The pressure P and volume V of a gas are connected by the relation PV1/4 = constant. The percentage increase in the pressure corresponding to a deminition of $\frac{1}{2}$ % in the volume is

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KollegeApply · Accepted Answer

\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 } 1/PdP/dV + $\frac{1}{4}$V = 0 \Rightarrow dP/P = - dV/4V = - $\frac{1}{4}$ $\times$ - $\frac{1}{2}$ = $\frac{1}{8}$ \text { Hence, the increase in the pressure is } $\frac{1}{8}$ \% .

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