Question

A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g mol⁻¹) is moving on a floor at a speed of 50 m s⁻¹. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

Answer 1

Number of moles of the ideal gas, n = 1 mole
Molecular weight of the gas, W = 20 g/mole
Mass of the gas, m =20 g
Velocity of the vessel, V = 50 m/s
Decrease in K.E. of the vessel = Internal energy gained by the gas
$$\begin{gathered} KE=\frac{1}{2}m(u^2-v^2) \\ KE=\frac{1}{2}\times 20\times {10}^{-3}\times (0-50\times 50) \\ KE=-25\mathrm{J}=\mathrm{gain}\mathrm{in}\mathrm{internal}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{gas} \\ \text{C}\mathrm{hange}\mathrm{in}\mathrm{internal}\mathrm{energy}\mathrm{of}\mathrm{a}\mathrm{gas}=\frac{d}{2}nR(∆T), \\ \mathrm{where}d\mathrm{is}\mathrm{the}\mathrm{degre}\mathrm{of}\mathrm{freedom}\mathrm{of}\mathrm{the}\mathrm{gas} \\ \text{F}\mathrm{or}\text{a}\mathrm{monoatomic}\mathrm{gas},d=3. \\ \mathrm{So},25=\frac{3}{2}nR(∆T) \\ \Rightarrow 25=1\times \frac{3}{2}\times 8.31\times ∆T \\ \Rightarrow ∆T=\frac{50}{(3\times 8.3)}=2\mathrm{K} \end{gathered}$$