Question

If the mean kinetic energy per unit volume of a gas is n times its pressure, then the value of n is
(A) 4.5
(B) 3.5
(C) 2.5
(D) 1.

Answer 1

Hint: Kinetic energy is mean when the velocity of the gas is in terms of \[{v_{rms}}\]. Pressure in terms of mean velocity is given by $P = \dfrac{{\rho {v^2}_{rms}}}{3}$. Now, mean kinetic energy per unit volume is n times pressure of the gas where value of n is 1.5.

Complete step by step answer:
Kinetic energy is given by
 $K = \dfrac{1}{2}m{v^2}_{rms}$
Where, \[{v_{rms}}\] is the root mean square velocity.
Pressure with mean velocity is given by,
 $P = \dfrac{{\rho {v^2}_{rms}}}{3}$
Substituting in place of mean velocity in terms of pressure
 \[\rho \] is the density which can be written as mass by volume,
 $\rho = \dfrac{m}{V}$
 $K = \dfrac{1}{2} \times m \times \dfrac{{3P}}{\rho }$
Since mean kinetic energy is per unit volume \[V = 1\] .
 \[\rho = m\]
 \[
  \dfrac{{{K_{mean}}}}{V} = \dfrac{{3PV}}{{2V}} \\
  \dfrac{{{K_{mean}}}}{V} = \dfrac{3}{2}P \\
 \]
Hence, Kinetic energy in terms of pressure is $K = \dfrac{3}{2}P$ that is 1.5 times the pressure.

Therefore, the correct option is D.

Note: We can solve this question by another method also. We know that the average velocity of a gas is nothing but \[{v_{rms}}\] which is equal to \[\dfrac{{3PV}}{m}\]. Substituting this equation in the equation for kinetic energy we get,
 \[
  {K_{mean}} = \dfrac{1}{2}m{v^2}_{rms} \\
  {K_{mean}} = \dfrac{1}{2}xmx{(\sqrt {\dfrac{{3PV}}{m}} )^2} \\
  {K_{mean}} = \dfrac{{3PV}}{2} \\
 \]
This is the mean kinetic energy, to find energy per unit volume:
 \[
  \dfrac{{{K_{mean}}}}{V} = \dfrac{{3PV}}{{2V}} \\
  \dfrac{{{K_{mean}}}}{V} = \dfrac{3}{2}P \\
 \]