Show that in an isothermal expansion of an ideal gas, a `DeltaU = 0` and b. `DeltaH = 0`.

Explanation

(a) For one moole of an ideal gas, `C_(v) = ( dU)/( dT) ` or ` dU =C_(v) dT`
For a finite change, `DeltaU = C_(v) DeltaT`
For isothermal process, `T=` constant so that`DeltaT =0`. Hence, `DeltaU = 0`
(b) `DeltaH = DletaU + Delta(PV) =DeltaU+ Delta(RT) =DeltaU+ R DeltaT`
But `DeltaU =0` ( proved above) and `DeltaT = 0` ( for isothermal process) , `:. DeltaH =0`