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In an adiabatic process, the pressure is increased by $\frac{2}{3} \% .$ If $\gamma=\frac{3}{2},$ then the volume decreases by nearly
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$\frac{4}{9} \%$
$\mathrm{PV}^{3 / 2}=\mathrm{K}$
$\log \mathrm{P}+\frac{3}{2} \log \mathrm{V}=\log \mathrm{K}$
$\frac{\Delta \mathrm{P}}{\mathrm{P}}+\frac{3}{2} \frac{\Delta \mathrm{V}}{\mathrm{V}}=0$
$\frac{\Delta \mathrm{V}}{\mathrm{V}}=-\frac{2}{3} \frac{\Delta \mathrm{P}}{\mathrm{P}}$ or $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\left(-\frac{2}{3}\right)\left(\frac{2}{3}\right)=-\frac{4}{9}$
$\log \mathrm{P}+\frac{3}{2} \log \mathrm{V}=\log \mathrm{K}$
$\frac{\Delta \mathrm{P}}{\mathrm{P}}+\frac{3}{2} \frac{\Delta \mathrm{V}}{\mathrm{V}}=0$
$\frac{\Delta \mathrm{V}}{\mathrm{V}}=-\frac{2}{3} \frac{\Delta \mathrm{P}}{\mathrm{P}}$ or $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\left(-\frac{2}{3}\right)\left(\frac{2}{3}\right)=-\frac{4}{9}$
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