States Of Matter - Class 11 Engineering Chemistry - Extra Questions

At $$27^0C$$ temperature 20 gm. $$H_2$$, 220 gm $$CO_2$$ and 140 gm $$N_2$$ are filled in a vessel having volume 2 litre. Find the total pressure in bar unit which gas is removed from the vessel so that pressure can be reduced by 50%.

No. of moles of $$H_{2}=\cfrac{20}{2}=10$$ moles
No. of moles of $$CO_{2}=\cfrac{220}{44}=5$$ moles
No. of moles of $$N_{2}=\cfrac{140}{28}=5$$ moles
So, according to Dalton's law of partial pressure,
pressure $$\propto$$ mole fraction of $$H_{2}$$ 
So, as mole fraction is 0.5, when hydrogen is removed half of the pressure is removed.


What temperature would be necessary to double the volume of a gas STP if the pressure is decreased by 50%?


$$\displaystyle \dfrac {PV}{T}=\dfrac {P'V'}{T'} $$
$$\displaystyle P=2P'$$ as pressure is decreased by 50%
$$\displaystyle V=0.5V'$$ as volume is doubled
$$\displaystyle T=273.15 \ K$$ under STP conditions
$$\displaystyle \dfrac {PV}{T}=\dfrac {P'V'}{T'} $$
$$\displaystyle \dfrac {2P' \times 0.5V'}{273.15 \ K}=\dfrac {P'V'}{T'} $$
$$\displaystyle \dfrac {2 \times 0.5}{273.15 \ K}=\dfrac {1}{T'} $$
$$\displaystyle \dfrac {1}{273.15 \ K}=\dfrac {1}{T'} $$
$$\displaystyle T'=273.15 \ K$$
Hence, temperature of 273.15 K would be necessary to double the volume of a gas STP if the pressure is decreased by $$50\%$$.


$$22.4$$ litres of a gas weighs $$70$$g at S.T.P. Calculate the weight (gm) of the gas if it occupies a volume of $$20$$ litres at $$27^o$$C and $$700$$mm Hg of pressure.

Let us consider the problem:
$$PV = nRT-----1$$
Given that:
$$R = 0.082$$
$$P = \frac{{700}}{{760}} = 0.921{\text{ }}atm$$
$$T = 273 + 27 = 300{\text{ }}{\text{ }}$$
Put the all value in equation 1
$$0.921 \times 20 = n \times 0.082 \times 300$$
$$n = 0.748{\text{ }}moles$$
$$1$$ mole contains $$70g$$
$$0.748{\text{ }}contains{\text{ }}-----?$$
$$0.748 \times \frac{{70}}{1} = 52.414$$


An evacuated glass vessel weighs $$50.0$$g when empty, $$148.0$$ gm when filled with a liquid of density $$0.98$$ g/mL and $$50.5$$ g when filled with an ideal gas at $$760$$ mm Hg at $$300$$k. Determine the molecular weight of the gas?

Weight of Liquid $$= 148 - 50 = 98g$$
Volume of liquid $$= \frac{{98}}{{0.98}} = 100mL = $$Volume of vessel
Thus a vessel of $$100mL$$ contains ideal gas at $$760mm$$of $$Hg$$ at $$300K$$
Weight of gas $$= 50.5 - 50 = 0.5g$$
Using $$PV = nRT$$
$$\frac{{760}}{{760}} \times \frac{{100}}{{1000}} = \frac{{0.5}}{m} \times 0.0821 \times 300$$
Since,
Molecular weight of gas $$(m) = 123$$


The molecular weights of oxygen and sulphur dioxide are $$32$$ and $$64$$ respectively. If $$1$$ litre of oxygen at $$15^o$$C and $$740$$ mm pressure contains $$N$$ molecules, the number of molecules in $$2$$ litre of sulphur dioxide under the same conditions of temperature and pressure will be:

By using Avogadro Law,
$$V \propto n$$ at constant pressure & temperature
Thus, $$\cfrac {V_{O_2}}{V_{SO_2}}=\cfrac {n_{O_2}}{n_{SO_2}}$$
$$\Rightarrow \cfrac {1L}{2L}=\cfrac {N/N_A}{n_{SO_2}}$$
$$\Rightarrow n_{SO_2}=\cfrac {2N}{N_A}$$
Thus, molecules of $$SO_2$$ is $$2N$$ .


500 mL of nitrogen at $$27^oC$$ are cooled to - $$5^oC$$ at the same pressure. Calculate the new volume.

Given,
$$V_1=500mL$$
$$T_1=27^0C=300K$$
$$T_2=-5^0C=268K$$
Now using Charle's Law,
$$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$$
$$\implies \dfrac{500}{300}=\dfrac{V_2}{268}$$
$$\implies V_2=446.66mL$$


Nitrogen gas is filled in a container of volume $$2.32L$$ at $$32^oC$$ and at $$4.7 atm$$ pressure. Calculate the number of moles of gas.[Write up to 3 decimal places]

$$ PV = nRT$$
$$4.7 * 2.32 = n * 0.0821 *305$$
$$\therefore  n = \dfrac{10.904}{25.040} = 0.435$$


The volume of given mass of a gas is $$0.6\ dm^3$$ at a pressure of $$101.325K\ Pa$$. Calculate the volume of the gas if its pressure is increased to $$142.860K\ Pa$$ at the same temperature. (Write answer up to 3 decimal places).

$$P_1V_1 = P_2V_2$$
$$ 101325 * 0.6 = 142860 * V_2$$
$$\therefore V_2 = \dfrac{101325*0.6}{142860} = 0.425$$


Calculate the volume occupied by $$5.25\ g$$ of nitrogen at $$26^{o}C$$ and $$74\cdot 2\ cm$$ of pressure.

