The ratio of the molar heat capacities of an ideal gas is Cp/Cv = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K (a) keeping the pressure constant, (b) keeping the volume constant and (c) adiabatically.

Sol. C base P/C base V = 7.6, n = 1 mole, ∆T = 50K (a) Keeping the pressure constant, dQ = du + dw, ∆T = 50 K, γ = 7/6, m = 1 mole, dQ = du + dw ⇒ nC base VdT = du + RdT ⇒ du = nCpdT – RdT = 1 * Rγ/γ – 1 * dT – RdT = R *7/6/7/6-1 dT – RdT = DT – RdT = 7RdT = 6 RdT = 6 * 8.3 * 50 = 2490 J. (b) Kipping Volume constant, dv = nC base VdT = 1 * R/γ – 1 * dT = 1 * 8.3/7/6-1 * 50 = 8.3 * 50 * 6 = 2490 J (c)Adiabetically dQ = 0, = [n * R/γ – 1(T base 1 – T base 2)] = 1 * 8.3/7/6 -1 (T base 1 – T base 2)] = 8.3 * 50 * 6 = 2490 J

Sol. C base VH base 2 = 2.4 Cal/g°C, C base PH^2 = 3.4 Cal/g°C M = 2 g/ Mol, R = 8.3 × 10^7 erg/mol-°C We know, C base P – C base V = 1 Cal/g°C So, difference of molar specific heats = C base P × M – CV × M = 1 × 2 = 2 Cal/g°C Now, 2 × J = R ⇒ 2 × J = 8.3 × 107 erg/mol-°C ⇒ J = 4.15 × 107 erg/cal.