Question

The pressure P and volume V of a gas are connected by the relation PV^1/4 = 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

Answer 1

(c) $\frac{1}{8}$ %

We have
$$\begin{gathered} \frac{△V}{V}=\frac{-1}{2}\% \\ P{V}^{\frac{1}{4}}=\mathrm{constant}=k\left(\mathrm{say}\right) \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log \left(P{V}^{\frac{1}{4}}\right)=\log k \\ \Rightarrow \log P+\frac{1}{4}\log V=\log k \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.x,\mathrm{we}\mathrm{get} \\ \frac{1}{P}\frac{dP}{dV}+\frac{1}{4V}=0 \\ \Rightarrow \frac{dP}{P}=-\frac{dV}{4V}=-\frac{1}{4}\times \frac{-1}{2}=\frac{1}{8} \\ \mathrm{Hence},\mathrm{the}\mathrm{increase}\mathrm{in}\mathrm{the}\mathrm{pressure}\mathrm{is}\frac{1}{8}\%. \end{gathered}$$