Kinetic Energy Of Ideal Gas MCQ - Practice Questions & Answers

NEET Sample Paper 2025 PDF by NTA - Model Question Paper With Solution

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Kinetic energy of ideal gas is considered one of the most asked concept.
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Directions: In this question, a word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as in the two matrices, given below. The columns and rows of Matrix (I) are numbered from 0 to 4 and that of Matrix (II) are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, e.g. 'G' can be represented by 01, and 'P' can be represented by 10, 44, etc. Similarly, you have to identify the set for the word.
BARK

32, 55, 69, 24

32, 03, 66, 57

67, 42, 66, 00

23, 30, 66, 97

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Concepts Covered - 1

Kinetic energy of ideal gas

The kinetic energy of ideal gas-
In ideal gases, the molecules are considered point particles. The point particles can have only translational motion and thus only translational energy. So for an ideal gas, the internal energy can only be translational kinetic energy.

Hence kinetic energy (or internal energy) of $n$ mole ideal gas

$E = \frac{1}{2} n M v_{m s}^{2} = \frac{1}{2} n M \times \frac{3 R T}{M} = \frac{3}{2} n R T$

1. kinetic energy of 1 molecule

$E = \frac{3}{2} k T$

where $k =$ Boltzmann's constant
and $k = 1.38 \times 10^{- 23} J / K$
i.e. Kinetic energy per molecule of gas does not depend upon the mass of the molecule but only depends upon the temperature of the gas.
2. kinetic energy of 1 -mole ideal gas

$E = \frac{3}{2} R T$
i.e. Kinetic energy per mole of gas depends only upon the temperature of the gas.

3. At T = 0, E = 0 i.e. at absolute zero the molecular motion stops.

The relation between pressure and kinetic energy

As we know $P$

$P = \frac{1}{3} \frac{m N}{V} v_{m s}^{2} = \frac{1}{3} \frac{M}{V} v_{m s}^{2} \Rightarrow P = \frac{1}{3} ρ v_{m s}^{2}$

And K.E. per unit volume=

$E = \frac{1}{2} (\frac{M}{V}) v_{m s}^{2} = \frac{1}{2} ρ v_{m s}^{2}$

So from equation (1) and (2), we can say that $P$

$P = \frac{2}{3} E$

i.e. the pressure exerted by an ideal gas is numerically equal to the two-thirds of the mean kinetic energy of translation per unit volume of the gas.

Law of Equipartition of Energy-

I.e Each degree of freedom is associated with energy $E = \frac{1}{2} k T$
1. At a given temperature $T_{, all}$ ideal gas molecules will have the same average translational kinetic energy as $\frac{3}{2} k T$
2. Different energies of a system of the degree of freedom $f$ are as follows
(i) Total energy associated with each molecule $= \frac{f}{2} k T$
(ii) Total energy associated with $N$ molecules $= \frac{f}{2} N k T$
(iii) Total energy associated with 1 mole $= \frac{f}{2} R T$
(iv) Total energy associated with $n$ mole $= \frac{n f}{2} R T$