Question

The pressure $p$ and the volume $v$ of a gas are connected by the relation $p{v}^{1.4}=$constant. The percentage error in $p$ corresponding to a decrease of $\frac{1}{2}\%$ in $v$.

• A. $0.5$
• B. $0.6$
• C. $0.7$
• D. $0.8$

Answer 1

The correct option is C $0.7$

Given ${pv}^{1.4}=$ constant ($k$ say), $\frac{dv}{v}\times 100=-\frac{1}{2}$
$\Rightarrow {pv}^{1.4}=k$
Applying $\ln$ on both sides, we get
$\ln p+1.4\ln v=\ln k$
Differentiating on both sides, we get
$\frac{dp}{p}+1.4\frac{dv}{v}=0$
$\Rightarrow \frac{dp}{p}=-1.4\frac{dv}{v}$
$\Rightarrow \frac{dp}{p}\times 100=-1.4\frac{dv}{v}\times 100=-1.4(-\frac{1}{2})=0.7$
$\therefore$ Percentage error in $p=0.7$