MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 6

The temperature of the sun, if pressure is

$1.4 \times 10^9$ atm, density is

$1.4 gcm^{-3}$ and average molecular weight is 2, will be

$[Given\ R = 8.4 J mol^{-1} K^{-1}]$

0%
$1.2 \times 10^7 K$
0%
$2.4 \times 10^7 K$
0%
$0.4 \times 10^7 K$
0%
$0.2 \times 10^7 K$

Explanation

$PV = nRT n = \displaystyle \frac{m}{M}$ and

$\rho = \displaystyle \frac{m}{V}$
or

$\displaystyle T = \frac{PV}{nR} = \frac{PM}{\rho R}$

$= \displaystyle \frac{1.4 \times 10^9 \times 1.01 \times 10^5 \times 2 \times 10^{-3}}{1.4 \times 10^3 \times 8.4}$

$= 2.4 \times 10^7 K$

What is new gauge pressure?

0%
1.82 atm
0%
1.70 atm
0%
1.92 atm
0%
none

Explanation

$\displaystyle \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$
or

$\displaystyle P_2 = \frac{2.72 \times 150 \times 318}{278 \times 159}$

$= 2.94 atm$
Gauge pressure

$= 2.94 - 1.02$

$= 1.92 atm$

A barometer tube 90 cm long contains some air above mercury. The reading is 74.8 cm when true atmospheric pressure is 76 cm and temperature is 30

$^o$ C. If the reading is observed to be 75.4 cm on a day when temperature is 10

$^o$ C, then find the true pressure.

0%
76.07 cm
0%
75.6 cm
0%
76.57 cm
0%
77.123 cm

Explanation

Let A be the area of the cross-section

$V_1 = (90 - 74.8) A = 15. 2 A cm^3$

$P_1 = 76 - 74.8 = 1.2$ cm of Hg

$P_2 = (P - 75.4)$ cm of Hg

$V_2 = (90 - 75.4) A = 14.6 A cm^3$

$\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2 V_2}{T_2} = \frac{1.2 \times 15.2 \times 283}{303 \times 14.6} = 75.4 + 1.17$

$= 76.57$ cm of Hg

Equal masses of N

$_2$ and O

$_2$ gases are filled in vessels A and B. The volume of vessel B is double of A. The ratio of pressure in vessel A and B will be

0%
16 : 7
0%
16 : 14
0%
32 : 7
0%
32 : 28

Explanation

Moles of

$N_2=n_A\dfrac{m}{28}$

Moles of

$O_2=n_B\dfrac{m}{32}$

Given that

$V_B=2V_A$

$P\propto \dfrac{n}{V}$

$\implies \dfrac{P_A}{P_B}=\dfrac{n_A}{n_B}\dfrac{V_B}{V_A}=\dfrac{16}{7}$

At what temperature the average translational KE of the molecules of a gas will become equal to the KE of an electron accelerated from rest through 1 V potential difference?

0%
$10^4 K$
0%
$2.34 \times 10^4 K$
0%
$7.73 \times 10^3 K$
0%
none of these

Explanation

$\displaystyle \frac{3}{2} KT = 1 eV = 1.6 \times 10^{-19} J$
or

$\displaystyle T = \frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} = 7730 K=7.73\times10^3K$

One mole of an ideal gas undergoes a process

$P = \displaystyle \frac{P_o}{1 + \left ( \displaystyle\frac{V}{V_o} \right )^2}$ , where

$P_o$ and

$V_o$ are constants. Find the temperature of the gas when

$V = V_o$ .

0%
$\displaystyle \frac{P_o V_o}{R}$
0%
$\displaystyle \frac{2P_o V_o}{3R}$
0%
$\displaystyle \frac{2 P_o V_o}{R}$
0%
$\displaystyle \frac{P_o V_o}{2R}$

Explanation

Since the gas is ideal, therefore

$P_o V_o = RT$

Using the relation given when

$V = V_o$

$P = \displaystyle \frac{P_o}{2}$

Thus we get

$\displaystyle \frac{P_o}{2} (V_o) = RT$

$T = \displaystyle \frac{P_oV_o}{2R}$ .

