Show that in the isothermal expansion of an ideal gas , DeltaU = 0 and

1.	Show that in the isothermal expansion of an ideal gas , DeltaU = 0 and DeltaH = 0
Answer» Solution :(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`

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