Question

The pressure P and volume V of a gas are connected by the relation $P{V}^{1/4}=constant$. The percentage increase in the pressure corresponding to a deminition of $\frac{1}{2}\%$ in the volume is

• A. $\frac{1}{2}$ %
• B. $\frac{1}{4}$ %
• C. $\frac{1}{8}$ %
• D. none of these

Answer 1

The correct option is C $\frac{1}{8}$ %

$P{V}^{1/4}=constant$

$\Rightarrow P=\frac{k}{V^{1/4}}$

$\Rightarrow \frac{dP}{dV}=-\frac{k}{4}{V}^{-5/4}$

Percentage error in V $=-\frac{1}{2}\%$

$\Rightarrow \frac{\Delta V}{V}=-\frac{1}{200}$

$\Rightarrow \Delta V=-\frac{V}{200}$

Approximate change in $P=dP=\left(\frac{dP}{dV}\right)\Delta V$

$=\frac{1}{800}k{V}^{-1/4}=\frac{1}{8}\%$ of P

Percentage increase in $V=\frac{1}{8}\%$