A light container having a diatomic gas enclosed within is moving with velocity $v$. Mass of the gas is $M$…
A light container having a diatomic gas enclosed within is moving with velocity $v$. Mass of the gas is $M$ and number of moles is $n$. The kinetic energy of gas w.r.t. ground is
$\frac{1}{2} M^2+\frac{3}{2} n R T$
$\frac{1}{2} M v^2$
$\frac{1}{2} M^2+\frac{5}{2} n R T$
$\frac{5}{2} n R T$
Solution
According to kinetic theory of gases, Energy of diatomic gas having $f$ number of degree of freedom is given as,
$
B_1=n \frac{f}{2} R T \quad \text { [For nmole gas] }
$
Since, degree of freedom of a diatomic gas,
$
f=5
$
$\therefore$ Kinetic energy of diatomic gas with respect to centre of mass,
$
B_2=n \frac{f}{2} R T
$
$
B_2=\frac{5}{2} n R T
$
Kinetic energy of gas with respect to ground $=\mathrm{KE}$ of centre of mass with respect to ground $+\mathrm{KE}$ with respect to centre of mass $=\frac{1}{2} M v^2+\frac{5}{2} n R T$