A gas is expanded from volume V 0 to 2V 0 under three different processes. Process 1 is isobaric, process 2 is isothermal and process 3 is isothermal and process 3 is adiabatic. Let Δ U 1 , Δ U 2 and Δ U 3 be the change in internal energy of the gas in these three processes. Then –
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Process 2 is an isothermal process
Hence, Δ U 2 = 0
Process 1 is an isobaric (P = constant) expansion.
Hence, temperature of the gas will increase
or Δ U 1 = positive
Process 3 is an adiabatic expansion. Hence, temperature will decrease
or Δ U 3 = negative
Therefore, Δ U 1 > Δ U 2 > Δ U 3 is the correct option.
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