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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/36c/d/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 36C / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

In the

(p, V)

plane, the Otto cycle consists of:

$A \to B$ : an adiabatic compression along $p V^{γ} = constant$ from $V_{1}$ to $V_{2}$ ;
$B \to C$ : a vertical constant-volume rise at $V_{2}$ ;
$C \to D$ : an adiabatic expansion back to $V_{1}$ ;
$D \to A$ : a vertical constant-volume fall at $V_{1}$ .

It is a clockwise loop, so its enclosed area is the positive net work output. In the

(T, S)

plane, the two reversible adiabats

A \to B

and

C \to D

are vertical constant-entropy segments. Constant-volume heating

B \to C

moves upward and to the right, while constant-volume cooling

D \to A

moves downward and to the left.

Let

r = \frac{V _{1}}{V _{2}} > 1.

The two adiabatic relations give

T_{B} = T_{A} r^{γ - 1}, T_{C} = T_{D} r^{γ - 1} .

Because heat is exchanged at constant volume,

Q_{1} = C_{V} (T_{C} - T_{B}), Q_{2} = C_{V} (T_{D} - T_{A}),

where

Q_{2}

denotes the positive amount rejected. Hence

\frac{Q _{2}}{Q _{1}} = \frac{T _{C} / r ^{γ - 1} - T _{A}}{T _{C} - T _{A} r ^{γ - 1}} = \frac{1}{r ^{γ - 1}} .

The thermal efficiency is therefore

η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{1}{r ^{γ - 1}} .

For a fixed gas,

γ

is fixed and the idealized efficiency increases monotonically with the compression ratio

r

, so it is maximized by making

r

as large as the physical constraints allow.

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