Physics - Heat & Thermo Question with Solution | TestHub

KE of the vessel$=\frac{1}{2}m{v}^2$

When the vessel is suddenly stopped, the ordered motion of the gas molecules is converted into disordered motion of the molecules increasing thereby the internal energy of the gas. Thus,

$∆U=n{C}_v∆T=\frac{1}{2}m{v}^2=\frac{1}{2}\left(nM\right)v^2$

Where$n$is number of moles of the gas in the vessel and$M$is molecular weight of the gas.

$\therefore ∆T=\frac{M{v}^2}{2C_v}$

As$C_v=\frac{R}{\mathrm{\gamma }-1}$

$\therefore ∆T=\frac{M{v}^2(\mathrm{\gamma }-1)}{2R}=\frac{M{v}^2(1.4-1)}{2R}=\frac{M{v}^2}{5R}$