen in flasks-
      $$n_1 = 1.2\times \dfrac{1}{3}= 0.4$$
      $$n_2 = 0.6\times \dfrac{1}{3} = 0.2$$


      $$20\ mL$$ of a solution containing $$0.2\ g$$ of impure sample of $$H_{2}O_{2}$$ reacts with $$0.316\ g$$ of $$KMnO_{4}$$ (acidic). Calculate:
      (i) Purity of $$H_{2}O_{2}$$.
      (ii) Volume of dry $$O_{2}$$ evolved at $$27^{\circ}C$$ and $$750\ mm\ P$$.

      1.The redox reaction is as follows:

      $$2{MnO_4}^- + 5H_2O_2 \rightarrow 5O_2+ 2Mn^{2+}$$

      The number of moles of $$KMnO_4$$ present $$=\displaystyle \frac {0.316}{158}=2 \times 10^{-3}$$ mol

      The number of moles of pure $$H_2O_2 = \dfrac{5}{2} \times 2 \times 10^{-3}=5 \times 10^{-3}$$

      $$0.2$$ g of sample contains $$0.17$$ g of pure $$H_2O_2$$.

      Therefore, percentage purity of $$H_2O_2$$ $$=\dfrac {0.17}{0.2} \times 100= 85\%$$

      2. The redox changes are as follows:

      $$Mn^{7+}+5e\longrightarrow Mn^{2+}$$

      $${O_2}^-\longrightarrow O_2+2e$$ $$\displaystyle\left[E_{O_2}=\frac{M}{2}\right]$$

      Eq. of $$O_2=$$ Eq. of $$KMnO_4$$

      $$\displaystyle\frac{w}{\displaystyle\frac{32}{2}}=\frac{0.316\times5}{158}$$ $$\implies w_{O_2}=0.16$$ g

      We know, $$PV=nRT$$.

      Substituting the values, we get

      $$\displaystyle\frac {750}{760}\times V=\frac {0.16}{32}\times 0.0821\times 300$$

      $$\therefore V_{O_2}=0.12479$$ L $$=124.79$$ mL

      So, the nearest integer is $$125$$ mL.


      There are n connected container having volume V, 2V, 3V, ......, nV separated by stopcock. All container have same moles of gas at same temperature. If pressure of first container is P, then final pressure when all stop cocks are opened is ?

      The number of moles present in one container is PV/RT.
      Total number of moles present in n containers is  nPV/RT.
      Letn $$ P_f$$ be the final pressure. The final volume is $$\dfrac{n(n+1)V}{2}$$.
      Total number of moles when all the stopcocks are opened is $$\dfrac{P_fn(n+1)v/2}{RT}$$.
      Since the number of moles remains same,
      $$\dfrac{P_fn(n+1)V/2}=n(PV)$$
      Hence, $$P_f=\dfrac{2P}{n+1}$$. This is the final pressure when all the stopcocks are opened.


      A mixture in which the mole ration of $${H}_{ 2 }$$ and $${O}_{ 2 }$$ is 2:1 is used to prepare water by the reaction
      $${ 2H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow { 2H }_{ 2 }O\left( g \right)$$
      The total pressure of the container is 0.8 atm at 20 before the reaction. Determine the final pressure at 120 after reaction assuming 80% yield of water.

      Concept :
      The total pressure and partial pressure are proportional to the number
      of moles in the system.
      Explanation : Total pressure = 0.8 atm at $$ 20^{\circ}C$$
      initial partial pressure of these gases
      pp of $$O_2 = \dfrac{0.8}{3} = 0.26\, atm $$
      pp of $$ H_{2} = 0.26 \times 2 = 0.52$$ atm
      Given:- yield = 80%
      therefore, 80% of original molecules left unreacted
      $$ \therefore $$ Partial pressure of original gases after reaction.
      $$ H_{2} = 0.52 \times 0.20 = 0.104\, atm $$
      $$ O_{2} = 0.36 \times 0.20 = 0.052 $$ atm
      Or the Basis of strichiometry, amount of $$ H_{2}O $$ formed will be equal to 
      the amount of $$ H_{2}$$ that realized therefore,
      $$ 0.52 \times 0.80 = 0.416\, atm $$
      Total pressure at $$ 20^{\circ}C = $$
      Sum of partial pressure of gases
      $$ PT (20^{\circ}C) = 0.104+0.052 + 0.416$$
      $$ = 0.572\, atm $$
      Pressure at $$ 120^{\circ}C :-$$
      $$ P(120^{\circ}C) = \dfrac{0.57 \times 3.93}{293} = 0.76\, atm $$


