Show that in an Isothermal Expansion of an Ideal Gas ∆U=0 and ∆H=0

(a) We know that for one mole of an ideal gas,
C_V = (∂U/∂T) _V
Therefore, dU = C_V dT
For a finite change, ∆U = C_V ∆T
As for an isothermal process, T is constant so that ∆T = 0.
Hence for an isothermal expansion of an ideal gas ∆U=0.

(b) We know that ,
∆H = ∆U + ∆ (PV)
As for ideal gas, PV =RT
Therefore, ∆H = ∆U + ∆ (RT)
i.e. ∆H = ∆U + R∆T = R∆T (from (a) ∆U=0)

Since T is constant, ∆T = 0. Hence, ∆H = 0.