Question

$m$ grams of a gas of molecular weight $M$ is flowing in an isolated tube with velocity $V$. If the gas flow is suddenly stopped the rise in its temperature is ($\gamma =$ratio of specific heats, $R$$=$ universal gas constant, $J$ $=$ mechanical equivalent of heat):

• A. $\frac{M{V}^2(\gamma -1)}{2RJ}$
• B. $\frac{m{V}^2(\gamma -1)}{2RJ}$
• C. $\frac{m{V}^2\gamma }{2RJ}$
• D. $\frac{M{V}^2\gamma }{2RJ}$

Answer 1

The correct option is A $\frac{M{V}^2(\gamma -1)}{2RJ}$

Since the gas flow is suddenly stopped, we will consider it to be an adiabatic process.
Work done in an adiabatic process:$W=\frac{\mu R\Delta T}{(\gamma -1)}=\frac{mR\Delta T}{M(\gamma -1)}$ ;where, $\mu$ is the no. of moles of the gas.
Energy available after the gas flow is suddenly stopped: $\frac{1}{2}m{V}^2$
Equating both, $\frac{1}{2}m{V}^2=J\frac{\mu R\Delta T}{\gamma -1}=J\frac{mR\Delta T}{M(\gamma -1)}$

$\therefore \Delta T=\frac{M{V}^2(\gamma -1)}{2RJ}$; where, J is the mechanical equivalent of heat.