      1115650_867169_ans_5d63fcf4ab8646c6a48b655714e5cfa3.png


      The kinetic molecular theory attributes an average kinetic energy of $$3/2\ KT$$ to each particle. What rms speed would a mist particle of mass $$10^{12}\ gm$$ have at room temperture $$(27^0{\circ}C)$$ according to the kinetic molecular theory. 

      Ans: $$V_{rms}=3.525\times 10^{-15}ms^{-1}$$
      Root mean square velocity
      $$V_{rms}=\left (\dfrac {3\ RT}{M}\right)^{1/2}$$ pen mole
      $$\therefore \ RT=KT$$ per molecule / particle
      $$K=1.380\times 10^{-23}m^2 kg^{5^{-2}}K^{-1}$$
      $$M=10^{12} gm=10^9 kg$$
      $$T=27^o C=(20+273)K=300\ K$$
      $$\therefore \ V_{rms}=\left (\dfrac {3\ KT}{M}\right)^{1/2}$$
      $$=\left (\dfrac {3\times 138064\times 10^{-23}m^2 kg\ s^{-2}\times 300\ K}{10^9 kg}\right)^{1/2}$$
      $$=\left (\dfrac {12.4258}{10} \times 10^{-21}\right)^{1/2} m\ s^{-1}$$
      $$=3.525\times 10^{-15}\ m\ s^{-1}$$


      At $$817^0C, K_p$$ for the reaction between pure $$CO_2$$ and excess hot graphite to form $$2CO(g)$$ is 10 atm.
      What is the analysis of the gases at equilibrium at $$817^0C$$ & a total pressure of 4.0 atm? What is the partial pressure of $$CO_2$$ at equilibrium?

      $$CO_{2}(g)+C$$ (hot graphite) $$\rightarrow 2CO$$
      $$P_{0}$$                                  $$0$$
      $$P_{0}-P$$                        $$2P$$
      total pressure $$=4\ atm$$
      $$P_{0}-P+2P=4$$
      $$P_{0}+P=4\Rightarrow P_{0}=4-P$$
      $$K_{P}=\dfrac{(P_{CO})^{2}}{P_{CO_{2}}}=\dfrac{(2P)^{2}}{P_{0}-P}\dfrac{4P^{2}}{P_{0}-P}$$
      $$\dfrac{4P^{2}}{P_{0}-P}=10$$
      $$4P^{2}=10P_{0}-10P$$
      $$4P^{2}=10)4-P)-10P$$
      $$4P^{2}+20P-40=0$$
      $$P^{2}+5P-10=0$$
      $$P=\dfrac{-5+\sqrt{25+41}}{2}=1.531$$
      $$P_{0}=4-1.53=2.469$$
      $$P_{CO_{2}}=P_{0}-P=2.468-1.531=0.937$$


      Calculate the value of $$R$$ in S.I unit for one mole of a gas.
      (Given $$P={10}^{5}N/{m}^{2}\,V=0.0227098{m}^{3},\,T=273.15K,\,n=1mole$$)

      ideal gas equation $$PV = nRT$$
                     $$\dfrac{PV}{nT} = R$$
                     $$\dfrac{10^5\times0.0227098} {1\times273.15K} = R$$
                     $$R =  8.314 Joule{mol^-1}{K^-1}$$



