Kinetic Energy Of Ideal Gas - Practice Questions & MCQ

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} 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 $\mathrm{k}=$ Boltzmann's constant
and $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$
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
  $

  
  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
  $

  
  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 $\frac{3}{2} k T$
  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 $=\frac{n f}{2} R T$