What is C

$_2$ in terms of R?

0%
$\displaystyle \frac{3}{2} R$
0%
2R
0%
$\displaystyle \frac{5}{2} R$
0%
3R

Explanation

Number of degrees of freedom

$=$ 2 translation + 2 rotation

$=$ 4

$\therefore \displaystyle C = \frac{R}{2}$ (number of degrees of freedom)

$= 2R$

The African bombardier beetle stenaptinus insignis can emit a jet of defensive spray from the movable tip of its abdomen. The beetle's body has reservoirs of two different chemicals. When the beetle is disturbed, these chemicals are combined in a reaction chamber producing a compounds that is warmed from

$20^o$ C to

$100^o$ C by the heat of reaction. The high pressure produced allows the compound to the sprayed out at speeds 19 ms

$^{-1}$ scaring away predators of all kinds. Assume specific heat of two chemicals and the spray to be same as that of water

$[4.19 \times 10^3 J (kgK)^{-1}]$ and initial temperature of chemicals to be

$20^o$ C. How many times does the pressure increase?

0%
1.5
0%
2.3
0%
1.9
0%
1.28

Explanation

$\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$
or

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{373}{293} = 1.28$

The correct graph between PV and P of one mol of gas at constant temperature will be

Explanation

For an ideal gas

$PV=nRT$

Thus for a constant temperature,

$PV=constant$

Thus

$PV$ does not change even when P changes.

A gas is filled in a container at any temperature and at pressure 76 cm of Hg. If at the same temperature the mass of gas is increased by 50% then the resultant pressure will be

0%
38 cm of Hg
0%
76 cm of Hg
0%
114 cm of Hg
0%
152 cm of Hg

Explanation

$PV=nRT$

$PV=\dfrac{m}{M}RT$

$P=\dfrac{m}{M}\dfrac{RT}{V}$ also given that P=76cm of Hg

Given that mass of the gas is increased by 50%

Now

$P_{1}=\left(\dfrac{m+\dfrac{m}{2}}{M}\right)\dfrac{RT}{V}$

$=\dfrac{3}{2}\dfrac{m}{M}\dfrac{RT}{V}$

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

$\dfrac{3}{2}\times76$

$114cm\ of\ Hg$

In the gas equation

$PV= RT$ , V is the volume of

0%
1 mol of gas
0%
1 g of gas
0%
gas
0%
1 litre of gas

Explanation

For an Ideal Gas,

$PV=nRT$

Here

$V$ is the volume of n moles of gas.

Thus for

$PV=(1)RT$ ,

$V$ is the volume of 1 mol of gas.

The mean kinetic energy of gas molecules is zero at

0%
0 $^o$ C
0%
-273 $^o$ C
0%
100 K
0%
100 $^o$ C

Explanation

$KE_{avg} =\dfrac{3}{2}kT$ where T is in Kelvin.

$KE_{avg} =0$ at

$T=0^{\circ}K=-273^{\circ}C$

If number of gas molecules in a cubical vessel is increased from N to 3N, then its pressure and total energy will become

0%
four times
0%
three times
0%
double
0%
half

Explanation

For an ideal gas

$P \times V = N \times K\times T$

where P is the pressure, V is volume of the cubical vessel,
N is no. of molecules, K Boltzmann's constant and T is temperature

So as N increases to 3N, pressure also increases by three times.

And the total energy for an ideal gas

$E = \dfrac {3}{2} \times N \times K \times T$
Hence the total energy also increases by three times.

At what temperature will the mean molecular energy of a perfect gas be one-third of its value of 27

$^o$ C?

