#### 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

\[\frac{1}{2} \%\]

\[\frac{1}{4} \%\]

\[\frac{1}{8} \%\]

none of these

#### Solution

\[\frac{1} {8} \]%

We have

\[\frac{\bigtriangleup V}{V} = \frac{- 1}{2} % \]

\[P V^\frac{1}{4} = \text { constant }= k \left( \text { say } \right)\]

\[\text { Taking log on both sides, we get }\]

\[\log \left( P V^\frac{1}{4} \right) = \log k\]

\[ \Rightarrow \log P + \frac{1}{4}\log V = \log k\]

\[\text { Differentiating both sides w . r . t . x, we get }\]

\[\frac{1}{P}\frac{dP}{dV} + \frac{1}{4V} = 0\]

\[ \Rightarrow \frac{dP}{P} = - \frac{dV}{4V} = - \frac{1}{4} \times \frac{- 1}{2} = \frac{1}{8}\]

\[\text { Hence, the increase in the pressure is } \frac{1}{8} \% .\]