The internal energy change when a system goes from state $^{'} P^{'}$ to $^{'} Q^{'}$ is $40 k J / m o l e$ . If the system goes from $^{'} P^{'}$ to $^{'} Q^{'}$ by a reversible path and returns to state $^{'} P^{'}$ by an irreversible path, what would be the net change in internal energy?

40 k J

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> 40 k J

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< 40 k J

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zero

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Solution

The correct option is C zero
The internal energy of a system is a state function. For a cyclic process, it is zero irrespective of the process and path followed.
Therefore, as the process has same initial and final points, it is a cyclic process.
Therefore, $Δ U_{n e t} = 0$

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The internal energy change when a system goes from state ′P′ to ′Q′ is 40 kJ/mole. If the system goes from ′P′ to ′Q′ by a reversible path and returns to state ′P′ by an irreversible path, what would be the net change in internal energy?

The correct option is C zeroThe internal energy of a system is a state function. For a cyclic process, it is zero irrespective of the process and path followed.Therefore, as the process has same initial and final points, it is a cyclic process.Therefore, ΔUnet=0

The internal energy change when a system goes from state $^{'} P^{'}$ to $^{'} Q^{'}$ is $40 k J / m o l e$ . If the system goes from $^{'} P^{'}$ to $^{'} Q^{'}$ by a reversible path and returns to state $^{'} P^{'}$ by an irreversible path, what would be the net change in internal energy?

The correct option is C zero
The internal energy of a system is a state function. For a cyclic process, it is zero irrespective of the process and path followed.
Therefore, as the process has same initial and final points, it is a cyclic process.
Therefore, $Δ U_{n e t} = 0$