0%
10 $^o$ C
0%
10 $^1$ K
0%
10 $^2$ K
0%
10 $^3$ J

Explanation

Mean

$KE \propto T$
therefore at T=27 which in kelvin becomes 300 K
So KE will be 1/3 when temperature becomes 1/3 which is 300/3=100K
option (C)

The pressure of a gas in a container is 10

$^{-11}$ pascal at 27

$^o$ C. The number of molecules per unit volume of vessel will be

0%
$6 \times 10^{23} cm^{-3}$
0%
$2.68 \times 10^{19} cm^{-3}$
0%
$2.5 \times 10^{6} cm^{-3}$
0%
$2400 cm^{-3}$

Explanation

$10^5 Pa \equiv 6.023 \times 10^{23}$ (equivalence equation according to information given in question)

$10^{-11} Pa \equiv 6.023 \times 10^7$

$22400 cc \Rightarrow 6.023 \times 10^7$

$1 cc \Rightarrow \displaystyle \frac{6.023 \times 10^7}{2400} =2400 cm^{-3}$

The mean kinetic energy of a gas molecule at 27

$^o$ C is 6.21

$\times$ 10

$^{-21}$ Joule. Its value at 227

$^o$ C will be

0%
$12.35 \times 10^{-21} Joule$
0%
$11.35 \times 10^{-21} Joule$
0%
$10.35 \times 10^{-21} Joule$
0%
$9.35 \times 10^{-21} Joule$

Explanation

Mean KE=

$\dfrac{3kT}{2}$

$6.21\times 10^{-21}=3k\times 300/2$

$k=1.38\times 10^{-23}$

KE at 500 K ,

$KE=3\times 1.38\times 10^{-23}\times 500/2=10.35\times 10^{-21}\ Joule$

The value of

$\gamma$ of triatomic gas (linear arrangement) molecules is

0%
$5/3$
0%
$7/5$
0%
$8/6$
0%
$9/7$

Explanation

The number of degrees of freedom for a triatomic gas with linear arrangement is 7.

We have the relation

$\gamma$

$=1+\dfrac{2}{f}$ (where

$f$ is number of degrees of freedom)

$\therefore \gamma=1+\dfrac{2}{7}=\dfrac{9}{7}$

At what temperature will the linear kinetic energy of a gas molecule be equal to that of an electron accelerated through a potential difference of 10V?

0%
273 K
0%
$19 \times 10^3 K$
0%
$38.65 \times 10^3$
0%
$11.3 \times 10^3 K$

Explanation

$The\quad kinetic\quad energy\quad of\quad gas\quad molecule\quad =\quad 3/2\quad kT\\ Energy\quad of\quad the\quad accelerated\quad electron\quad =eV\\ Hence\quad eV=\quad 3/2\quad kT\quad For\quad the\quad given\quad case\\ k={ 1.38 }\times 10^{ -23 }\\ 1.6\times { 10 }^{ -19 } \times 10 = 1.5 kT\\ T=\dfrac { 1.6\times { 10 }^{ -19 }\quad \times 10\quad \times 2 }{ 3k } \\ T={ 11.3\times 10 }^{ 3 } K$

An enclosure of volume 3 litre contains 16g of oxygen, 7g of nitrogen and 11g of carbondioxide at 27

$^o$ C. The pressure exerted by the mixture is approximately

0%
9 atmosphere
0%
8.3 atmosphere
0%
3 atmosphere
0%
1 atmosphere

Explanation

Total number of moles = number of moles of oxygen + number of moles of nitrogen + number of moles of carbon-di-oxide

$\dfrac { 16 }{ 32 } +\dfrac { 7 }{ 14 } +\dfrac { 11 }{ 44 } =1mole$

Molecular weight of Carbon-di-oxide = 44g

Molecular weight of Oxygen =32g

Molecular weight of nitrogen=14g

Using ideal gas equation,

$P=\dfrac { nRT }{ V } =\dfrac { 1 \times8.3 \times300 }{ (3\times{ 10 }^{ -3 }) } =8.3atm(nearly)$

Hence, Option B is correct.

The law of equipartition of energy is applicable to the system whose constituents are :

0%
in random motion
0%
in orderly motion
0%
at rest
0%
moving with constant speed

The pressure of a gas filled in a closed vessel increases by 0.4%. When temperature is increases by 1

$^o$ C the initial temperature of the gas is

0%
250 $^o$ C
0%
250 K
0%
250 $^o$ F
0%
2500 $^o$ C

Explanation

Since volume is constant i.e number of moles is constant.

$P_1$ and

$T_1$ are the initial pressure and temperature.

