One mole of a monatomic ideal gas (gamma = 5/3, Cv = 12.5 J/(mol K)) is compressed adiabatically so its volume halves. (a) By what factor does the temperature rise? (b) If it started at 300 K, how much work was done on the gas? Solution: (a) Adiabatic: T V^(gamma-1) = constant, so T2/T1 = (V1/V2)^(gamma-1) = 2^(2/3) = 1.587. (b) T2 = 300 x 1.587 = 476 K. Adiabatic: W_by = -delta_U, so work done ON the gas = delta_U = n Cv delta_T = (1)(12.5)(476 - 300) = 2200 J.