P$$=\quad \dfrac { 74.2 }{ 76 }atm$$

V$$=?$$

m$$=5.25g$$

M$$=28g$$

T$$=\quad 273+26=299K$$

$$PV=nRT$$

$$  \dfrac { 74.2 }{ 76 } \times V\quad =\dfrac { m }{ M } \times 0.0821\times 299$$

$$  V=\quad \dfrac { 5.25 }{ 28 } \times 0.0821\times 299\times \dfrac { 76 }{ 74.2 } $$

$$ V=\quad 4.7107\quad L$$


$$1112\ dm^{3}$$ of dry carbon monooxide contains $$P$$ molecules. Calculate the number of molecules in $$336\ dm^{3}$$ of dry chlorides gas. Assume all measurement are made at the same temperature and pressure.

$$1112$$  $$d{m}^{3}$$ contain $$P$$ molecules
therefore $$336$$ $${m}^{3}$$ contain molecules$$=\dfrac { P\times 336 }{ 1112 } =0.30P$$


How much thermal energy should be added to $$3.45\ g\ Ne$$ in a $$10$$ liter flask to raise the tempereture from $$0^{o}C$$ to $$100^{o}C$$. (Atomic  weight $$Ne=20.18$$)?

By using the formula $${ nC }_{ Process }(dt)$$ which is equal to heat, and here heat is converted into thermal energy.
Here process is isochoric.

So, $$C_{ Process }\quad changes\quad to\quad { C }_{ isochoric }$$

$$=\dfrac { 3 }{ 2R } $$

 $$nC_{ isochoric }(dt)=\left( \dfrac { 3.45 }{ 20.18 }  \right) \times \left( \frac { 3 }{ 2 } \times 2 \right) \times (100-0)$$

 $$=5.146\quad cal$$


The mean kinetic energy of a molecule at $${ O }^{ 0 }C$$ is $$5.621\times { 10 }^{ -14 }erg$$. Calculate Boltzmann's constant. If the value of $$R=8.314\times { 10 }^{ 7 }erg$$, then calculate the no. of molecules present in one mole of gas.

$$\dfrac{3}{2} k \times  T  = 5.621 \times 10^{-14}$$
$$\Rightarrow \dfrac{3}{2} \times k \times 273 = 5.621 \times 10^{-14}$$
$$\Rightarrow k = \dfrac{5.621 \times 2 \times 10^{-14}}{3 \times 273}$$
$$\Rightarrow k = 1.372 \times 10^{-16} $$ erg
$$\Rightarrow \dfrac{R}{N_A} = 1.372 \times 10^{-16} erg $$
$$\Rightarrow N_A = 6.059 \times 10^{23}$$


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

$$PV=nRT$$
$$\Rightarrow 0.5 \times V= 0.6 \times R \times 300$$
$$\therefore V= 360R$$

In $$2^{nd}$$ case, the pressure will be same
Flask 1: $$PV=aRT$$
$$\Rightarrow P(360R)=aR(300)$$
$$\Rightarrow a= 1.2P$$
Flask 2: $$PV=bRT$$
$$\Rightarrow P(360R)=bR(600)$$
$$\Rightarrow b= 0.6P$$
$$\Rightarrow a+b= 1.2P+0.6P= 1.8P= 0.6$$
$$\Rightarrow P= 0.33$$
$$\therefore a= 0.396, b=0.198$$ .


Write down the reason for the following:
a) Ink can be blotted with chalk.
b) Sweat can be blotted with tissue paper.

a) Due to capillarity
b) Due to capillarity


How many oxygen atoms combine with one carbon atom?

$$2$$ Oxygen atoms


How many oxygen atoms combine with $$1000$$, carbon atom?

$$2000$$ Oxygen atoms.


What is the law called which deals with the ratios of the volumes of the gaseous reactants and products?

Gay Lussac's law of gaseous volumes is the law called which deals with the ratios of the volumes of the gaseous reactants and products.


Take water in a vessel and put a needle on the surface of water carefully.
a. Will the needle be immersed in water. Justify your answer.
b. Why do the water drops assume spherical shape?

a. No, because of the surface tension of water.
b. The surface tension reduces the surface area.


Fill in the blanks
The average kinetic energy of the molecules of gas is directly proportional to the _________

The average kinetic energy of the molecules of gas is directly proportional to the Absolute temperature.


What is the ideal gas equation? How can it be derived? Also, express it in terms of the density of the gas.


1929961_1865906_ans_100e0bd57e8f4017ab962d3d6e9d1f3f.jpg


In the equation $$PV=RT$$, the value of $$R$$ in SI unit is (as nearest integer) :

As we know,
value of R is 8.314 in SI unit so nearest integer value is 8.


At room temperature, ammonia gas at one atmosphere pressure and hydrogen chloride at $$P$$ atmospheric pressure allowed to effuse through identical pinholes from opposite ends of a glass tube of one metre length and uniform section. $$NH_{4}Cl$$ is first formed at a distance of $$60\ cm$$ from the end through which $$HCl$$ gas is sent in. What is the value of?

Using Graham's law of diffusion,
$$\dfrac {r_{HCl}}{r_{NH_{3}}} = \sqrt {\dfrac {M_{NH_{3}}}{M_{HCl}}} \times \dfrac {P_{HCl}}{P_{NH_{3}}}$$
=$$\dfrac{60}{40}=0.682\times P_{HCl}$$
$$P_{HCl}=2.198\ atm$$


A polythene bag of $$3$$ litre capacity is filled by Helium gas (occupying $$Li$$ at $$0.3atm$$ and $$300K$$) at $$300K$$. Subsequently enough $$Ne$$ gas is filled to make total pressure $$0.4atm$$ at $$300K$$. Calculate ratio of moles of $$Ne$$ to $$He$$ in container.

$${ x }_{ He }=\dfrac { { p }_{ He } }{ { p }_{ He }+{ p }_{ Ne } } =\dfrac { 0.3 }{ 0.3+0.1 } =\dfrac { 3 }{ 4 } $$
$${ x }_{ Ne }=\dfrac { 0.1 }{ 0.1+0.3 } =\dfrac { 1 }{ 4 } $$
now, $$\dfrac { { n }_{ Ne } }{ { n }_{ He } } =\dfrac { { x }_{ Ne } }{ { x }_{ He } } =\dfrac { 1 }{ 3 } $$


At a pressure of 3 atm air (treated an ideal gas) is pumped into the tubes of a cycle rickshaw. The volume of each tube at a given pressure is $$0.004{m}^{3}$$. One of the tubes gets punched and the volume of the tube reduces to $$0.0008{m}^{3}$$. Find the number of moles of air that have leaked out? 
[Assume that the temperature remains constant at 300K. $$ R=25/3J{ mol }^{ -1 }{ K }^{ -1 }  $$]

$$ P_{1} = 3 atm \, P_{2} = ?$$
$$ V_{1} = 0.004 m^3 = 4 L,  V_{2} = 0.0008 m^3=0.8 L$$
$$ T = 300 K $$

