Question

A box of negligible mass containing $2$ moles of an ideal gas of molar mass $M$ and adiabatic exponent $\gamma$ moves with constant speed $v$ on a smooth horizontal surface. If the box suddenly stops, then change in temperature of gas will be

• A. $\frac{(\gamma -1)M{v}^2}{4R}$
• B. $\frac{\gamma M{v}^2}{2R}$
• C. $\frac{M{v}^2}{2(\gamma -1)R}$
• D. $\frac{(\gamma -1)M{v}^2}{2R}$

Answer 1

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

Kinetic energy of molecules is given by $\frac{1}{2}nM{v}^2$

Relation between $c_v$ and internal energy $U$is given by

$U=n{c}_v\Delta T$

$c_v=\frac{U}{n\Delta T}........\left(1\right)$

$\because \frac{c_p}{c_v}-1=\lambda -1$

$\frac{c_p-c_v}{c_v}=\lambda -1$

$\frac{R}{c_v}=\lambda -1$

$c_v=\frac{R}{\lambda -1}........\left(2\right)$

From (1) and (2)$\frac{U}{n\Delta T}=\frac{R}{\lambda -1}$

$U=\frac{Rn\Delta T}{\lambda -1}$

Now the internal energy is equal to the change in KE

$\frac{Rn\Delta T}{\lambda -1}=\frac{1}{2}nM{v}^2$

$\frac{R\Delta T}{\lambda -1}=\frac{1}{2}nM{v}^2$

$\Delta T=\frac{M{v}^2}{2R}(\lambda -1)$If the box suddenly stops, then change in temperature of gas will be $\Delta T=\frac{M{v}^2}{2R}(\lambda -1)$