According to the question,

${ P }_{ 2 }=1.004{ P }_{ 1 }\\ { T }_{ 2 }={ T }_{ 1 }+1$

$\therefore \dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac{P_2}{T_2}=\dfrac { (1+0.004){ P }_{ 1 } }{ { T }_{ 2 } } \\ \Rightarrow { T }_{ 2 }=1.004{ T }_{ 1 }=({ T }_{ 1 }+1)$

$0.004T_1 = 1$

$\\ \Rightarrow { T }_{ 1 }=250K$

Hence, Option B is correct

The most probable velocity for monoatomic gas is

0%
$\displaystyle \sqrt{\frac{3 kT}{m}}$
0%
$\displaystyle \sqrt{\frac{8 kT}{ \pi m}}$
0%
$\displaystyle \sqrt{\frac{2 kT}{m}}$
0%
zero

Explanation

$KE_p=\dfrac{mv_p^2}{2}=kT$ [standard result]

$v_p=\sqrt{\dfrac{2kT}{m}}$

The dimension of universal gas constant R are

0%
$M^2 L^2 T^{-2}$
0%
$ML^2 T^{-2} \theta^{-1}$
0%
$M^2 L^2 T^{-2} \theta^{-2}$
0%
$MLT^{-2} \theta^{-2}$

Explanation

using ideal gas equation

$R=\dfrac { PV }{ nT }$

dimensions of pressure

$M{ L }^{ -1 }{ T }^{ -2 }$

dimensions of Volume

${L}^{3}$

number of moles is dimensionless

dimensions of Temperature

$\theta$

so, dimensions of R

$\dfrac { (M{ L }^{ -1 }{ T }^{ -2 })({ L }^{ 3 }) }{ \theta } =M{ L }^{ 2 }{ T }^{ -2 }{ \theta }^{ -1 }$

Hence,Option B is correct

In outer space there are 10 molecules per cm

$^3$ on an average and the temperature there is 3K. The average pressure of this light gas is

0%
$10^5 Nm^{-2}$
0%
$5 \times 10^{-14} Nm^{-2}$
0%
$0.4 \times 10^{-16} Nm^{-2}$
0%
$4.14 \times 10^{-16} Nm^{-2}$

Explanation

$P= \displaystyle \frac{nKT}{V} = \frac{10 \times 1.38 \times 10^{-23} \times 3}{10^{-6}}$

$= 4.14 \times 10^{-16} Nm^{-2}$

If the mean kinetic energy per unit volume of a gas is n times its pressure, then the value of n is

0%
4.5
0%
3.5
0%
2.5
0%
1.5

Explanation

Mean kinetic energy of a gas =

$\dfrac{1}{2}MV_{rms}^{2}$

$=\dfrac{1}{2}\times M\times\dfrac{3PV}{M}$

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

Mean kinetic energy per unit volume=

$\dfrac{3}{2}P$

A cylinder contains 2kg of air at a pressure of 10

$^5$ Pa. If 2 kg more air is pumped into it, keeping the temperature constant, the pressure will be

0%
$10^{10} Pa$
0%
$2 \times 10^5 Pa$
0%
$10^5 Pa$
0%
$0.5 \times 10^5 Pa$

Explanation

Since both volume and Temperature is constant,

$\dfrac { P }{ n } =constant=\dfrac { P }{ m } \\ n=no.of\quad moles=\dfrac { mass\quad of\quad air(m) }{ Molecular\quad weight\quad of\quad water(M) }$

$\dfrac { { 10 }^{ 5 } }{ 2 } =\dfrac { P }{ (2+2) } \\ \Rightarrow P=2\times{ 10 }^{ 5 }Pa$

Hence, Option B is correct

One mole of a gas at a pressure 2 Pa and temperature 27

$^o$ C is heated till both pressure and volume are doubled. What is the temperature of the gas?

0%
1200 K
0%
900 K
0%
600 K
0%
300 K

Explanation

We have the relation PV=nRT (ideal gas equation)

$\dfrac{PV}{nR}$

Given, that pressure and volume are doubled i.e,

$P_1=2P$ ,

$V_1=2V$ .

