Question

The pressure p and the volume v of a gas are connected by the relation pv^1.4 = const. Find the percentage error in p corresponding to a decrease of 1/2% in v.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have} \\ p{v}^{1.4}=\mathrm{constant}=k\left(\mathrm{say}\right) \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log \left(p{v}^{1.4}\right)=\log k \\ \Rightarrow \log p+1.4\log v=\log k \\ \mathrm{Differentiating}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.x,\mathrm{we}\mathrm{get} \\ \frac{1}{p}\frac{dp}{dv}+\frac{1.4}{v}=0 \\ \Rightarrow \frac{dp}{p}=\frac{-1.4dv}{v} \\ \mathrm{Now},dp=\frac{dp}{dv}dv=\frac{-1.4p}{v}dv \\ \Rightarrow \frac{dp}{p}\times 100=-1.4\left(\frac{dv}{v}\times 100\right)=-1.4\times \left(\frac{-1}{2}\right)=0.7\left[\mathrm{Since}\mathrm{we}\mathrm{are}\mathrm{given}\frac{1}{2}\%\mathrm{decrease}\mathrm{in}v\right] \\ \mathrm{Hence},\mathrm{the}\mathrm{error}\mathrm{in}\mathit{p}\mathrm{is}0.7\%. \end{gathered}$$