Thermodynamics: Common Mistakes and Fixes (7)

easy • 3 min read
Tags Thermodynamics

Question

One mole of an ideal monatomic gas is taken through the cycle: A→BA \to B isothermal expansion at T=300 KT = 300 \text{ K} from VV to 2V2V; B→CB \to C isochoric cooling to T/2T/2; C→AC \to A adiabatic compression back to AA. Find the work done per cycle and the heat absorbed in the isothermal step. Use R=8.314 J/(mol⋅K)R = 8.314 \text{ J/(mol·K)}.

Solution — Step by Step

For an isothermal expansion of an ideal gas:

WAB=nRTln⁡VBVA=(1)(8.314)(300)ln⁡2≈1729 JW_{AB} = nRT \ln\frac{V_B}{V_A} = (1)(8.314)(300)\ln 2 \approx 1729 \text{ J}

Since ΔU=0\Delta U = 0 for isothermal (ideal gas, TT constant), QAB=WAB≈1729 JQ_{AB} = W_{AB} \approx 1729 \text{ J} — heat absorbed.

Volume is constant, so WBC=0W_{BC} = 0. Temperature drops from 300 K300 \text{ K} to 150 K150 \text{ K}.

For monatomic ideal gas, Cv=32RC_v = \tfrac{3}{2}R:

QBC=nCvΔT=(1)(32)(8.314)(−150)≈−1871 JQ_{BC} = nC_v\Delta T = (1)(\tfrac{3}{2})(8.314)(-150) \approx -1871 \text{ J}

Heat released.

By definition QCA=0Q_{CA} = 0. Use first law: ΔU=−W\Delta U = -W. Going from TC=150T_C = 150 to TA=300T_A = 300:

ΔUCA=nCvΔT=(1)(32)(8.314)(150)≈1871 J\Delta U_{CA} = nC_v\Delta T = (1)(\tfrac{3}{2})(8.314)(150) \approx 1871 \text{ J}

So WCA=−1871 JW_{CA} = -1871 \text{ J} (work done on the gas during compression).

Wnet=WAB+WBC+WCA=1729+0−1871=−142 JW_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 1729 + 0 - 1871 = -142 \text{ J}

The negative sign tells us this cycle runs clockwise on VV-TT but counterclockwise on PP-VV — net work is done on the gas, so this is a refrigerator, not an engine.

Why This Works

Each process has its own first-law shortcut: isothermal ⇒ΔU=0\Rightarrow \Delta U = 0, isochoric ⇒W=0\Rightarrow W = 0, adiabatic ⇒Q=0\Rightarrow Q = 0. Knowing which term vanishes saves time.

For a complete cycle, ΔUcycle=0\Delta U_{\text{cycle}} = 0, so Qnet=WnetQ_{\text{net}} = W_{\text{net}}. We can verify: Qnet=1729−1871+0=−142 JQ_{\text{net}} = 1729 - 1871 + 0 = -142 \text{ J}. Matches. Always cross-check with this identity.

Alternative Method

Use TT-SS diagrams for cycle problems — the area enclosed equals the heat exchanged per cycle, with sign matching the cycle direction. For this problem, since the cycle runs counterclockwise in TT-SS, heat is absorbed from the cold reservoir and rejected to the hot reservoir, confirming refrigerator behaviour.

For monatomic ideal gas: Cv=32RC_v = \tfrac{3}{2}R, Cp=52RC_p = \tfrac{5}{2}R, γ=5/3\gamma = 5/3. For diatomic at room temp: Cv=52RC_v = \tfrac{5}{2}R, Cp=72RC_p = \tfrac{7}{2}R, γ=7/5\gamma = 7/5. JEE Main almost always specifies “monatomic” or “diatomic” — read carefully.

Common Mistake

The two killer mistakes in cyclic thermodynamics:

  1. Sign of work. Convention: WW done by the gas is positive in the first law ΔU=Q−W\Delta U = Q - W (Indian textbooks) or ΔU=Q+W\Delta U = Q + W (some Western books with WW as work on gas). Pick one and stick to it for the whole problem.

  2. Forgetting that ΔU\Delta U depends only on temperature for an ideal gas. Many students try to integrate PdVPdV for ΔU\Delta U on an adiabat — wrong. ΔU=nCvΔT\Delta U = nC_v\Delta T always, regardless of process.

Final answer: Wnet≈−142 JW_{\text{net}} \approx -142 \text{ J} (work on gas), QAB≈1729 JQ_{AB} \approx 1729 \text{ J} absorbed.

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