$T_1=\dfrac{P_1V_1}{nR}$

$=\dfrac{\left( 2P \times 2V\right)}{nR}$

$=4T$

Given that T=27

$^o$ C i.e, 300K

$T_1=4\times 300$

$=1200K$

What is number of degrees of freedom of an ideal diatomic molecule at ordinary temperature?

0%
7
0%
6
0%
5
0%
3

If the value of R = 2/5 C

$_v$ for a gas, then the atomicity of the gas will be:

0%
monoatomic
0%
diatomic
0%
polyatomic
0%
any of these

Explanation

We have

$C_v=\displaystyle\frac{5}{2}R$ which is the

$C_v$ value for diatomic molecules.

The internal energy of air in a room of volume 50 m

$^3$ at atmospheric pressure will be

0%
2.5 X 10 $^7$ erg
0%
2.5 X 10 $^7$ Joule
0%
5.255 X 10 $^7$ Joule
0%
1.25 X 10 $^7$ Joule

Explanation

The internal energy of the room is given by the molecular motion and is called as the kinetic internal energy given by

$\displaystyle\frac{f}{2}nRT$ . Using the gas law we have

$PV=nRT$ and as the gas is mainly comprised of nitrogen and oxygen both being diatomic we take the degree of freedom as 5. Thus we calculate internal energy as

$\displaystyle\frac{5}{2}PV=\frac{5}{2}\times 1.013\times 10^5\times 50=1.25\times 10^7 J$

Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?

0%
Kinetic energy
0%
Momentum
0%
Density
0%
Speed

The total kinetic energy of 8 litres of helium molecules at 5 atmosphere pressure will be

0%
$6078 \ erg$
0%
$6078 \ Joule$
0%
$607 \ erg$
0%
$607 \ Joule$

The Avogadro's number when to tends to infinity, the phenomenon of Brownian motion would

0%
remain completely unaffected.
0%
become more vigorous, than that observed with present finite value of Avogadro's number, for all sizes of the Brownian particles.
0%
becomes more vigorous, than that observed with the present finite value of Avogadro's number, only for relatively large Brownian particle.
0%
become practically unobservable as the molecular impact would tend to balance one another, for practically.

A product from a chemical industry passes through three states - gas, liquid, solid. The product is initially formed in gaseous state which is then liquefied and finally solidified. For same mass (say 5g) which state has maximum internal energy?

0%
solid
0%
liquid
0%
gas
0%
all have equal internal energy

The temperature of an ideal gas is increased from 27

$^o$ C to 927

$^o$ C. Then, the mean kinetic energy of gas molecules :

0%
becomes $\displaystyle \frac{927}{27}$ times
0%
becomes $\displaystyle \sqrt{\frac{927}{27}}$ times
0%
become double
0%
become four times

Explanation

$T_{1}=27^{0}C=300K$

$T_{2}=927^{0}C=1200K$

$V_{rms}$ ( root mean square velocity )

$=\sqrt\frac{3RT}{M}$

$KE_{1}=\dfrac{1}{2}\times{m}\times{V_{rms}^{2}}$

$KE_{2}=\dfrac{1}{2}\times{m}\times{ \dfrac{3R\times{1200}}{m} }$

$=4KE_{1}$

In two vessels of the same volume, atomic hydrogen and helium with pressure 1 atm and 2 atm are filled. If temperature of both the same is the same, then the average speed of hydrogen atom

$v_H$ will be related to helium

$v_{He}$ as

0%
$v_{H}$ $= \sqrt{2}$ $v_{He}$
0%
$v_H$ $=$ $v_{He}$
0%
$v_H$ $=$ 2 $v_{He}$
0%
$v_H$ $=$ $\dfrac{v_{He}}{2}$

Explanation

By Maxwell's speed distribution,

$<v>\alpha \sqrt { \dfrac { RT }{ M } }$ . Since the temperature of two gases is same, hence

$<v>\alpha \sqrt { \dfrac { 1 }{ M } }$

Also,

${ M }_{ He }=4{ M }_{ H }$

Hence,

$<{ v }_{ H }>=2<{ v }_{ He }>$

Answer is option C.

