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:

Always track the sign convention carefully: - Work done by system = positive. - Work done on system = negative. This avoids errors in multi-step thermodynamic cycles.

Answer 1

The Correct Option is A

In thermodynamics, the first law states that the change in internal energy of a system, ΔU, is equal to the heat added to the system, Q, minus the work done by the system, W. Mathematically, this is expressed as ΔU = Q - W.

We analyze the given processes as follows: 

For process 1-2:

Heat absorbed, Q_1-2 = 150 kJ

Work done by the system, W_1-2 = 90 kJ

Change in internal energy, ΔU_1-2 = Q_1-2 - W_1-2 = 150 kJ - 90 kJ = 60 kJ

For process 2-3:

Work done on the system, W_2-3 = -80 kJ (since work is done on the system, it's negative)

Heat rejected, Q_2-3 = -60 kJ (as it's rejected, it's negative)

Change in internal energy, ΔU_2-3 = Q_2-3 - W_2-3 = -60 kJ - (-80 kJ) = 20 kJ

Total change from state 1 to state 3:

ΔU_1-3 = ΔU_1-2 + ΔU_2-3 = 60 kJ + 20 kJ = 80 kJ

For an adiabatic process (no heat exchange, Q = 0) from state 3 back to state 1:

ΔU_3-1 = Q_3-1 - W_3-1 = 0 - W_3-1

Since the total change in internal energy around the full cycle must be zero, ΔU_1-3 + ΔU_3-1 = 0, it follows that:

80 kJ - W_3-1 = 0

W_3-1 = 80 kJ

Thus, the work interaction needed to restore the system to the initial state by an adiabatic path is 80 kJ.