The pressure p and the volume V of a gas are related as $pV^{1/4} = a$, constant. Find the percentage increase in the volume corresponding to a diminution of $\dfrac{1}{2} \%$ in the pressure.
Percentage decrease in pressure $= \dfrac{1}{2} \%$
i.e. $\dfrac{\Delta p}{p} = -\dfrac{1}{2} \% = -\dfrac{1}{200}$ $\;\; \cdots$ (1)
Now, $pV^{1/4} = a$
$\therefore$ $\log p + \dfrac{1}{4} \log V = \log a = 0$ [since a is a constant, $\log a =0$]
i.e. $\log V = -4 \log p$ $\;\; \cdots$ (2)
Differentiating equation (2) w.r.t p gives
$\dfrac{1}{V} \dfrac{dV}{dp} = -\dfrac{4}{p}$
i.e. $\dfrac{dV}{dp} = -\dfrac{4V}{p}$ $\;\; \cdots$ (3)
Now, $\Delta V = \left(\dfrac{dV}{dp}\right) \times \Delta p = -\dfrac{4V}{p} \times \Delta p = -4V \times \left(\dfrac{\Delta p}{p}\right) = -4V \times \left(-\dfrac{1}{200}\right) = \dfrac{V}{50}$
Now, $\%$ Increase in volume $= \dfrac{\Delta V}{V} \times 100 \% = \dfrac{1}{V} \times \dfrac{V}{50} \times 100 \% = 2 \%$