      Pure $$PCl_5$$ is introduced into an evacuated chamber and it comes to equilibrium at $$250^oC$$ and $$2\ atm$$. The equilibrium contains $$40.7\%$$ chlorine by volume. Calculate the partial pressures of all the gases present at equilibrium.

      $$V_{sol} = \dfrac{nRT}{P} = {1\times .0821\times 523}{2} = .05L$$ or $$50ml$$
      Now in $$100ml$$ of solution chlorine is $$40.7ml$$
      $$\therefore$$ in $$50 ml$$ chlorine is $$= \dfrac{40.7\times 50}{100} = 20.35ml$$
      $$PCl_5 \longrightarrow PCl_3 +Cl_2$$
      As the stochiometry coffiecient of $$PCl_3$$ and $$Cl_2$$ is same so  the volumes will be same.
      So $$V_{PCl_3} = V_{Cl_2} = 20.35ml$$  $$Moles(n) = \dfrac{V}{22.4}$$
      $$n_{PCl_3} or n_{Cl_2} = \dfrac{20.35}{22400} = 9\times 10^{-4}$$

      $$V_{PCl_5} = 100- 40.7 = 59.3ml$$
       so $$n_{PCl_5} = \dfrac{59.3}{22400}=2.6\times 10^{-3}$$
      Now
      $$P_{PCl_5} = \dfrac{2.6\times 10^{-3}}{4.4\times 10^{-3}}$$
      $$P_{PCl_3} or P_{Cl_2} = \dfrac{.9\times 10^{-3}}{4.4\times 10^{-3}}$$



      A quantity of an ideal gas is collected in a graduated tube over the mercury in a barometer type arrangement. The volume of gas at $$20^o$$C is $$50$$ ml and the level of mercury is $$100$$ mm above the outside of the mercury level. The atmospheric pressure is $$750$$ mm. Volume of gas at STP is :(Take $$R=0.083$$ it. atm/K/mole).

      $$P_1= 650$$ torr, $$V_1=50ml$$, $$T_1=20^oC=293K$$
      $$P_2=760$$torr, $$V_2=?$$, $$T_2=273K$$
      Ideal gas Law: $$\cfrac {P_1V_1}{T_1}= \cfrac {P_2V_2}{T_2}$$ rearranges to $$V_2= \cfrac {P_1V_1T_2}{T_1P_2}$$
      $$\Rightarrow V_2= \cfrac {(650)(50)(273)}{(760)(293)}$$
      $$\Rightarrow 39.84ml$$
      $$\Rightarrow$$ Volume of gas at STP is $$39.84ml$$


      Calculate the total pressure in a 10 litre cylinder which contains o.4g He,1.6g oxygen and 1.4g nitrogen at$${ 27 }^{ 0 }C$$. Also calculate the partial pressure of He gas in the cylinder. Assume ideal behavious for gases :

      PV = nRT = 10 litre T = $$27^0$$ C = 300 K
      nHe = $$\frac{0.4}{4} = 0.1$$

      $$N_2$$ =  $$\frac{1.4}{28} = 0.05$$

      $$O_2$$ =  $$\frac{1.6}{32} = 0.05$$

      So the total number of moles = 0.1 + 0.05 + 0.05 $$\times 0.2$$
      P = 0.492 atm
      Partial Pressure = Total Pressure$$ \times$$ mole fraction

      $$P_{He}$$= 0.492 $$(\frac {0.1}{0.2})$$
      = 0.246 atm is the partial pressure of He gas in then cylinder. 

      $$P_{N_2}$$ = 0.492 $$(\frac {0.05}{0.2})$$
      = 0.123 atm  is the partial pressure of nitrogen gas in then cylinder.
       
      $$P_{O_2}$$ = 0.492 $$(\frac {0.05}{0.2})$$
      = 0.123 atm is the partial pressure of oxygen gas in then cylinder. 


      An open container of volume $$V$$ contains air at temperature $$27^oC$$ or $$300K$$. The container is heated to such a temperature so that amount of gas coming out is $$2/3$$ of
      (a) amount of gas initially present in the container. 
      (b) amount of gas finally remaining in the container.
      Find the temperature to which the container should be heated.

