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One mole of a diatomic ideal gas (rigid molecules, CV = 52 R) is taken round the cycle shown in the…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

One mole of a diatomic ideal gas (rigid molecules, $C_V = \tfrac52 R$ ) is taken round the cycle shown in the figure. From state A, at $400\ \text{K}$ , it expands at constant temperature to twice its volume (state B). It is then compressed at constant pressure back to its first volume (state C). Finally it is heated at constant volume back to A. Take $R = 8.314\ \text{J/(mol K)}$ and $\ln 2 = 0.693$ .

A pressure-volume graph of a cycle. Point A is at volume V0 and pressure P0. A curve (an isotherm) runs down to point B at volume 2V0 and pressure P0 over 2. A horizontal line runs back left to point C at volume V0 and pressure P0 over 2, and a vertical line runs up from C to A. Arrows show the order A, B, C.

What is the temperature at C?
Find the work done by the gas, and the heat it takes in, in each of the three steps. (Count work done by the gas and heat taken in by the gas as positive, so a negative value means work done on the gas, or heat given out.)
What is the efficiency of the cycle as a heat engine?
What would be the efficiency of a Carnot engine working between the highest and lowest temperatures of this cycle?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the corners with $PV = RT$ , then the work and heat for each step, then the efficiency.

Step 1. The corners (a). A: $(P_0, V_0, 400\ \text{K})$ . B: same temperature, twice the volume, so $P_0/2$ . C: pressure $P_0/2$ and volume $V_0$ , so the temperature is halved: $200\ \text{K}$ .

Step 2. A to B, isothermal. $W = RT_0\ln 2 = 3325.6 \times 0.693\ \text{J} = 2305\ \text{J}$ , and since $\Delta U = 0$ , $Q = 2305\ \text{J}$ taken in.

Step 3. B to C, isobaric. $W = \dfrac{P_0}{2}(V_0 - 2V_0) = -\dfrac{RT_0}{2} = -1663\ \text{J}$ (work is done on the gas). $Q = C_P\Delta T = \tfrac72 R \times (200 - 400)\ \text{J} = -5820\ \text{J}$ : heat is given out.

Step 4. C to A, isochoric. $W = 0$ and $Q = C_V\Delta T = \tfrac52 R \times 200\ \text{J} = 4157\ \text{J}$ taken in.

Step 5. Efficiency (c), (d). Net work: $2305 - 1663 = 642\ \text{J}$ . Heat taken in: $2305 + 4157 = 6462\ \text{J}$ . So $\eta = \dfrac{642}{6462} = 9.9\%$ . A Carnot engine between $400\ \text{K}$ and $200\ \text{K}$ would reach $1 - \dfrac{200}{400} = 50\%$ .

Answer: (a) $200\ \text{K}$ . (b) A to B: $W = 2305\ \text{J}$ , $Q = 2305\ \text{J}$ ; B to C: $W = -1663\ \text{J}$ , $Q = -5820\ \text{J}$ ; C to A: $W = 0$ , $Q = 4157\ \text{J}$ . (c) $9.9\%$ . (d) $50\%$ .

Check. Over a whole cycle $\Delta U = 0$ , so the net heat must equal the net work: $2305 + 4157 - 5820 = 642\ \text{J}$ . It does.

Watch out. The heat taken in at constant volume (C to A) counts as heat supplied, even though no work is done in that step. Leaving it out would make the efficiency look far too high.

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