 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-3/34d/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

On the straight sloping leg,

p = p_{0} (3 - V / V_{0})

and

d p = - p_{0} d V / V_{0}

. Put

v = V / V_{0}

, which increases from

1

2

. The reversible heat transfer is

d Q = p_{0} V_{0} (\frac{15}{2} - 4 v) d v .

Thus heat is absorbed for $1 < v < 15/8$ and emitted for $15/8 < v < 2$ , with zero instantaneous heat at

v = 15/8

. The switching point has

(p, V) = (9 p_{0} /8, 15 V_{0} /8)

. In particular the net heat on this leg,

\frac{3}{2} p_{0} V_{0}

, is not the total positive heat on it.

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