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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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In kinetic theory of gases, a molecule of mass m of an ideal gas collides with a wall of vessel with velocity V. The change in the linear momentum of the molecule is

2mV

– mV

Zero

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Explanation

$1 . ∆ ρ = mV - (- mV) = 2 mV$

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The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is

1 : 16

1 : 8

1 : 4

1 : 1

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Explanation

4. Kinetic energy is function of temperature.

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A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from 300 K to 600 K, the average K.E. of helium atoms is

Halved

Unchanged

Doubled

Increased by factor $\sqrt{2}$

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Explanation

3. Kinetic energy ∝ Temperature. Hence if temperature is doubles, Kinetic energy will also be doubled

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The kinetic energy of translation of 20 gm of oxygen at 47°C is (molecular wt. of oxygen is 32 gm/mol and R = 8.3 J/mol/K)

2490 joules

2490 ergs

830 joules

124.5 joules

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Explanation

1. Kinetic energy for m gm gas $E = \frac{f}{2} mrT$

If only translational degree of freedom considered then

$f = 3 \Rightarrow E_{Trans} = \frac{3}{2} mrT = \frac{3}{2} m (\frac{R}{M}) T = \frac{3}{2} \times 20 \times \frac{8.3}{32} \times (273 + 47) = 2490 J$

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The translatory kinetic energy of a gas per gm is

$\frac{3}{2} \frac{RT}{N}$

$\frac{3}{2} \frac{RT}{M}$

$\frac{3}{2} RT$

$\frac{3}{2} NKT$

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Explanation

2. Kinetic energy for $1 gm \Rightarrow E_{Trans} = \frac{3}{2} rT = \frac{3}{2} \frac{RT}{M}$

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The kinetic energy of one gram molecules of a gas at standard temperature and pressure: $(R = 8.31 J / mol - K)$

$0.56 \times 10^{4} J$

$1.3 \times 10^{2} J$

$2.7 \times 10^{2} J$

$3.4 \times 10^{3} J$

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Explanation

4. Kinetic energy per gm mole $E = \frac{f}{2} RT$

If nothing is said about gas , then we should calculate the translational kinetic energy

i.e.,

$E_{Trans} = \frac{3}{2} RT = \frac{3}{2} \times 8.31 \times (273 + 0) = 3.4 \times 10^{3} J$

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The mean kinetic energy of a gas at 300 K is 100 J. The mean energy of the gas at 450 K is equal to

100 J

3000 J

450 J

150 J

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Explanation

4. Average kinetic energy $\propto$ Temperature

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{100}{E_{2}} = \frac{300}{450} \Rightarrow E_{2} = 150 J$

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Read the given statements and decide which is/are correct on the basis of kinetic theory of gases

(I) Energy of one molecule at absolute temperature is zero

(II) r.m.s. speeds of different gases are same at same temperature

(III) For one gram of all ideal gas kinetic energy is same at same temperature

(IV) For one mole of all ideal gases mean kinetic energy is same at same temperature

All are correct

I and IV are correct

IV is correct

None of these

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Explanation

4. Mean kinetic energy of any ideal gas is given by $E = \frac{f}{2} T$ which is different for different gases. (f is not same for all gases)

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The translational kinetic energy of gas molecule for one mole of the gas is equal to

$\frac{3}{2} RT$

$\frac{2}{3} RT$

$\frac{1}{2} RT$

$\frac{2}{3} KT$

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Explanation

1. Kinetic energy for 1 mole gas

$E = \frac{f}{2} RT \Rightarrow E_{Translation} = \frac{3}{2} RT (∵ For all gases translation degree of freedom f = 3)$

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At 27°C temperature, the kinetic energy of an ideal gas is $E_{1}$ . If the temperature is increased to 327°C, then kinetic energy would be

$2 E_{1}$

$\frac{1}{2} E_{1}$

$\sqrt{2} E_{1}$

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

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Explanation

1. As temperature changes from 27°C (300 K) to 327°C(600 K) i.e. doubled so kinetic energy also

doubled because E ∝ T.

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