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

Practice free Kinetic Theory of Gases (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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The pressure P, volume V and temperature T of gas in the jar A and the other gas in the jar B at pressure 2P, volume V/4 and temperature 2T, then the ratio of the number of molecules in the jar A and B will be :

1 : 1

1 : 2

2 : 1

4 : 1

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Explanation

$\frac{N_{a}}{N_{B}} = \frac{P_{A} V_{A}}{P_{B} V_{B}} \times \frac{T_{B}}{T_{A}} \Rightarrow \frac{N_{A}}{N_{B}} = \frac{P \times V \times (2 T)}{2 P \times \frac{V}{4} \times T} = \frac{4}{1}$

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A balloon contains $500 m^{3}$ of helium at 27°C and 1 atmosphere pressure. The volume of the helium at – 3°C temperature and 0.5 atmosphere pressure will be :

$500 m^{3}$

$700 m^{3}$

$900 m^{3}$

$1000 m^{3}$

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Explanation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{1 \times 500}{300} = \frac{0.5 \times V_{2}}{270} \Rightarrow V_{2} = 900 m^{3}$

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A vessel contains 1 mole of $O_{2}$ gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at temperature 2T has a pressure of

P/8

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Explanation

$PV = μRT \Rightarrow P \propto μT (∵ V and R = constant) \Rightarrow \frac{P_{2}}{P_{1}} = \frac{μ_{2}}{μ_{1}} \times \frac{T_{2}}{T_{1}} \Rightarrow \frac{P_{He}}{P} = \frac{1}{1} \times \frac{2 T}{T} \Rightarrow P_{He} = 2 P .$

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A given mass of a gas is allowed to expand freely until its volume becomes double. If $C_{b}$ and $C_{a}$ are the velocities of sound in this gas before and after expansion respectively, then $C_{a}$ is equal to :

$2 C_{b}$

$\sqrt{2} C_{b}$

$C_{b}$

$\frac{1}{\sqrt{2}} C_{b}$

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Explanation

3. For ideal gas, on free expansion there is no change in temperature. Hence $C_{a} = C_{b}$

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One litre of Helium gas at a pressure 76 cm of Hg and temperature 27° is heated till its pressure and volume are doubled. The final temperature attained by the gas is :

927°C

900°C

627°C

327°C

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Explanation

1. $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{76 \times 1}{300} = \frac{152 \times 2}{T_{2}}$

$\Rightarrow T_{2} = 1200 K = 927 ° C$

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A gas at 27°C temperature and 30 atmospheric pressure is allowed to expand to the atmospheric pressure. If the volume becomes 10 times its initial volume, then the final temperature becomes

100°C

173°C

273°C

– 173°C

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Explanation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow T_{2} = \frac{P_{2} V_{2}}{P_{1} V_{1}} \times T_{1} \Rightarrow T_{2} = \frac{1}{30} \times \frac{10}{1} \times 300 = 100 K = - 173 ° C$

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In the relation $n = \frac{PV}{RT}, n =$

Number of molecules

Atomic number

Mass number

Number of moles

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Explanation

4. Number of moles

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The equation of state corresponding to 8 g of $O_{2}$ is :

PV = 8RT

PV = RT/4

PV = RT

PV = RT/2

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Explanation

$PV = nRT = m (\frac{R}{M}) T = \frac{8}{32} RT [∵ M_{o_{2}} = 32] \Rightarrow PV = \frac{RT}{4} .$

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Three containers of the same volume contain three different gases. The masses of the molecules are $m_{1}, m_{2}$ and $m_{3}$ and the number of molecules in their respective containers are $N_{1}, N_{2}$ and $N_{3}$ . The gas pressure in the containers are $P_{1}, P_{2}$ and $P_{3}$ respectively. All the gases are now mixed and put in one of the containers. The pressure P of the mixture will be :

$P < (P_{1} + P_{2} + P_{3})$

$P = \frac{P_{1} + P_{2} + P_{3}}{3}$

$P = P_{1} + P_{2} + P_{3}$

$P > (P_{1} + P_{2} + P_{3})$

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Explanation

3. According to Dalton’s law of partial pressures, the total pressure will be $P_{1} + P_{2} + P_{3}$ .

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At a given volume and temperature, the pressure of a gas :

Varies inversely as its mass

Varies inversely as the square of its mass

Varies linearly as its mass

Is independent of its mass

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Explanation

3. $PV = mrT \Rightarrow P \propto m, (V, T \to constant)$

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