Question

A closed system undergoes a process 1–2 in which it absorbs 150 kJ of energy as heat and does 90 kJ of work. Then it follows another process 2–3 in which 80 kJ of work is done on it while it rejects 60 kJ as heat. If it is desired to restore the system to the initial state (state 1) by an adiabatic path, the work interaction (in kJ) in this process will be

• A. 80
• B. 100
• C. 50
• D. 70

Hint:

In an adiabatic process, the change in internal energy is equal to the work done by or on the system.

Answer 1

The Correct Option is A

Given:
- Process 1–2: Heat absorbed \( Q_{1-2} = 150 \, {kJ} \), Work done \( W_{1-2} = 90 \, {kJ} \)
- Process 2–3: Work done on system \( W_{2-3} = -80 \, {kJ} \), Heat rejected \( Q_{2-3} = -60 \, {kJ} \)
- Process 3–1: Adiabatic process (no heat exchange)
From the first law of thermodynamics:
\[ \Delta U = Q - W \] For process 1–2:
\[ \Delta U_{1-2} = 150 - 90 = 60 \, {kJ} \] For process 2–3:
\[ \Delta U_{2-3} = -60 - (-80) = 20 \, {kJ} \] Now, the total change in internal energy from state 1 to state 3 is:
\[ \Delta U_{{total}} = \Delta U_{1-2} + \Delta U_{2-3} = 60 + 20 = 80 \, {kJ} \] Since the process 3–1 is adiabatic, \( Q = 0 \), so the work interaction is equal to the negative of the change in internal energy:
\[ W_{3-1} = - \Delta U_{{total}} = -80 \, {kJ} \] Thus, the work interaction in the adiabatic process is \( 80 \, {kJ} \).