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

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 \\\$$