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Kinetic Theory

JEE Main level, single correct, medium

80 of 168

Question 80 of 168

Match List I with List II: List I List II (A) $3$ Translational degrees of freedom (I) Monoatomic gases (B) $3$ Translational, $2$ rotational degrees of freedoms (II) Polyatomic gases (C) $3$ Translational, $2$ rotational and $1$ vibrational degrees of freedom (III) Rigid diatomic gases (D) $3$ Translational, $3$ rotational and more than one vibrational degrees of freedom (IV) Nonrigid diatomic gases Choose the correct answer from the options given below:

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Answer: (A) $\left(A\right)-\left(I\right),\left(B\right)-\left(III\right),\left(C\right)-\left(IV\right),\left(D\right)-\left(II\right)$

The formula for degree of freedom is given by

f=3N-k

where

N

is the number of particles,

k

is the individual relationship in the system.
For a monoatomic gas particle,

N=1,k=0

. So,

f=3

Hence, monoatomic gas has

3

translational degree of freedom.
For a rigid diatomic gas molecule, there are

5

degrees of freedom,

3

translational and

2

rotational. A diatomic gas can have two extra degrees of freedom due to rotation along two independent axes.
A non rigid diatomic molecule has one extra degree of freedom than a rigid one. This is the vibrational degree of freedom. Hence, it has

6

degrees of freedom.
So, a non rigid diatomic gas has

3

translational,

2

rotational and

1

vibrational degrees of freedom.
A polyatomic gas has more than two atoms per molecule. Hence, the total number of degrees of freedom is

3

translational,

3

rotational and more than one vibrational degree of freedom.

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JEE Main 2025 Apr 2 Shift 1 question paper

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