Question 1.5: An ideal gas is characterised by the relation pV = NR T as i...

An ideal gas is characterised by the relation pV = NR T as in § 1.2 where the pressure p (T, V) is a function of T and V, the temperature T (p, V) is a function of p and V and the volume (T, p) is a function of T and p. Calculate,

\frac{∂p (T, V)}{∂T} \frac{∂T (p, V)}{∂V} \frac{∂V (T, p)}{∂p}.

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The partial derivative of the pressure, the temperature and the volume of an ideal gas that satisfies the relation pV = NR T are given by,

\frac{∂p (T, V)}{∂T} = \frac{∂}{∂T} \biggl(\frac{NR T}{V}\biggr) =\frac{NR}{V}.

\frac{∂T (p, V)}{∂V} = \frac{∂}{∂V} \biggl(\frac{ pV}{NR}\biggr) =\frac{p}{NR}.

\frac{∂V (T,p)}{∂p} = \frac{∂}{∂p} \biggl(\frac{NR T}{p}\biggr) =-\frac{NR}{p^2} = -\frac{V}{p}.

Thus,

\frac{∂p (T, V)}{∂T} \frac{∂T (p, V)}{∂V} \frac{∂V (T, p)}{∂p} = -1.

This result can be generalised, as shown in § 4.7.2.

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