No. of moles before the leakage of the air, 

 $$n_1= \dfrac{PV}{RT} = \dfrac{3\times 4}{0.082 \times300}$$

$$ n_1= 0.488\ mole$$

No. of moles after the leakage of the air, 

Assuming the pressure becomes equal to atmospheric pressure.

$$ P_2V_2 = n_2RT $$

$$ n_2 = \dfrac{PV}{RT} = \dfrac{1\ atm \times 0.8 L}{0.082 L-atm-mol^{-1} K^{-1} \times300K}=0.0325\ mol$$

$$ \therefore $$ leaked mole of gas = $$ 0.488- 0.0325=0.455\ mol$$

No. of moles of air leaked out $$= 0.455\ mol$$


A sample of gas occupies $$10 L$$ at $${ 127 }^{ o }C$$ and 1 bar pressure . The gas is cooed to $${ -73 }^{ o }C$$ at the same pressure. What will be the volume of the gas? 

Solution:- 
According to Charle's law 
$$V \propto T$$
$$\cfrac{{V}_{1}}{{V}_{2}} =  \cfrac{{T}_{1}}{{T}_{2}}$$
Let the initial volume be V.
$$\therefore$$ Initial volume, $$\left( {V}_{1} \right) = 10 L$$
Final volume, $$\left( {V}_{2} \right)= ?$$
Initial temperature, $$\left( {T}_{1} \right) = 127 ℃ = \left( 273 + 127 \right) K = 400 K$$
Final temperature, $$\left( {T}_{2} \right) = -73 ℃ = \left( 227 + \left( -73 \right) \right) K = 200 K$$
$$\cfrac{10}{{V}_{2}} = \cfrac{400}{200}$$
$$\Rightarrow \; {V}_{2} = \cfrac{10}{2} = 5 L$$
Hence, at $$73^0C$$ the volume of the gas will be $$5 L$$.


Helium gas at $$27^oC$$ is subjected to a pressure change from 5 atm to 2 atm at constant temperature. The initial volume of the gas was $$10 dm^3$$. Calculate the change in volume of gas in $$dm^3$$.

$$PV=nRT \Rightarrow PV=constant =$$

$$\therefore P_{1}V_{1}=P_{2}V_{2}$$

$$\Rightarrow 5\times 10=2\times V_{2}\Rightarrow V_{2}=25$$

$$\therefore $$ change in volume = $$15 dm^{3}$$


One mole of an ideal gas is subjected to a process in which $$P = \dfrac{1}{8.21}$$ $$V$$ where $$P$$ is in atm and a $$ V$$ in litre. If the process is operating from $$1$$ atm to finally $$10$$ atm (no higher pressure achieved during the process) then that would be the maximum temperature obtained & at what instant will it occur in the process.

$$ \text{As,     PV=nRT}   \\  \Rightarrow \text{PV=RT     [$\because$ n=1 ]} \\  \text{Given,  P(8.21) = V} \Rightarrow P (8.21) = \cfrac{RT}{P} \Rightarrow P^{2}(8.21) = (0.0821)T \\ \Rightarrow T= 100 P^{2} \\ \text{At 10 atm T will be highest,     T = (100)(100) = 10$^{4}$ K}$$


A compound exists in the gaseous state both as a monomer $$(A)$$ and dimer $$(A_2)$$. The molecular weight of the monomer is $$48$$. In an experiment, $$96$$ g of the compound was confined in a vessel of volume $$33.6$$ litres and heated to $$273^0$$C. Calculate the pressure developed, if the compound exists as a dimer to the extent of $$50$$ per cent by weight, under these conditions. $$(R = 0.082)$$

$$A$$ & $$A_2$$ are two states in gaseous phase having ratio of $$50$$% i.e $$1:1$$ .
$$\therefore$$ Mole of $$A=\cfrac {96}{2}\times \cfrac {1}{48}=1$$
We know $$n=\cfrac {W}{m}$$
$$\therefore$$ Mole of $$A_2=\cfrac {96}{2}\times \cfrac {1}{96}=\cfrac {1}{2}$$
Total moles of $$A$$ & $$A_2$$ are=$$1+\cfrac {1}{2}=\cfrac {3}{2}$$
$$PV=nRT$$
$$\Rightarrow P \times 33.6=\cfrac {3}{2}\times 0.0821\times 546$$
$$\Rightarrow$$ Pressure= $$2 atm$$


Fixed mass of a gas is subjected to the changes as shown is diagram, calculate $$T_3, T_4, P_1, P_2$$ and $$V_1$$ as shown is diagram. Considering gas obeys $$PV = nRT$$ equation.
1093570_053645c78849483fbf8fb91fcb87ca53.png

$$ 10(4) = 8(P_{1}) \text{          [Boyle's Law $\rightarrow$ constant temperature]} \\ P_{1} = 5 atm \\ \text{for constant volume      P $\propto$ T} \\ \Rightarrow \cfrac{P_{1}}{3} = \cfrac{T_{2}}{T_{3}} \Rightarrow \cfrac{5}{3} = \cfrac{600}{T_{3}} \Rightarrow T_{3} = 360 K \\ \text{At constant pressure,    V$\propto$ T} \\ \Rightarrow \cfrac{8}{V_{1}} = \cfrac{300}{600} \Rightarrow V_{1} = 16 litres \\ \cfrac{(3)(V_{1})}{T_{3}} = \cfrac{P_{2}(8)}{T_{4}} \Rightarrow \cfrac{P_{2}}{T_{4}} = \cfrac{6}{360} \\ 60 P_{2} = T_{4}$$


One mole of $$NH_4Cl(s)$$ is kept in an open container & then covered with a lid. The container is now heated to $$600$$K where all $$NH_4Cl$$(s) dissociates into $$NH_3$$ and $$HCl(g)$$. If volume of the container is $$24.63$$ litres, calculate what will be the final pressure of gases inside the container. Also find whether the lid would stay or bounce off if it can with stand a pressure difference of $$5.5$$ atm. Assume that outside air is at $$300$$K and $$1$$ atm pressure.

