Question:
The pressure P and volume V of a gas are connected by the relation PV1/4 = constant. The percentage increase in the pressure corresponding to a deminition of 1/2 % in the volume is
(a) $\frac{1}{2} \%$
(b) $\frac{1}{4} \%$
(c) $\frac{1}{8} \%$
(d) none of these
Solution:
(C) $\frac{1}{8} \%$
We have
$\frac{\Delta V}{V}=\frac{-1}{2} \%$
$P V^{\frac{1}{4}}=$ constant $=k \quad$ (say)
Taking $\log$ on both sides, we get
$\log \left(P V^{\frac{1}{4}}\right)=\log k$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{P} \frac{d P}{d V}+\frac{1}{4 V}=0$
$\Rightarrow \frac{d P}{P}=-\frac{d V}{4 V}=-\frac{1}{4} \times \frac{-1}{2}=\frac{1}{8}$
Hence, the increase in the pressure is $\frac{1}{8} \%$.