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