Given that $$\underset {1 mole}{NH_4Cl}\longrightarrow \underset {1 mole}{NH_3}+\underset {1 mole}{Cl}$$
$$PV=nRT$$
        $$=\cfrac {2\times \cfrac {82.1}{1000}\times 600}{24.63}$$
        $$=\cfrac {821\times 3}{25\times 24.63}=4$$
Pressure= $$4-1=3 $$ atm
So, the lid does not bounce.


Pressure of gas contained in a closed vessels is increased by $$0.4\%$$, when heated by $$^{ o }{ C }$$. Calculate its final temperature. Assume ideal nature.

Initial pressure in the vessel $$1atm$$ $$={ P }_{ 1 }$$
After increasing $$0.4$$% the pressure $${ P }_{ 2 }=1.004{ P }_{ 1 }$$.
and $${ T }_{ 2 }={ T }_{ 1 }+1$$
$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \Rightarrow \dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { T }_{ 2 } } $$    (As volume constant).
$${ T }_{ 1 }=\dfrac { { P }_{ 1 }{ T }_{ 2 } }{ { P }_{ 2 } } $$
     $$=\dfrac { { P }_{ 1 }\left( { T }_{ 1 }+1 \right)  }{ 1.004{ P }_{ 1 } } $$
$${ T }_{ 1 }=\dfrac { { T }_{ 1 }+1 }{ 1.004 } $$
$${ T }_{ 1 }=\dfrac { 1 }{ 0.004 } =250K$$


The density of a gas is $$30^{o}C$$  and $$1.3\ atm$$ is $$0.027\ g/ cc$$. Find its $$M.W$$.

$$ P= \cfrac{dRT}{M} \text{;     d=0.027 g/cc = 27 g/L ;    T=30+273 = 303 K ;     P = 1.3 atm} \\ M= \cfrac{dRT}{P} = \cfrac{(27)(0.0821)(303)}{1.3}  = 516.6 gms \\ \text{So molecular weight of gas = 516.6 gms}$$


What volume of oxygen gas $$(O_2)$$ measured at $$0^0$$C and $$1$$ atm, is needed to burn completely $$1$$L, of propane gas $$(C_3H_8)$$ measured under the same conditions?

$$C_3H_8+5O_2 \longrightarrow 3CO_2+4H_2O$$

$$1$$ mole of $$C_3H_8$$ requires $$5$$ mole of $$O_2$$

$$\therefore 1$$ mole $$= 22.4dm^3$$

$$1\times 22.4dm^3$$ propane needs $$5\times 22.4dm^3$$

$$\therefore 5L$$ of oxygen is required to burn $$1L$$ of propane completely.


What is the density of helium at $$500^{o}\ C$$ and $$100\ mm$$ pressure?

$$ \text{Ideal Gas Equation } \Rightarrow P= \cfrac{dRT}{M} \Rightarrow d = \cfrac{PM}{RT}  = \cfrac{(\cfrac{100}{760})(4)}{(0.0821)(773)} = 8.293 g/cc$$


A flask of $$2\ dm^{3}$$ capacity contains $$O_{2}$$ at $$101.325\ k Pa$$ and $$300\ K$$. The gas pressure is reduced to $$0.1\ Pa$$. Assuming ideal behaviour, answer the following:
i) What will be the volume of the gas which is left behind?
ii) What amount of $$O_{2}$$ and $$t$$ he corresponding number of molecules are left behind in the flask?
iii) In now $$2g$$ of $$N_{2}$$ is introduced, what will be the pressure of the flask

Given $$V_1=2dm^3$$             $$P_t=101.325 kP_a ,P_2=0.1 P_a, T=300K$$
(i) Volume of $$O_2$$ left behind will be the same i.e $$2dm^3$$
(ii) The amount of $$O_2$$ left behind
$$n=\cfrac {P_2V_1}{RT}=\cfrac {0.1 \times 10^{-3}kP_a\times 2dm^3}{8.314 \times kP_adm^3/k/mol\times 300}$$
$$=8.019 \times 10^{-8} mol$$
(iii) $$2g$$ of $$N_2= \cfrac {1}{14} mol$$
Total amount of gases in the flask
$$=\cfrac {1}{14}+8.125\times 10^{-8}\simeq \cfrac {1}{14}mol$$
Thus the pressure of the flask is given by,
$$P=\cfrac {nRT}{V}$$
     $$=\left[\left(\cfrac{1}{14}\right)\times (8.315 kP_adm^3k^{-1}mol^{-1}\times 300K)\right]/2dm^3$$
     $$=89.08kP_a$$ .


Plot between P vs PV.
1110876_6e29d1f9d99f4676868a2ea1b1dea26b.PNG

For ideal gas, $$PV=nRT$$
When $$T=constant$$, $$PV=constant$$
Hence $$P$$ vs $$PV$$ will be a straight line plot parallel to $$P$$ axis.

1220724_1110876_ans_04772324aa8247f5bb593a573162d359.png


At 300 K certain mass of a gas occupies 1 x $$10^{-1}dm^3$$ volume. Calculate its volume at 450K and at the same pressure.


1201383_1122291_ans_a8db9ce8586942d49edb5a0df3863237.jpg


One litre of a gas weighs $$2g$$ at $$300K$$ and $$1 atm$$ pressure. If the pressure is made $$0.75 atm$$ at which of the following temperatures will one litre of the same gas, weigh one gram?

$$PV=nRT$$ (Ideal gas equation.)
$$\Longrightarrow \cfrac{P_1V_1}{P_2V_2}=\cfrac{n_1T_1}{n_2T_2}\Longrightarrow \cfrac{1\times 1}{0.75\times 1}=\cfrac{\cfrac{2}{M_w}\times 300}{\cfrac{1}{M_w}\times T}$$
$$\Longrightarrow T=\cfrac{2\times 300\times 0.75}{1}=450K$$
$$\therefore$$ $$\text{Temperature }= 450K.$$


300 ml of gas at $$27^0$$C is cooled to $$-3^0$$C at constant pressure, the final volume is:

$$\cfrac {V_1}{V_2}=\cfrac {T_1}{T_2}$$   {as Pressure is constant}
$$\Rightarrow \cfrac {300}{V_2}=\cfrac {300}{270}$$
$$\Rightarrow \cfrac {300 \times 270}{300}=V_2$$
$$\Rightarrow V_2=270 ml$$


Tempreture at the foot of a mountain is $$30^{0}C$$ and pressure is 760mm whereas at the top of the mountain these are $$0^{0}C$$ and 710mm.compair the densities of the air at the foot and the topof the mountain.