      (a) Here, $$P$$ & $$V$$ are constant, $$n$$ & $$T$$ are changing. Let, initially the amount of gas present be $$n$$ & temp is $$27^oC$$ or $$300K$$. Finally amount of gas present in container $$=n-\dfrac{2}{3}n = \left(\dfrac{1}{3}\times n\right)$$ & final temperature be $$T$$.
      Then using $$n_1T_1 = n_2T_2$$, we have, $$n\times 300=\dfrac{n}{3}\times T_2 \Rightarrow T_2 = 900K$$ i.e., final temp $$= 900K$$

      (b) Let there be $$x$$ moles of gas remaining in the container, $$\dfrac{2}{3}$$ of $$x$$ come out
      $$\therefore \left(\dfrac{2}{3} x + x\right) = n \Rightarrow \dfrac{5x}{3} = n$$  $$\therefore x = \dfrac{3n}{5}$$
      $$\therefore$$ Using $$n_1\ T_1 = n_2\ T_2$$      $$n\times 300 K = \dfrac{3n}{5} \times T_2$$ 
      $$\therefore T_2 = 500K$$
      Final temperature $$=500K$$


      Find out the temperature at which vessel was heated.

      Let the number of moles of air present in vessel= $$n$$
      $$60$$% means $$3/5^{th}$$ of the original moles of gas has been expelled.
      $$\therefore$$ Remaining number of moles of air in vessel= $$(n-\dfrac35n)=\cfrac {2n}{5}$$
      At constant pressure and volume,
      $$nT_1= n_2T_2$$
      $$\Rightarrow (n)(300)=(2n/5)(T_2)$$
      $$\therefore$$ Vessel's temperature, $$T_2= 750K= 477^oC$$


      A gaseous mixture containing equal masses of $$He$$ and $$CH_4$$ is a total pressure of $$2 \ atm$$. Calculate partial pressure of $$He$$.

      Let $$x \ gm$$ of $$He$$ and $$CH_4$$ be present.

      Given, $$P_{total}=2 \ atm$$

      Number of moles of $$He= \cfrac x4$$

      Number of moles of $$CH_4= \cfrac x{16}$$

      According to Dalton’s law of partial pressure,

      $$P_A= \chi _A P_{total}$$

      $$\chi _A$$ is mole fraction of $$A$$

      $$\chi _{CH_4} = \cfrac {\cfrac x4}{\cfrac x4 + \cfrac x{16}}=\cfrac {0.25}{0.3125}=0.8$$

      $$P_{CH_4}=0.8 \times 2=1.6 \ atm$$

      $$\therefore P_{He}=0.4\ atm$$



      A gaseous mixture containing equal amount of nitrogen and methane has total pressure $$7.86$$ mm Hg at $$300$$K. Calculate the partial pressure (in mm Hg) of methane in the mixture.

      Total pressure $$=P$$
      Partial pressure of methane $$={ P }_{ m }$$
      Partial pressure of nitrogen $$={ P }_{ n }$$
      Partial pressure of methane $$=$$ mole fraction of methane $$\times$$ $$P$$
      mol fraction of methane $$=\dfrac { x }{ 16 } / (\dfrac { x }{ 14 } +\dfrac { x }{ 16 }) =\dfrac { 14 }{ 30 } $$
      $$\Rightarrow \dfrac { 14 }{ 30 } \times 7.86=\dfrac { 110.04 }{ 30 } =3.668$$ $$mm{ H }_{ g }$$.


      Two litres of an ideal gas at a pressure of $$10$$ atm expands isothermally to a final volume of $$10$$ litres against a constant external pressure of $$1$$ atm. What will be the work done if process is done reversibly? 

      Heat absorbed for an isothermal expansion is given as, $$nRTln(\dfrac{V_2}{V_1})$$ where, n is the number of moles, T is the constant temperature associated with the process, $$V_2$$ is the final volume and $$V_1$$ was the initial volume.

