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