Thermodynamics: Tricky Questions Solved (9)

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Tags Thermodynamics

Question

One mole of an ideal monatomic gas is taken through a cycle: (A→B) isobaric expansion at P0P_0 from V0V_0 to 2V02V_0, (B→C) isochoric cooling at 2V02V_0 until pressure is P0/2P_0/2, (C→A) returns to A along a straight line on the PP-VV diagram. Find the net work done by the gas in one cycle and the efficiency of the cycle. Treat P0P_0, V0V_0 as known.

Solution — Step by Step

A→B (isobaric): WAB=P0(2V0−V0)=P0V0W_{AB} = P_0(2V_0 - V_0) = P_0 V_0.

B→C (isochoric): WBC=0W_{BC} = 0.

C→A (straight line on PP-VV): area under the line from C(2V0,P0/2)C(2V_0, P_0/2) to A(V0,P0)A(V_0, P_0). Direction of motion is decreasing volume, so work is negative.

∣WCA∣=12(P0+P0/2)(2V0−V0)=3P0V04|W_{CA}| = \tfrac{1}{2}(P_0 + P_0/2)(2V_0 - V_0) = \tfrac{3 P_0 V_0}{4}

So WCA=−3P0V04W_{CA} = -\tfrac{3 P_0 V_0}{4}.

 Wnet=P0V0+0−3P0V04=P0V04W_{\text{net}} = P_0 V_0 + 0 - \tfrac{3 P_0 V_0}{4} = \tfrac{P_0 V_0}{4}

The area enclosed (a triangle with base V0V_0 and height P0/2P_0/2) gives the same 12⋅V0⋅P02=P0V04\tfrac{1}{2} \cdot V_0 \cdot \tfrac{P_0}{2} = \tfrac{P_0 V_0}{4} — sanity check passes.

For monatomic gas, CV=32RC_V = \tfrac{3}{2}R, CP=52RC_P = \tfrac{5}{2}R.

A→B: TA=P0V0/RT_A = P_0V_0/R, TB=2P0V0/RT_B = 2P_0V_0/R. QAB=nCPΔT=52R⋅P0V0R=5P0V02Q_{AB} = nC_P\Delta T = \tfrac{5}{2}R \cdot \tfrac{P_0V_0}{R} = \tfrac{5 P_0 V_0}{2}. Positive — heat in.

B→C: TC=(P0/2)(2V0)/R=P0V0/RT_C = (P_0/2)(2V_0)/R = P_0V_0/R. ΔT=TC−TB=−P0V0/R\Delta T = T_C - T_B = -P_0V_0/R. QBC=nCVΔT=−3P0V02Q_{BC} = nC_V\Delta T = -\tfrac{3 P_0 V_0}{2}. Heat out.

C→A: needs careful analysis. Using ΔU=nCV(TA−TC)=0\Delta U = nC_V(T_A - T_C) = 0 (since TA=TCT_A = T_C). So QCA=WCA=−3P0V04Q_{CA} = W_{CA} = -\tfrac{3 P_0 V_0}{4}. Heat out.

Total heat absorbed Qin=QAB=5P0V02Q_{\text{in}} = Q_{AB} = \tfrac{5 P_0 V_0}{2}.

η=WnetQin=P0V0/45P0V0/2=110=10%\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{P_0 V_0/4}{5 P_0 V_0/2} = \frac{1}{10} = 10\%

Net work =P0V0/4= P_0V_0/4, efficiency =10%= 10\%.

Why This Works

For any closed cycle on a PP-VV diagram, WnetW_{\text{net}} equals the enclosed area (positive if traversed clockwise). ΔU=0\Delta U = 0 over a cycle, so Qnet=WnetQ_{\text{net}} = W_{\text{net}}. Efficiency uses only the heat absorbed in the denominator — heat rejected does not enter.

The temperature check TA=TCT_A = T_C is the elegant move. It makes C→A an isothermal-like leg in terms of ΔU\Delta U, even though it is not actually isothermal on the diagram.

Alternative Method

Compute total heat rejected Qout=∣QBC∣+∣QCA∣=3P0V02+3P0V04=9P0V04Q_{\text{out}} = |Q_{BC}| + |Q_{CA}| = \tfrac{3P_0V_0}{2} + \tfrac{3P_0V_0}{4} = \tfrac{9 P_0 V_0}{4}. Then Wnet=Qin−Qout=10P0V04−9P0V04=P0V04W_{\text{net}} = Q_{\text{in}} - Q_{\text{out}} = \tfrac{10P_0V_0}{4} - \tfrac{9P_0V_0}{4} = \tfrac{P_0V_0}{4}. Same answer through energy balance.

Common Mistake

Students put Qin=QAB+QBC+QCAQ_{\text{in}} = Q_{AB} + Q_{BC} + Q_{CA} (all three, with signs) in the denominator. That gives Qin=WnetQ_{\text{in}} = W_{\text{net}} and η=100%\eta = 100\%, which is impossible for a non-Carnot cycle. The denominator is only the heat absorbed (positive Q legs). This is a recurring JEE Advanced trap.

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