The air density at Mount Everest is less than that at the sea level. It is found by mountaineers that for one trip lasting a few hours, the extra oxygen needed by them corresponds to 30,000cc at sea level (pressure 1 atmosphere, temperature 27

$^o$ C). Assume that the temperature around Mount Everest is -

$73^o$ C and that the oxygen cylinder has a capacity of 5.2 litters. The pressure at which oxygen be filled (at site) in the cylinder is

0%
3.86 atm
0%
5.00 atm
0%
5.77 atm
0%
1 atm

Explanation

Since moles of gas must remain constant,

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

$\dfrac { 1(30000) }{ 300 } =\dfrac { P(5200) }{ 143 }$

$P=3.86atm$

Which of the following quantities is the same for all ideal gases at the same temperature?

0%
The kinetic energy of 1 molecule
0%
The kinetic energy of 1 g
0%
The number of the molecules in 1 mole
0%
The number of molecules in 1 g

Explanation

Translational KE of a molecule

$K=\dfrac{3}{2}kT$
At constant temp KE of a molecule is same for all molecules.
If we include rotational KE, it will not be same for all molecules since poly-atomic molecules will have rotational energy in contrast to mono-atomic ones.

Which of the following assumptions is/are true according to kinetic theory?

0%
All matter is composed of small particles
0%
The particles of matter are in constant motion
0%
All collisions between the particles of matter are perfectly elastic
0%
All of the above

Helium gas is filled in a closed vessel (having negligible thermal expansion coefficient). When it is heated from 300 K to 600 K, then average kinetic energy of helium atom will be :

0%
$\sqrt{2}$
0%
$2$ times
0%
unchanged
0%
half

Explanation

Kinetic energy of gas molecules is

$\dfrac{3}{2}RT$ .

Thus

$\dfrac{E_f}{E_i}=\dfrac{T_f}{T_i}=\dfrac{600}{300}=2$

The top surface of a liquid behaves like stretched membrane because

0%
The molecules below the top surface exert upward pressure
0%
The molecules are attracted more to the sides of the containing vessel.
0%
The molecules below the top surface attract the molecules above
0%
none of the above

Explanation

The property of

$surface$

$tension$ :
The molecules in the liquid experience a force which acts equally for its neighboring molecules.
However this cohesive force pulls the molecules on the surface inwards.
Due to this unbalanced force the top surface of a liquid behaves like

$stretched$

$membrane.$

A gas mixture consists of 2moles of oxygen and 4moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

0%
4 RT
0%
15 RT
0%
9 RT
0%
11 RT

Explanation

The internal energy of the system is given as

$U=n(\displaystyle\frac{f}{2}R)T$ . Here f is the degree of freedom which is 5 for oxygen(diatomic molecule) and 3 for argon(monoatomic molecule). Thus we get the total internal energy as

$2(\displaystyle\frac{5}{2})RT+4(\frac{3}{2})RT=11RT$

Choose the correct answer from the given options below.

Kinetic energy of the molecules ------------- with an increase in temperature

0%
increases
0%
decreases
0%
remains the same
0%
increases then decreases

A sample of an ideal gas occupies a volume of 'V' at a pressure 'P' and absolute temperature 'T'. The mass of each molecule is 'm'. The equation for density is

0%
mkT
0%
P/kT
0%
P/(kTV)
0%
Pm/kT

Explanation

We know that PV = nRT =(m/M)RT
Where M = Molecular weight
Now

$P\times \left(\displaystyle \frac{m}{d}\right)$ =

$\left(\displaystyle\ \frac{m}{M} \right)RT$
where d = density of gas

$\displaystyle\ \frac{p}{d}$ =

$\displaystyle\ \frac{RT}{M}$ =

$\displaystyle\ \frac{kN_{A}T}{M}$
Where

$R$ =

$kN_{A}$ , k is Boltzmann constant
But

$\displaystyle\ \frac{ M}{N_{A}}$ = m = mass of each molecule so
So,

$d$ =

$\displaystyle\ \frac{P\times m}{kT}$

If 2 mole of an ideal monatomic gas at temperature

$T_{0}$ is mixed with 4 moles of another ideal monatomic gas at temperature

$2T_{0}$ , then the temperature of the mixture is

0%
$\displaystyle\ \frac {5}{3} T_{0}$
0%
$\displaystyle\ \frac {3}{2} T_{0}$
0%
$\displaystyle\ \frac {4}{3} T_{0}$
0%
$\displaystyle\ \frac {5}{4} T_{0}$

Explanation

Let T be the temperature of the mixture, then

$U$ =

$U_{1}+U_{2}$

where,

$U_1$ and

$U_2$ are the energies of the gas molecules.

