Show that for an isothermal expansion of an ideal gas (i) ∆U = 0 and (ii) ∆H = 0.
(i) For one mole of an ideal gas, Cv = (∆U / ∆T) v
For an isothermal process, T is constant so that ∆T = 0 .-. ∆U = 0
(ii) ∆H =∆U + ∆(pV)
For an ideal gas, pV = RT
∆H = ∆U + ∆(RT) = ∆U + p∆T
Since T is constant, ∆T = 0
.-. ∆H =0