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Kinetic Energy Of Ideal Gas - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • Kinetic energy of ideal gas is considered one of the most asked concept.

  • 38 Questions around this concept.

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A flask contains hydrogen and oxygen in the ratio of 2: 1 by mass at temperature 27^{\circ} \mathrm{C}. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is:

The average kinetic energy of a molecule of the gas is :

A flask contains Hydrogen and Argon in the ratio 2: 1 by mass. The temperature of the mixture is 30\degree\, C. The the ratio of average kinetic energy per molecule of the two gases (K argon/K hydrogen) is :
(Given: Atomic Weight of Ar = 39.9)

Concepts Covered - 1

Kinetic energy of ideal gas

The kinetic energy of ideal gas-

In ideal gases, the molecules are considered as 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} nM v_{m s}^{2}=\frac{1}{2} nM \times \frac{3 R T}{M}=\frac{3}{2} nR 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 depends 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=\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} \rho v_{m s}^{2}  ......(1)

And K.E. per unit volume= E=\frac{1}{2}\left(\frac{M}{V}\right) v_{m s}^{2}=\frac{1}{2} \rho v_{m s}^{2}...... (2)

So from equation (1) and (2), we can say that P=\frac{2}{3}E

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

  • Law of Equipartition of Energy-

According to this law, for any system in thermal equilibrium, the total energy is equally distributed among its various degrees of freedom.

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}kT

2. Different energies of a system of the degree of freedom f are as follows

    \begin{array}{l}{\text { (i) Total energy associated with each molecule }=\frac{f}{2} k T} \\ \\ {\text { (ii) Total energy associated with } N \text { molecules }=\frac{f}{2} N k T} \\ \\ {\text { (iii) Total energy associated with } 1 \text { mole }=\frac{f}{2} R T} \\ \\ {\text { (iv) Total energy associated with } n \text { mole }=\frac{nf}{2} R T} \end{array}

 

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Kinetic energy of ideal gas

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Kinetic energy of ideal gas

Physics Part II Textbook for Class XI

Page No. : 325

Line : 14

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