$\Rightarrow$

$\displaystyle\ \frac{f}{2}(n_{1}+n_{2})RT$

$n_1$ and

$n_2$ are the number of gas molecules.

$R$ is the universal gas constant.

$=\displaystyle\ \frac{f}{2}(n_{1})(R)(T_{0})$ +

$\displaystyle\ \frac{f}{2}(n_{2})(R)(2T_{0})$

$\Rightarrow$

$(2+4)T$ =

$2T_{0}+8T_{0}$

(

$\because$

$n_{1}=2$

$n_{2} =4)$

$\therefore$

$T$ =

$\displaystyle\ \frac{5}{3}T_{0}$

How many degrees of freedom are associated with 2grams of He at NTP?

0%
3
0%
$3.01\times10^{23}$
0%
$9.03\times10^{23}$
0%
6

Explanation

Moles of He =

$\displaystyle\ \frac{2}{4}$ =

$\displaystyle\ \frac{1}{2}$
Molecules =

$\displaystyle\ \frac{1}{2}\times6.02\times10^{23}$ =

$3.01\times10^{23}$
As there are 3 degrees of freedom corresponding of 1 molecule of a monatomic gas.

$\therefore$ Total degrees of freedom =

$3\times3.01\times10^{23}$

$= 9.03\times10^{23}$

Kinetic theory of matter states about :

0%
three states of matter
0%
differences between the three states of matter
0%
fourth state of matter
0%
all of the above

Explanation

The kinetic theory explains the differences between the three states of matter. It states that all matter is made up of moving particles which are molecules or atoms.

In solids, the particles are so tightly bound to each other that they can only vibrate but not move to another location.

In liquids, the particles have enough free space to move about, but they still attract one another.

In gases, the particles are far apart and can move about freely since there is much free space.

Solids change into liquids, and liquids into gases when the particles gain more kinetic energy,

So, the correct option is

$B$

State whether true or false:

Linear molecules have

$3N-5$ vibrational degrees of freedom, whereas non linear molecules have

$3N-6$ vibrational degrees of freedom, where N is no. of atoms present in a molecule.

0%
True
0%
False

Maxwell's laws of distribution of velocities shows that

0%
the number of molecules with most probable velocity is very large
0%
the number of molecules with most probable velocity is small
0%
the number of molecules with most probable velocity is zero
0%
the number of molecules with most probable velocity is exactly equal to 1

A vessel contains air at a temperature of

$15^{0}C$ and 60% R.H. What will be the R.H if it is heated to

$20^{0}C$ ? (S.V.P at

$15^{0}C$ is 12.67 & at

$20^{0}C$ is 17.36mm of Hg respectively)

0%
262%
0%
26.2%
0%
44.5%
0%
46.2%

Explanation

By definition,

$\displaystyle\ \frac{V.P }{S.V.P }\times100$ =

$\displaystyle\ \frac{60}{100}$ (both VP and SVP are in room temperature)

$= \displaystyle\ \frac{(V.P)_{15}}{12.67}$

$(V.P)_{15} = 7.6 mm$

$of$

$Hg$
Now since unsaturated vapour obeys gas equation & mass of water remains constant so

$P$ =

$\left (\displaystyle\ \frac{nRT}{V} \right)$

$\Rightarrow$

$P\ \alpha \ T$

$\displaystyle\ \frac{(V.P)_{15}}{(V.P)_20}$

$= \displaystyle\ \frac{273+15}{273+20}$

$\Rightarrow$

$(RH)_{20}$ = 7.73 mm of Hg
So

$(R.H)_{20} =\displaystyle\ \frac{(V.P)_{20}}{(S.V.P)_{20}}\times 100$

$= 44.5$ %

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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