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 = C_{P} / C_{V})

(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} $