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A cylinder holds 1.00 mol of an ideal gas at 8.00 bar, kept at 300 K by a large water bath.

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

A cylinder holds $1.00\ \text{mol}$ of an ideal gas at $8.00\ \text{bar}$ , kept at $300\ \text{K}$ by a large water bath. The gas is to expand until its pressure is $1.00\ \text{bar}$ . The figure shows the gas's isotherm from the start A to the end B, and a way of doing it in two steps: the external pressure is first dropped to $4.00\ \text{bar}$ (the gas expands to point M on the isotherm), then to $1.00\ \text{bar}$ . Take $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$ and $1\ \text{L bar} = 100\ \text{J}$ . Give each work as the work done by the gas (in NCERT's convention, $w$ is minus this).

A pressure-volume graph for one mole of gas at constant temperature. A falling curve, the isotherm, runs from point A at 8.00 bar down to point B at 1.00 bar, through point M at 4.00 bar. A stepped path is drawn: from A straight down to 4.00 bar, then along at 4.00 bar to M; from M straight down to 1.00 bar, then along at 1.00 bar to B. The area under the stepped path is shaded. The volumes are not marked.

What work does the gas do if the external pressure is dropped straight to $1.00\ \text{bar}$ , in one step?
What work does it do in the two steps of the figure?
What work does it do if it expands reversibly?
The pressure of the first step may be chosen freely. Which pressure gives the most work in two steps?
What is that most work?

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.

Volumes. $nRT = 2494.2\ \text{J} = 24.942\ \text{L bar}$ . So $V_A = \dfrac{24.942}{8.00} = 3.118\ \text{L}$ , $V_M = 6.236\ \text{L}$ and $V_B = 24.942\ \text{L}$ .

(a) One step. $W = p_2(V_B - V_A) = 1.00 \times (24.942 - 3.118)\ \text{L bar} = 2182\ \text{J}$ .

(b) Two steps. $4.00 \times (6.236 - 3.118) = 1247\ \text{J}$ , then $1.00 \times (24.942 - 6.236) = 1871\ \text{J}$ ; in all $3118\ \text{J}$ . In the figure, this is the area under the two steps.

(c) Reversibly. $W = nRT\ln\dfrac{p_1}{p_2} = 2494.2 \times 2.0794\ \text{J} = 5187\ \text{J}$ , the whole area under the isotherm, the most of all.

(d) Best first step. With a first step at $p$ : $W(p) = p\left(\dfrac{nRT}{p} - \dfrac{nRT}{p_1}\right) + p_2\left(\dfrac{nRT}{p_2} - \dfrac{nRT}{p}\right) = nRT\left(2 - \dfrac{p}{p_1} - \dfrac{p_2}{p}\right)$ . Setting $\dfrac{dW}{dp} = -\dfrac{1}{p_1} + \dfrac{p_2}{p^2} = 0$ gives $p = \sqrt{p_1 p_2} = \sqrt{8 \times 1}\ \text{bar} = 2.83\ \text{bar}$ .

(e) Most work in two steps. $W = nRT\left(2 - 2\sqrt{\dfrac{p_2}{p_1}}\right) = 2494.2 \times 1.2929\ \text{J} = 3225\ \text{J}$ .

Check. A scan over every first-step pressure from $1$ to $8\ \text{bar}$ finds the largest work at $2.83\ \text{bar}$ , and the order is one step < two steps < best two steps < reversible: $2182 < 3118 < 3225 < 5187\ \text{J}$ . The more steps, the closer the path comes to the reversible one.

Watch out. In each step the gas works against the external pressure of that step, not against its own falling pressure.

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