Question

A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma$. It is moving with speed $v$ and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by
$($consider all physical quantities in $SI$ units and $R$ is universal gas constant$)$

• A. $\frac{(\gamma -1)}{2\gamma R}M{v}^2$
• B. $\frac{\gamma M{v}^2}{2R}$
• C. $\frac{(\gamma -1)}{2R}M{v}^2$
• D. $\frac{(\gamma -1)}{2(\gamma +1)R}M{v}^2$

Answer 1

The correct option is C $\frac{(\gamma -1)}{2R}M{v}^2$

Assuming that no heat is lost to the surrounding, then all the kinetic energy of the vessel will be utilized in increasing its internal energy when the vessel suddenly stop.
$\Rightarrow$ Loss in $K.E=$ Gain in internal energy
$\Rightarrow \frac{1}{2}m{v}^2=n{C}_v\Delta T$ ...........$\left(1\right)$
Suppose, $m=$ mass of gas in the vessel
$\therefore n=\frac{m}{M}$
Also,
$C_v=\frac{R}{\gamma -1}$
From Eq. $\left(1\right)$, we get,
$\frac{1}{2}m{v}^2=\frac{m}{M}\times \frac{R}{(\gamma -1)}\times \Delta T$
$\Rightarrow \Delta T=\frac{(\gamma -1)M{v}^2}{2R}$

Why this question? Tips: Apply principle of energy conservation for vessel.