A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g mol ⁻¹ ) travels on a floor at a speed of 50 m s ⁻¹ . The vessel is abruptly brought to rest. Assuming that the lost mechanical energy goes entirely into the internal energy of the gas, determine the rise in its temperature. (Take R = 8.31 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

The kinetic energy of the gas due to the motion of the vessel is given by K = 1 2 m v^2 . The mass of the gas is m = n M = 1 20 10⁻³ kg = 0.02 kg . Substituting the values, we get: K = 1 2 0.02 (50)^2 = 25 J When the vessel is abruptly brought to rest, this kinetic energy is entirely converted into the internal energy of the gas. Thus, the change in internal energy is U = 25 J . For a monatomic ideal gas, the change in internal energy is given by U = n C_v T = n ( 3 2 R ) T . Equating the energies: 25 = 1 3 2 8.31