      Considering, $$n=1$$ applying ideal gas law, we can find the temperature,

      so, $$P_1V_1=nRT$$

      or, $$T=\dfrac{P_1V_1}{nR}$$

      or, $$RT=P_1V_1$$, for $$n=1$$

      so, $$RT=10\times2=20\ atm.L$$

      So, heat absorbed is, $$1\times 10\times ln(\dfrac{10}{2})=32.2\ atm.L$$

      And we know from $$1^{st}$$ law of thermodynamics, change in internal energy for an isothermal process is zero. 

      So, work done $$=-$$ heat absorbed $$=-32.2 \ atm.L$$



      An evacuated bulb of known volume is filled with $$H_2$$ gas at room temperature $$(30^oC)$$. The pressure of the gas in the bulb is 750 mm Hg. A portion of the gas is transferred to a different flask and found to occupy a volume of 50.0 mL at 1 atm pressure and at the same temperature. The pressure of the $$H_2$$ gas remaining in the original bulb drops to 600 mm Hg. What is the volume of the bulb assuming $$H_2$$ gas is an ideal gas ?

      For ideal gas, equation of state is given by $$PV=nRT$$
      Let $$V$$ be the volume of bulb and $${ n }_{ 1 }$$ moles are there initially then we have the initial state equation as,
      $$756\times V={ n }_{ 1 }RT\quad \longrightarrow \left( I \right) $$
      let $${ n }_{ 2 }$$ moles are transferred to flask, then we have 
      $$760\times 40={ n }_{ 2 }\times RT\quad \longrightarrow \left( II \right) $$
      Now we have remaining $$\left( { n }_{ 1 }-{ n }_{ 2 } \right) $$ moles in the bulb at pressure $$625$$ m of $$Hg$$.
      Hence we have,
      $$625\times V=\left( { n }_{ 1 }-{ n }_{ 2 } \right) RT={ n }_{ 1 }RT-{ n }_{ 2 }\times R\times T\quad \longrightarrow \left( III \right) $$
      by substituting for $${ n }_{ 1 }RT$$ from $$(I)$$ and for $${ n }_{ 2 }RT$$ from $$(II)$$ and in equation $$(III)$$, we can solve the equation $$(III)$$ we get, $$V=232ml$$


      Match the description in Column I with graph provided Column II. For $$n$$ moles of ideal gas at temperature $$T$$.

      A=4 
      In $$\dfrac { P }{ V } $$vs P curve P is increased
      B=3
      In $$\dfrac { P }{ V } $$vs V curve the value of V decreased with increasing $$\dfrac { P }{ V } $$ value
      C=1
      In$$\dfrac { V }{ P } $$vs $${ P }^{ -2 }$$curve value of $${ P }^{ -2 }$$ is increased
      D=2
      In $$\dfrac { P }{ V } $$vs log P curve there will be a constant value.



      If the pressure and absolute temperature of $$2$$ liter of carbon dioxide are doubled, the volume of carbon dioxide would become (in litres)?

      The pressure and absolute temperature are doubled.
      Therefore, $${ P }_{ 2 }=2{ P }_{ 1 }$$ and $${ T }_{ 2 }=2{ T }_{ 1 }$$ and $${ V }_{ 1 }=2$$ litre 
      now, $$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } $$
      $$\Rightarrow \dfrac { { P }_{ 1 }\times 2lit }{ { T }_{ 1 } } =\dfrac { 2{ P }_{ 1 }\times { V }_{ 2 } }{ 2{ T }_{ 1 } } $$
      $$\Rightarrow { V }_{ 2 }=2lit$$
      $$\therefore$$   The volume of carbon dioxide would not change i.e. $$2lit$$.


      The density of liquid nitrogen is $$0.807 \, g \, mL_{-3}$$. If a person acidently swallowed  a $$.028mL$$ drop of liquid nitrogen. What volume of nitrogen gas would be evolved in their body at $$100.0kPa$$ and $$27^oC$$?


