Answer
(a) $P=\frac{100}{T}$
(b) $-\frac{1}{900}$
Work Step by Step
(a) Given $PT=100$, we have $P=\frac{100}{T}$ which shows $P$ as a function of $T$.
(b) The instantaneous rate of change of P with respect to T when $T=300K$ is the derivative of $P$ with respect to $T$:
$P'=\lim_{h\to 0} \frac{P(300+h)-P(300)}{h}=\lim_{h\to 0} \frac{\frac{100}{300+h}-\frac{100}{300}}{h}=\lim_{h\to 0} \frac{100(300-300-h)}{300h(300+h)}=\lim_{h\to 0} \frac{-1}{3(300+h)}=-\frac{1}{900}$ $lb/in^2K$