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Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

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One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat $(in J K^{- 1} {mol}^{- 1})$ at constant volume is $(R = 8.3 J K^{- 1} {mol}^{- 1})$

18.7

18.9

19.2

None of the above

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Explanation

$1 . C_{V_{mix}} = \frac{μ_{1} C_{V_{1}} + μ_{2} C_{V_{2}}}{μ_{1} + μ_{2}} = \frac{1 \times \frac{3}{2} R + 3 \times \frac{5}{2} R}{1 + 3} = \frac{9}{4} R = \frac{9}{4} \times 8.3 = 18.7$

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The number of translational degrees of freedom for a diatomic gas is

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Explanation

2. Number of translational degrees of freedom (3) are same for all types of gases.

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For a gas if ratio of specific heats at constant pressure and volume is $γ$ then value of degrees of freedom is

$\frac{3 γ - 1}{2 γ - 1}$

$\frac{2}{γ - 1}$

$\frac{9}{2} (γ - 1)$

$\frac{25}{2} (γ - 1)$

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Explanation

$2 . γ = 1 + \frac{2}{f}, \Rightarrow γ - 1 = \frac{2}{f} \Rightarrow \frac{f}{2} = \frac{1}{γ - 1} \Rightarrow f = \frac{2}{γ - 1}$

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If a gas has n degrees of freedom ratio of specific heats of gas is

$\frac{1 + n}{2}$

$1 + \frac{1}{n}$

$1 + \frac{n}{2}$

$1 + \frac{2}{n}$

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Explanation

$4 . γ = \frac{C_{P}}{C_{V}} = \frac{\frac{f}{2} R + R}{\frac{f}{2} R} = 1 + \frac{2}{f} = 1 + \frac{2}{n}$

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A gaseous mixture consists of 16g of helium and 16g of oxygen. The ratio $\frac{C_{P}}{C_{V}}$ of the mixture is

1.4

1.54

1.59

1.62

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Explanation

$4 . μ_{1} = moles of helium = \frac{16}{4} = 4 μ_{2} = moles of oxygen = \frac{16}{32} = \frac{1}{2} \frac{\frac{4 \times 5 / 3}{\frac{5}{3} - 1} + \frac{1 / 2 \times 7 / 5}{\frac{7}{5} - 1}}{\frac{4}{\frac{5}{3} - 1} + \frac{1 / 2}{\frac{7}{5} - 1}} = 1.62$

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The pressure exerted by the gas on the walls of the container because

It loses kinetic energy

It sticks with the walls

On collision with the walls there is a change in momentum

It is accelerated towards the walls

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Explanation

3. Pressure, $P = \frac{F}{A} = \frac{1}{A} . \frac{∆ p}{∆ t} (∆ p = Change in momentum)$

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Gas at a pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to

$4 P_{0}$

$2 P_{0}$

$P_{0}$

$\frac{P_{0}}{2}$

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Explanation

$2 . v_{rms} = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 PV}{m}} \Rightarrow v_{rms} \propto \sqrt{\frac{P}{M}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{P_{1}}{P_{2}} \times \frac{m_{2}}{m_{1}}} \Rightarrow \frac{v}{2 v} = \sqrt{\frac{P_{0}}{P_{2}} \times \frac{m / 2}{m}} \Rightarrow P_{2} = 2 P_{0}$

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A box contains n molecules of a gas. How will the pressure of the gas be effected, if the number of molecules is made 2n?

Pressure will decrease

Pressure will remain unchanged

Pressure will be doubled

Pressure will become three times

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Explanation

(c)

$PV = nRT n = Molecules \times N_{A} \Rightarrow \frac{P_{2}}{P_{1}} = \frac{N_{2}}{N_{1}} = \frac{2}{1} \Rightarrow P_{2} = 2 P_{1}$

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At a given temperature, the pressure of an ideal gas of density $ρ$ is proportional to

$\frac{1}{ρ^{2}}$

$\frac{1}{ρ}$

$ρ^{2}$

$ρ$

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Explanation

$4 . v_{rms} = \sqrt{\frac{3 P}{ρ}} = P \propto ρ [v_{rms} is constant for fixed temperature]$

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Consider a gas with density $ρ$ and $\bar{c}$ as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity v, then the pressure exerted by the gas is

$\frac{1}{3} ρ {\bar{c}}^{2}$

$\frac{1}{3} ρ {(c + v)}^{2}$

$\frac{1}{3} ρ {(\bar{c} - v)}^{2}$

$\frac{1}{3} ρ {(c^{- 2} - v)}^{2}$

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Explanation

(a) Pressure of the gas will not affected by motion of the system, hence by

$v_{rms} = \sqrt{\frac{3 P}{ρ}} \Rightarrow {\bar{c}}^{2} = \frac{3 P}{ρ} \Rightarrow P = \frac{1}{3} ρ {\bar{c}}^{2}$

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