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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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If the ratio of vapour density for hydrogen and oxygen is $\frac{1}{16}$ , then under constant pressure the ratio of their rms velocities will be

$\frac{4}{1}$

$\frac{1}{4}$

$\frac{1}{16}$

$\frac{16}{1}$

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Explanation

$1 . v_{rms} = \sqrt{\frac{3 P}{ρ}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{ρ_{2}}{ρ_{1}}} = \sqrt{\frac{16}{1}} = \frac{4}{1}$

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Molecular motion shows itself as

Temperature

Internal Energy

Friction

Viscosity

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Explanation

1. Temperature

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What is the velocity of wave in monoatomic gas having pressure 1 kilopascal and density $2.6 kg / m^{3}$ ?

$3.6 m / s$

$8.9 \times 10^{3} m / s$

Zero

None of these

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Explanation

$4 . v = \sqrt{\frac{γP}{ρ}} = \sqrt{\frac{\frac{5}{3} \times 10^{3}}{2.6}} = 25 m / \sec$

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If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of the molecules

Is decreased

Is increased

Remains unchanged

Increases or decreases according to the nature of the gas

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Explanation

2. The distance travelled by a molecule in two consecutive collision is called free path. If pressure is reduced, a molecule will travel longer distance before collision, so mean free path is increased.

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At constant volume, temperature is increased. Then

Collision on walls will be less

Number of collisions per unit time will increase

Collisions will be in straight lines

Collisions will not change

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Explanation

2. As the temperature increases, the average velocity increases. So the collisions are faster.

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A cubical box with porous walls containing an equal number of $O_{2}$ and $H_{2}$ molecules is placed in a large evacuated chamber. The entire system is maintained at constant temperature T. The ratio of $v_{rms}$ of $O_{2}$ molecules to that of the $v_{rms}$ of $H_{2}$ molecules, found in the chamber outside the box after a short interval is

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

$\frac{1}{4}$

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

$\sqrt{2}$

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Explanation

2. Temperature remain constant so

$v_{rms} \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{v_{O_{2}}}{v_{H_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$

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The value of C_V for one mole of neon gas is

$\frac{1}{2} R$

$\frac{3}{2} R$

$\frac{5}{2} R$

$\frac{7}{2} R$

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Explanation

2. Neon gas is monoatomic and for monoatomic gases $C_{v} = \frac{3}{2} R$

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At constant volume, for different diatomic gases the molar specific heat is

Same and 3 cal/mole/°C approximately

Exactly equal and its value is 4 cal/mole/°C

Will be totally different

Approximately equal and its value is 5 cal/mole/°C

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Explanation

$4 . {(C_{V})}_{dia} = \frac{5}{2} R Joule / mol / ° C = \frac{5}{2} \frac{R}{J} \approx 5 \frac{cal}{mol - ° C}$

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At constant volume the specific heat of a gas is $\frac{3 R}{2}$ , then the value of $' γ'$ will be

$\frac{3}{2}$

$\frac{5}{2}$

$\frac{5}{3}$

None of the above

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Explanation

$3 . C_{V} = \frac{R}{(γ - 1)} \Rightarrow γ = 1 + \frac{R}{C_{V}} = 1 + \frac{R}{\frac{3}{2} R} = \frac{5}{3}$

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The relation between two specific heats of a gas is

$C_{P} - C_{V} = \frac{R}{J}$

$C_{V} - C_{P} = \frac{R}{J}$

$C_{P} - C_{V} = J$

$C_{V} - C_{P} = J$

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Explanation

1. When $C_{P}$ and $C_{V}$ are given with calorie and R with Joule then $C_{P} - C_{V} = R / J .$

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