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Find the temp. at which volume & pressure of $$28g$$ nitrogen gas becomes $${ 10dm }^{ 3 }$$ & $$2.46$$ atmosphere resp.


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How does kinetic energy of photoelectron change when the intensity of radiation is increased?


1342480_1137334_ans_3ac63cd1b6f3474184538bf475e23e0a.jpg


11.35 litres of a gas at STP are found to have to have a mass of 8.5g. the number of mole of gas present in 34 g of gas is ?


1350693_1139520_ans_d7cd57ad429942e1955fa050ebd3128b.jpg


Which law relates solubility of solvents with pressure?

Henry's law


Match the following gas laws with the equation representing them.

(i) Boyle's law: 
The pressure of the gas is inversely proportional to its volume ($$T$$ = constant).

(ii) Charles' law: 
Volume of the gas is directly proportional to its temperature ( $$p$$ = constant).

(iii) Dalton's law:
At constant $$V$$ and $$T$$, total pressure of the gaseous mixture of two or more non reacting gases is equal to the algebraic sum of their partial pressure.

(iv) Avogadro law: 
It states that equal volumes of all gases, under the same condition of temperature and pressure, contain equal number of molecules. $$V\propto n.$$


Give various forms of ideal gas equation.

  • $$PV=nRT$$

  • $$PV=\dfrac{m}{M}RT$$

  • $$PV=nR=constant$$

  • $$PM=\dfrac{m}{V}RT$$

  • $$PM=dRT$$

  • $$d = \dfrac{PM}{RT}$$

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


Two flasks $$A$$ and $$B$$ have equal volumes. Flask $$A$$ contains $$H_2$$ and is maintained at $$300 K$$ while $$B$$ contains equal mass of $$CH_4$$ gas and is maintained at $$600 K$$. In which flask pressure is greater? How many times more?

$$\text{Pressure }\alpha\ \text{number of moles}$$
$$\text{Pressure }\alpha\  \text{temperature}$$
If temperature had been equal, pressure in $$A$$ would have been $$8$$ times but in $$A$$ temperature is $$\dfrac{1}{2}$$ of that in $$B$$
$$\therefore  \text{Pressure will be }4 \text{ times more}$$

$$P_1 \times V = n_1 RT_1,\ P_2 \times V = n_2RT_2$$

$$\implies \dfrac{P_1}{P_2} = \dfrac{8 \times R \times 300K}{1 \times R \times 600K}$$

$$\implies \dfrac{P_1}{P_2} = \dfrac{4}{1}$$

$$\text{Pressure in flask A = }4 \times \text{ Pressure in flask B}$$.



Surface tension tends to minimise the surface area of a liquid. Suggest an experiment to prove this.
(Follow the pattern: Required materials, procedure, expected observation).

Materials required: metallic loop, thread, soap water, pin. 

Method of experiment: Tie a thread to a metallic loop, immerse it in soap water and create a soap film as shown in fig. a, b, c.
Prick a portion of film using a pin. The shape of remaining portion of soap film will be as such the surface area is reduced.



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Define Dalton's law of partial pressure. Using this law, how is the pressure of dry gas determined?

Dalton’s law of partial pressure is a gas law which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure exerted by each individual gas in the mixture.  

For example, the total pressure exerted by a mixture of two gases $$A$$ and $$B$$ is equal to the sum of the individual partial pressures exerted by gas $$A$$ and gas $$B$$. 

Mathematically, 

$$P_{Total} = P_1 + P_2 + P_3 + ... + P_n$$

where, $$P_{Total}$$ is the total pressure exerted by the mixture of gases and $$P_1 , P_2 ... P_n$$ are the partial pressures of the gases in the mixture of $$‘n’$$ gases.

Based on Dalton's law, the pressure of dry gas collected can be calculated by subtracting the pressure of the water vapour from the total pressure. This can be mathematically represented as  :

$$P_{dry\ gas}=P_{wet\ gas}-P_{water\ vapor}$$


Explain the following:
$$AlF_{3}$$ is a high melting solid whereas $$SiF_{4}$$ is a gas.

$$AlF_{3}$$ is an ionic solid due to large difference in electronegativities of $$Al$$ and $$F$$ whereas $$SiF_{4}$$ is a covalent compound and hence there are only weak van der Waal's forces among their molecules.


Estimate the average thermal energy of a helium atom at
(i) room temperature (27 $$\displaystyle ^{\circ}C$$),
(ii) the temperature on the surface of the Sun (6000 K),
(iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

(i) At room temperature, $$T=27^oC = 300K$$
Average thermal energy $$=(3/2)kT$$
Where, k is Boltzmann constant $$=1.38\times 10^{-23}m^2kg\,s^{-2}K^{-1}$$
$$\therefore (3/2)kT = (3/2)\times 1.38\times 10^{-38}\times 300$$
$$=6.21\times 10^{-21}J$$

Hence, the average thermal energy of a helium atom at room temperature of $$27^oC$$ is $$6.21\times 10^{-21}J$$.

(ii) On the surface of the sun, $$T = 6000K$$
Average thermal energy $$=(3/2)kT$$
$$=(3/2)\times 1.38\times 10^{-38}\times 6000$$
$$=1.241\times 10^{-19}J$$

Hence, the average thermal energy of a helium atom on the surface of the sun is $$1.241\times 10^{-19}J$$.

(iii) At temperature, $$T = 10^7K$$
Average thermal energy $$=(3/2)kT$$
$$=(3/2)\times 1.38\times 10^{-23}\times 10^7$$
$$=2.07\times 10^{-16}J$$

Hence, the average thermal energy of a helium atom at the core of a star is $$2.07\times 10^{-16}J$$. 


The heat absorbed during the isothermal expansion of an ideal gas against vacuum is 

$$(\partial U/\partial V)_T=0$$ (for ideal gas), because heat depends upon temperature.