      1343123_1134628_ans_360638fb96f74538842d1ef9e1047e02.jpg


      If the kinetic energy of an electron is increased $$9$$ times  then the wavelength associated with it would become.


      1352474_1281680_ans_4ac0a598d0db4356964a5aaa8327512e.jpg


      For a gas adsorbed on a particular adsorbent at
      $${O^ \circ }C,$$ the plot of $$\log \frac{x}{m}$$ versus log $$P$$ where $$P$$ is
      in atm has a slope and intercept as shown in the fig.
      Find the mass of the gas adsorbed by $$10$$ g of the adsorbent at $$0.2$$ atm.
      1290893_582f09d425e0441f9f63b76442148813.png

      $$\dfrac{x}{m} = k . p^{v_n}$$
      $$\log \left(\dfrac{x}{m} \right) = \dfrac{1}{n} [\log \ k.p]$$
      $$\Rightarrow \log \left(\dfrac{x}{m} \right) = \dfrac{1}{n} [\log p + \log k]$$
      $$\log \dfrac{x}{m} = \dfrac{1}{n} \log p + \dfrac{1}{n} \log k$$
      $$y = m x + c$$
      $$\Rightarrow m = \tan \theta $$
      $$\rightarrow \dfrac{1}{n} = \tan 45^{\circ}$$
      $$\rightarrow \dfrac{1}{n} = 1$$
      $$\rightarrow n = 1$$
      $$\log k = 0.30$$
      $$\log \dfrac{x}{10} = \log 0.2 + 0.30$$
      $$\log \dfrac{x}{10} = \log \dfrac{2}{10} + 0.30$$
      $$\log x = \log 2 + 0.30$$
      $$\log \dfrac{x}{2} = 0.30$$
      $$\dfrac{x}{2} = 2$$
      $$\Rightarrow x = 4$$

      1352662_1290893_ans_44d55ce327d9428a9db4e3efc2b25e9d.png


      1L flask containing vapour of methyl alcohol at a pressure of 1atm, and 25C was evacuated till the fill pressure was $${ 10 }^{ -3 }$$ mm.How many molecules of methyl alcohol were left in the flask?


      1323398_1327484_ans_a5b0fb49f48145e0b87cc0f82552e6b4.jpg


      At one bar pressure, the volume of a gas is 0.6 liter. If the gas receives 122 Joules of heat at one atmosphere pressure, the volume becomes 2 liters, then calculate its internal energy. (1 liter bar=101.32 Joule)

      Pressure P=1 bar


      A mixture of helium and neon gases is collected over water at $${28.0^0}C$$ and $$745mm$$ of Hg, If the partial pressure of helium is $$368mm$$ of Hg, what is the partial pressure of neon? (Vapour pressure of water at $${28.0^0}C = 28.33$$ mm of Hg)


      2041268_1360663_ans_3496fde0ef644775aaed8c5b71916910.jpg


      At $$400\ K$$ temperature in a closed vessel the $$\%$$ by volume of $$He,Ne$$ and $$Ar$$ are $$40\%, 40\%$$ and $$20\%$$ respectively. If the total pressure is $$25$$ bar, then find the partial pressure of each gas. (Total pressure is $$25$$ bar)


      1371024_1356369_ans_ac6f3972f09540388da941df4e580edd.jpg


      When the size of a soap bubble is increased by using more air in it, the surface area increases. Does it mean that the average separation between the surface molecules is increased?

      No. The average intermolecular distances do not increase with an increase in the surface area.

      A soap bubble's layer consists of several thousand layers of molecules. An increase in the surface area causes the surface energy to also increase. This in turn allows more and more molecules from the inner liquid layers of the bubble to attain potential energy, enabling them to enter the outer surface of the bubble. Hence, the surface area increases.



      All the substances are formed by small ............


      1990881_1819749_ans_72f43547080d4b2cb6b641ebb7578917.jpg


      What are the three states of matter ? Define each of them with two examples. 

