Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with constant velocity v.

Q: Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is (γ = CP / CV)

(a) $\displaystyle \frac{M v^2(\gamma -1)}{2R(\gamma+1)}$

(b) $\displaystyle \frac{M v^2(\gamma -1)}{2R}$

(c) $\displaystyle \frac{M v^2}{2R(\gamma+1)}$

(d) $\displaystyle \frac{M v^2}{2R(\gamma -1)}$

Ans: (b)

Sol: K.E of the vessel = $\displaystyle \frac{1}{2}mv^2$

When the vessel is suddenly stopped , it will increase the internal energy of the gas .

$\displaystyle \frac{1}{2}mv^2 = \Delta U$

$\displaystyle \frac{1}{2}(nM)v^2 = n C_v\Delta T$

Where n = no. of moles of the gas & M = Molecular weight of the gas

$\displaystyle \Delta T = \frac{Mv^2}{2C_v}$

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

$\displaystyle \Delta T = \frac{Mv^2(\gamma -1)}{2R}$