Match the Column-I with Column-II.

$$A.$$ Internal energy of a gas can be, 
$$U=\cfrac{3}{2}RT/\cfrac{5}{2}RT/371.16KJ$$ at $$298K$$
There are permittable in case of mono atomic, di atomic molecules cases.

$$B.$$ Translational Kinetic energy of gas molecules, $$K.E=\cfrac{3}{2}RT=3.716KJ$$ at  $$298K.$$

$$C.$$ At zero kelvin, there will be no molecular motion.
$$T=t°C+273\\0=t+273\\t=-273°C$$

$$D.$$ The lowest positive temperature at which molecules of gas has no heat is again $$0$$ Kelvin, $$\Longrightarrow t=-273°C.$$


The pressure exerted by $$12g$$ of an ideal gas at temperature $${1}^{o}C$$ in a vessel of volume $$V$$ litre is one atm. When the temperature is increased by $$10$$ degrees at the same volume, the pressure increases by $$10$$%. Calculate the temperature $$t$$ and volume $$V$$. (Molecular weight of the gas$$=120$$)

Case I : Given
P = 1 atm
w = 12 g
T = (t+273)K
V = V litre
Case ii :
T = (t+283)K
P = 1 + 1010010100 = 1.1 atm
w = 12 g
V = V l
Using gas equation
Case I : 1×V=12m×R(t+273)1×V=12m×R(t+273) ----(i)
Case II : 1.1×V=12m×R(t+283)1.1×V=12m×R(t+283) ----(ii)
By equation (i) and (ii)
1.11=t+283t+2731.11=t+283t+273
∴1.1t+300.3=t+283∴1.1t+300.3=t+283
0.1t = -17.3
t=−173∘Ct=−173∘C = 100K
Also from case I (Since m = 120)
1×V=12120×0.082×1001×V=12120×0.082×100
V = 0.82 l


Butane gas burns in air according to the following reaction,
 
$$2{ C }_{ 4 }{ H }_{ 10 }+13{ O }_{ 2 }\longrightarrow 10{ H }_{ 2 }O+8{ CO }_{ 2 }$$. 

Suppose the initial and final temperatures are equal and high enough so that all reactants and products act as perfect gases. Two moles of butane are mixed with $$13$$ moles of oxygen and then completely reached. Find the final pressure (if the volume remains unchanged and the pressure before the reaction is $${P}_{0}$$)?

$$ 2\, C_{4}H_{10}+ 13O_{2}  \rightarrow 10 H_{2} O+ 8CO_{2}$$
Pressure before reaction $$ = P_{0}$$
                   $$ 2\,C_{4}H_{10} + 13O_{2} \rightarrow 10H_{2}O + 8CO_{2}$$
At $$t =0$$       2 moles     13 moles
At $$t=t\,\,\, \,\, \, (2-2) \,\,\, (13-13) \,\,\, 10\, moles \,\,\,\,\, 8\, moles$$
                    = 0 mol    = 0 mol    of $$ H_{2}O(g) $$    of $$CO_{2}$$

At constant volume and temperature :$$\rightarrow$$
$$ PV= n R T$$
$$ \dfrac{P}{n} = \dfrac{RT}{V}$$ constant

$$ \dfrac{P}{n} =$$ constant

$$ \dfrac{P_{1}}{n_{1}} = \dfrac{P_{2}}{n_{2}}$$

So, As initial pressure was $$ P_{0}$$
$$ P_{1} = P_{0}$$

initial moles $$ n_{1} = 15$$  moles $$(2+13)$$

final mole $$ n_{2} = 18 $$ mole $$(10+8)$$
Now, 
$$ \dfrac{P_0}{15} = \dfrac{P_{2}}{18} \Rightarrow P_{2} = \dfrac{6}{5}P_{0}$$ 


A sample of calcium carbonate is $$80\%$$ pure, $$25$$ gm of this sample is treated with excess of $$HCl$$. How much volume of $$CO_2$$ will be obtained at $$1$$ atm & $$273$$ K?

The mass of pure $${ CaCO }_{ 3 }=\dfrac { 80 }{ 100 } \times 25=20g$$
$${ CaCO }_{ 3 }\left( s \right) +2HCl\left( aq \right) \rightarrow Ca{ Cl }_{ 2 }\left( s \right) +{ CO }_{ 2 }\left( g \right) +{ H }_{ 2 }O\left( l \right) $$
The mole ratio of calcium carbonate and $${ CO }_{ 2 }$$ is $$1:1$$
Moles of $${ CaCO }_{ 3 }=\dfrac { Mass\quad of\quad { CaCO }_{ 3 } }{ Molar\quad mass } =\dfrac { 20 }{ 100 } =0.2$$ moles
Moles of $${ CO }_{ 2 }=0.2$$ moles
Molar gas volume $$1$$ mole $$=22.4$$ litres
                               $$0.2$$ mole $$=2\times 22.4=4.48$$ litres.
$$4.48$$ litres of $${ CO }_{ 2 }$$ will be obtained.


At constant temperature, a gas is at a pressure of $$1080$$ mm Hg. If the volume is decreased by $$40$$%, find the new pressure of the gas.

According to boyle's law 
when $$T =cons$$
$$P_1V_1 = P_2V_2$$ now $$P_1 = 1000 V_1 = 1200$$ and $$P_2 = 700$$ ( as it decreased by $$30\%$$) 
So $$V_2 = \dfrac{P_1\times V_1}{P_2} = \dfrac{1000\times 1200}{700} = 1700cm^3$$


How many molecules of $$CO_2$$ gas is passed in $$900$$ml water at $$298$$ K temperature . The value of $$K_H$$ is $$6.02 \times 10^{-4}$$ bar and partial pressure of $$CO_2$$ gas is $$ 2 \times 10^{-8} bar $$:

  • According to Henry's law, $$p_{CO_2} = K_H.\chi$$
$$p_{CO_2} = \text{partial pressure of } CO_2 = 2\times10^{-8} bar$$
$$K_H = \text{Henry's law constant } = 6.02\times10^{-4}bar$$
$$\chi = \text{mole fraction of }CO_2 $$
  • $$\chi = \dfrac{\text{number of moles of }CO_2}{\text{number of moles of }CO_2 + \text{number of moles of }H_2O} = \dfrac{\text{number of moles of }CO_2}{\text{number of moles of }H_2O}$$ 
(since, the moles of $$CO_2$$ are negligible in comparison to $$H_2O$$)
number of moles of $$H_2O = \dfrac{\text{given weight of }H_2O}{\text{molecular weight of }H_2O} = \dfrac{900g}{18g} = 50$$
(As, $$1ml = 1g$$ so, $$900ml = 900g$$)
  • $$\text{Number of moles of }CO_2 = \dfrac{p_{CO_2}\times{\text{number of moles of }H_2O}}{K_H}$$
$$\text{Number of moles of }CO_2 = \dfrac{2\times{10^{-8}}\times{50}}{6.02\times{10^{-4}}} = \dfrac{10^{-2}}{6.02}$$
$$\text{Number of molecules of }CO_2 = \dfrac{10^{-2}}{6.02}\times{6.02\times{10^{23}}} = 10^{21} molecules$$


Calculate the total pressure of a mixture of 8 g of oxygen and 4g of hydrogen confined in vessel of $$1d{ m }^{ 3 }$$ at $${ 27 }^{ 0 }C$$.
$$\left( R=0.083\quad bar\quad d{ m }^{ 3 }{ k }^{ -1 }{ mol }^{ -1 } \right) $$

Moles of $$O_2$$ = $$\frac{8}{32}$$ = $$\frac{1}{4}$$
Moles of $$H_2$$ = $$\frac{4}{2}$$ = 2
Therefore total moles of both gas(n) = $$\frac{1}{4} + $$ = $$\frac{9}{4}$$
Given :
V = 1 $$dm^3$$
T = 300 K
$$\left(R=0.083\quad bar\quad d{ m }^{ 3 }{ k }^{ -1 }{ mol }^{ -1 } \right) $$
PV = nRT
P = $$ ( 9 \times 0.083 \times 300)$$ /  $$ (4 \times 1)=56.025\ bar$$


A certain mass of gas occupied $$850ml$$ at a pressure of $$760\ mm$$ of $$Hg$$. On increasing the pressure it was found that the volume of  the gas was $$75%$$ of its initial value. Assuming constant temperature, find the final pressure of the gas?

$$P_{ 1 }=760mm$$
$${ V }_{ 1 }=850ml$$
$${ V }_{ 2 }\quad ={ 75 }{ 100 }\times 850={ 75\times 85 }{ 10 }$$
$${ P }_{ 2 }=?$$
$$P_{ 1 }{ V }_{ 1 }={ P }_{ 2 }{ V }_{ 2 }$$
$$\therefore \quad 760\times 850={ P }_{ 2 }\times { 75\times 85 }{ 10 }$$
$$\therefore \quad { 760\times 10\times 10 }{ 75 }={ P }_{ 2 }$$
$${ P }_{ 2 }=1013.33mm$$


A gas cylinder containing cooking gas can withstand a pressure of $$14.9$$ atm. The pressure guage of the cylinder indicates $$12$$ atm at $$27^{o}C$$. Due to sudden fire in the building, its temperature stats rising. At what temperature cylinder will explode?

$${ P }_{ 1 }=12\quad bar,{ T }_{ 1 }=27+273=300K$$
$${ P }_{ 2 }=14.9\ bar,{ T }_{ 2 }=?$$
Applying pressure temperature law,
$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } $$
or $${ T }_{ 2 }=\dfrac { { T }_{ 1 }\times { P }_{ 2 } }{ { P }_{ 1 } } $$
Substituting the values in the above equation,
$${ T }_{ 2 }=\dfrac { 300\times 14.9 }{ 12 } =372.5K$$
Temperature in $$^{ 0 }C$$$$=372.5-273=$$$$99.5^\circ C$$

Hence the cylinder will explode at a temperature $$99.5^\circ C$$


$$500 ml$$ of a gas A exerts a pressure $$600 torr$$ in a vessel. $$400 ml$$ of a gas B exerts  a pressure $$200 torr$$ in another vessel. If the $$2$$ gases are mixed in a $$3 L$$ vessel then the total pressure of the gas is?

$$P_{ A }{ V }_{ A }+{ P }_{ B }{ V }_{ B }={ P }_{ T }{ V }_{ T }\\ 500\times 600+200\times 400={ P }_{ T }\times 3000\\ { P }_{ T }=126.66\quad torr\\ $$


The density of a mixture of $$O_2$$ and $$N_2$$ gases at $$1$$ atm and $$273$$K is $$0.0013$$ gm/ml. If partial pressure of $$O_2$$ in the mixture is A then calculate value of $$25$$A.

Given density= $$1.3gm/lit$$ at STP
$$1 lit= 1.3 gm$$
$$\therefore 22.4 lit= 22.4 \times 1.3=29.12 gm$$
$$\therefore H.W=29.12 gm$$
Total weight of given mixture at $$22.4 lit$$ STP is:
$$\Rightarrow 28x \neq 32(1-x)$$
$$\Rightarrow 29.12=28x+32-32x$$
$$\Rightarrow x= 0.72$$
Number of moles of $$N_2=0.72$$ & $$O_2=1-0.72=0.28$$
$$\therefore$$ Partial pressure of $$O_2$$ at STP= $$\cfrac {n_{O_2}}{n_{O_2}+n_{N_2}}\times P$$
                                                    $$=\cfrac {0.28}{(0.28+0.72)}\times 1$$
$$\Rightarrow P_{O_2}=0.28 atm$$ .