      Three states of matter are:
      (i) Solids
      (ii) Liquids
      (iii) Gases
      $$\bullet$$ The difference in the intermolecular force of attraction, intermolecular spaces and random motion of particles of matter leads to the formation of three states of matter.


                              $$\textbf{SOLIDS}$$

                            $$\textbf{LIQUIDS}$$

                          $$\textbf{GASES}$$

                             $$\textbf{Shape}$$

      Solids have definite shape

      Liquid has no definite shape (It takes the shape of the container in which it is stored.)

      Gases does not have definite shape

                           $$\textbf{Volume}$$

      Solids have definite volume

      Liquids have a definite volume

      Gases do not have definite volume either.

          $$\textbf{Intermolecular attractions}$$

      The particles present in the solids have strong intermolecular force of attraction between them

      The intermolecular force of attraction between the particles is less than solid particles.

      The intermolecular force of attraction between the gaseous particles is least 

             $$\textbf{Intermolecular space}$$

      There is little space between the particles of a solid

      These particles have a greater space between them.

      The space between gas particles is the greatest

                           $$\textbf{Examples}$$

      Iron rod, stone etc.

      Water, oil etc.

      Oxygen gas , carbon dioxide gas.



      State the following 
      (c) Gas equation.

      The gas equation is:

      $$\dfrac{P_1V_1}{T_1}= \dfrac{P_2V_2}{T_2}$$


      Give the assumptions of the kinetic molecular theory

      $$i)$$ Gases are made of tiny particles which move in all possible directions at all possible speeds. The size of molecules is small as compared to the volume of the occupied gas.

      $$ii)$$ There is no force of attraction between gas particles or between the particles and the walls of the container. So, the particles are free to move in the entire space available to them.

      $$iii)$$ The moving particles of gas collide with each other and with the walls of the container. Because of these collisions, gas molecules exert pressure. Gases exert the same pressure in all directions.

      $$iv)$$ There is large interparticle space between gas molecules, and this accounts for high compressibility of gases.

      $$v)$$ Volume of a gas increases with a decrease in pressure and increase in temperature.

      $$vi)$$ Gases have low density as they have large intermolecular spaces between their molecules.

      $$vii)$$ Gases have a natural tendency to mix with one and other because of large intermolecular spaces. So, two gases when mixed form a homogeneous gaseous mixture.

      $$viii)$$ The intermolecular space of a gas is reduced because of cooling. Molecules come closer resulting in liquefaction of the gas. 


      Deduce the relation $$PV = nRT$$, where $$R$$ is a constant called universal gas constant.

      According to Boyle’s law, $$V\ \alpha\ \dfrac{1}{P}$$  (at constant $$T$$ and $$n$$)

      According to Charle’s law, $$V\ \alpha\ \dfrac{1}{T}$$ (at constant $$P$$ and $$n$$)

      According to Avogadro's law, $$V\ \alpha\ n$$ (at constant $$T$$ and $$P$$)

      Combining the three laws, $$V\ \alpha\ \dfrac{nT}{P}$$ or $$PV\ \alpha\ nRT$$

      Or $$PV = nRT$$ where $$R$$ is a constant called universal gas constant.



      How is the partial pressure of a gas in a mixture related to the total pressure of the gaseous mixture?

      Partial pressure of a gas = Mole fraction of that gas $$\times$$ Total Pressure.


      Explain the following observation : The tyre of an automobile is inflated at lesser pressure in summer than in winter.

      The pressure of the air is directly proportional to the temperature. During summer due to high temperature the pressure in the tyre will be high as compared to that in water. The tube may burst under high pressure in summer. Therefore, it is advisable to inflate the tyres to lesser pressure in summer than in winter.


      Briefly explain Pressure-Temperature Law.


      1928998_1865902_ans_996a7998d5f2404297e251b81ab3bfd5.jpg


      State and explain Daltons law of partial pressures. How is this law applied in the determination of the pressure of dry gas from that of the moist gas?


      1928910_1865910_ans_bad6423f66d640ecb0d6db55e3eb3e18.jpg


      Class 11 Medical Chemistry Extra Questions