Thermodynamics: Numerical Problems Set (5)

Heat absorbed equals work done in isothermal: $Q_{AB} = 1728$ J.

Volume constant, so $W_{BC} = 0$. Heat released:

For adiabatic, $Q = 0$, so $W = -\Delta U = -nC_v(T_A - T_C) = -(1)(\tfrac{3}{2}R)(300 - 150) = -1871$ J. Wait — work done by gas in compression should be negative; let me redo signs.

In adiabatic compression, $\Delta U = nC_v\Delta T = 1871$ J (positive, gas heats up). $Q = 0 \Rightarrow W_{CA} = -\Delta U = -1871$ J (work done by gas is negative).

Negative — this cycle as drawn is actually a refrigerator-type cycle, not a heat engine. Re-check direction: the cycle described (isothermal expansion, isochoric cool, adiabatic compress) is not a closed engine for a monoatomic gas at these specific points, because the final volume after adiabatic compression must reach $V_A$.

Let’s verify: from C ($T = 150$ K, $V = 2V$) compressed adiabatically to volume $V$. Using $TV^{\gamma-1} =$ const with $\gamma = 5/3$:

This does not equal $300$ K, so the cycle as stated does not close. The problem is over-specified — a common JEE trap.

A valid cycle has constraints: choose any three of (initial $T$, final $T$, volume ratio, type of process) — the fourth is determined. If we keep $T_A = 300$ K, $V_A→2V$ isothermal, then isochoric cool to $T_C$, the adiabatic compression fixes $T_C = 300/2^{2/3} \approx 189$ K, not $150$ K.

Final answer: the cycle as stated is inconsistent. With $T_C \approx 189$ K, the cycle closes; net work and efficiency follow from valid values.

Why This Works

In thermodynamics, a closed cycle must return to its starting state — all three state variables ($P, V, T$) repeat. If you specify too many points, the constraint may be violated. Catching this early saves you from wrong answers.

Alternative Method

Always sanity-check a cycle by computing the third process’s endpoint independently and matching it to the given start.