Kinetic Theory of Gases - NEET Physics Questions
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Kinetic Theory of Gases

Practice Questions:
Question 21: easy

Match Column – I and Column – II and choose the correct match from the given choices.

Column-I
(A) Root mean square speed of gas molecules
(B) Pressure exerted by ideal gas
(C) Average kinetic energy of a molecule
(D) Total internal energy of 1 mole of a diatomic gas

Column-II
(P) \(\frac{1}{3} n m \bar{v}^2\)
(Q) \(\sqrt{\frac{3RT}{M}}\)
(R) \(\frac{5}{2} RT\)
(S) \(\frac{3}{2} k_B T\)

 

1. (A) - (R), (B) - (Q), (C) - (P), (D) - (S)
2. (A) - (R), (B) - (P), (C) - (S), (D) - (Q)
3. (A) - (Q), (B) - (R), (C) - (S), (D) - (P)
4. (A) - (Q), (B) - (P), (C) - (S), (D) - (R)
View Answer

By kinetic theory: \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\) -> (A)-(Q); Pressure \(P = \frac{1}{3} nm\bar{v}^2\) -> (B)-(P); Average KE \(= \frac{3}{2}k_BT\) -> (C)-(S); and Internal energy for diatomic gas \(= \frac{5}{2}RT\) -> (D)-(R).

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Question 22: easy

The internal energy of an ideal monoatomic gas increases by the same amount as work done on the gas, then:

1. The process should be adiabatic
2. The process should be isothermal
3. The process should be isochoric
4. The process should be isobaric
View Answer

By the first law, \(dQ = dU + dW\). If work is done on the gas, \(dW_{\text{by}} = -dW_{\text{on}}\). Given \(dU = dW_{\text{on}}\), so \(dQ = 0\). This represents an adiabatic process.

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Question 23: easy

The ratio of total K.E of a molecule of Argon and Oxygen gas at 27°C is equal to:

1. 3 : 5
2. 3 : 2
3. 2 : 3
4. 5 : 7
View Answer

Total K.E. per molecule is \(\frac{f}{2}k_B T\). For monoatomic Argon, \(f=3\), and for diatomic Oxygen, \(f=5\). At equal temperature, the ratio of K.E. is \(3:5\).

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