One mole of $$N_2O_4(g)$$ at $$300$$K is kept in a close container under one atmosphere. It is heated $$600$$K when $$20\%$$ by mass of $$N_2O_4(g)$$ decomposes to $$NO_2(g)$$. The resultant pressure is:

         $${ N }_{ 2 }{ O }_{ 4 }\left( g \right) \rightleftharpoons 2{ NO }_{ 2 }$$
Initial : $$1$$ mole         $$0$$ mole
At equilibrium moles of $${ N }_{ 2 }{ O }_{ 4 }=1-20$$% of $$1$$ mole $$=0.8$$ mole
At equilibrium moles of $${ NO }_{ 2 }=0.2\times 2=0.4$$ mole
$$\therefore$$   Total moles at equilibrium $$=0.8+0.4=1.2$$ mole
$${ P }_{ 1 }{ V }_{ 1 }={ n }_{ 1 }{ RT }_{ 1 }$$
$$\Rightarrow 1\times { V }_{ 1 }=1\times R\times 300\Rightarrow { V }_{ 1 }=300R\quad \longrightarrow \left( 1 \right) $$
$${ P }_{ 2 }{ V }_{ 2 }={ n }_{ 2 }R{ T }_{ 2 }$$
$$\Rightarrow { P }_{ 2 }\times { V }_{ 2 }=1.2\times R\times 600\quad \longrightarrow \left( 2 \right) $$
dividing (2) by (1),
$$\dfrac { { P }_{ 2 }\times { V }_{ 2 } }{ { V }_{ 1 } } =\dfrac { 1.2\times R\times 600 }{ 300\times R } \quad \left[ { V }_{ 1 }={ V }_{ 2 } \right] $$
$$\Rightarrow { P }_{ 2 }=2.4$$ atm
$$\therefore$$   Resultant pressure of mixture is $$2.4$$ atm.


28.32 litres of chorine were libereted at noramal condition of tempreture and pressure. calculate the volume of the gas at $$12^{0}C$$ and780mm pressure.


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One litre gas at 40 K and 300 atm pressure is compressed to a pressure of 600 atm and 200 K. the comressibillity factor is changed from 1.2 to 1.6 respectively.calculate the final volume of the gas ?


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A mixture of $$90$$ $$mole%$$ oh $${H}_{2}$$ $$10$$ $${%}_{2}$$ $${D}_{2}$$ at $$298\ K$$ and a pressure of $$1$$ atm is allowed to effuse through a small orifice of area $$03\ {mm}^{2}.$$ Calculate the composition of the initial gas that passes through.


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At constant temperature $$250\ mL$$ of argon at a $$760\ mm\ Hg$$ pressure and $$60\ mL$$ of nitrogen at $$500\ mm$$ pressure are put together in one litre flask. The final pressure is _________ $$mm\ Hg$$.

$$PV=nRT$$
For Argon, $$P=760 \ mm \ Hg=1 \ atm$$
$$V=0.25L$$
So, $$n=\dfrac{1\times{0.25}}{RT}=\dfrac{1}{4RT}$$
For Nitrogen, $$P=500 \ mm \ Hg=\dfrac{500}{760}atm$$
$$V=0.06L$$
So, $$n=\dfrac{500\times{0.06}}{760RT}=\dfrac{3}{76RT}$$
After mixing both the gases,
Total number of moles of both the gases, $$n=\dfrac{1}{4RT}+\dfrac{3}{76RT}=\dfrac{88}{304RT}$$
$$V=1L$$
$$P=\dfrac{nRT}{V}=\dfrac{88}{304RT}\times{RT}=\dfrac{88}{304}atm$$

$$P=\dfrac{88}{304}\times{760} \ mm \ Hg=220 \ mm \ Hg$$


An unspecified quantity of an ideal gas was at initial pressure of $$5$$ atm and temperature of $$303$$K. The gas is expanded at $$303$$K until the volume has increased by $$60\%$$ of the initial value. Next, the quantity of the gas in the vessel is increased by $$20\%$$ of the initial value while the volume is maintained constant. Finally, the temperature is adjusted at constant volume until the gas pressure is again $$5$$ atm. What is the final temperature, in Kelvin?


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At $$627^oC$$ and $$1\ atm, SO_3$$ partially dissociates into $$SO_2$$ and $$O_2$$. One litre of the equilibrium mixture weighs $$0.94\ g$$ under the above conditions. Calculate the partial pressures of the constituent gases in the mixture.

$$T=627^oC=627+273=900\ K$$
$$P=1\ atm$$
$$W_{cg}=1\ lit$$
Initially no. of moles $$=\dfrac{initial\ Wt}{Mol.\ wt\ of\ SO_3}$$ $$ \therefore$$ Wt remain conserved)=$$\dfrac{0.94}{80}=0.01175$$

                       $$2SO_3 \rightleftarrows 2SO_2+O_2$$
Initially            $$0.01175$$       $$0$$          $$0$$
At eq-b       $$0.01175-2x$$ $$2x$$       $$x$$

$$\therefore $$ No. of moles at $$T=900\ K$$ & $$P=1\ atm$$

$$=\dfrac{Volume}{molar\ volume}=\dfrac{1\ lit}{\dfrac{0.0821 \times 900\ lit}{1}}=0.0135$$
$$\therefore 0.01175+x=0.0135$$.

$$x=1.78 \times 10^{-3}$$

$$\therefore P_{SO_2}=\dfrac{2x}{0.01175+x}\times 1=\dfrac{2\times 1.78 \times 10^{-3}}{0.01353}=0.266\ atm$$

$$P_{SO_2}=\dfrac{0.01175-2x}{0.01175+x}\times \dfrac{819 \times 10^{-3}}{0.01358}=0.60\ atm$$

$$P_{O_2}=P_{total}-(P_{SO_2}+P_{SO_3})=1-0.266+0.60=0.1286\ atm$$


In a gravity free cabin, a non-conducting liquid of surface tension T and possessing uniform volume charge density p is sprayed into large number of drops. After spraying, the drops float around in the cabin and keep on breaking apart into smaller drops and coalescing into larger drops. While floating, the electrostatic forces between the drops are negligible. Estimate radius of the drops when their number reaches a stable value.

 Total energy comprising of electrostatic potential energy and surface energy due to surface tension of all the drops eventually acquires a minimum value. 
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The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $$r=2.5 \times 10^{-5}m$$. The surface tension of sap is $$T=7.28 \times 10^{-2}Nm^{-1}$$ and the angle of contact is $$0^{\circ}$$. Does surface tension alone account for the supply of water to the top of all trees?

The height to which the sap will rise is 
$$ h = \dfrac{2Tcos0^o}{\rho gr}$$ = $$\dfrac{2 \times 7.28 \times 10^{-2}}{10^{3} \times 9.8 \times 2.5 \times 10^{-5}}$$ = $$0.6 m $$
This is the maximum height to which the sap can rise due to surface tension. Since many trees have heights much more than this, capillary action alone cannot account for the rise of water in all trees.


Class 11 Engineering Chemistry Extra Questions