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(A) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^\gamma = k$,where $\gamma = C_P/C_V = 3/2$.Taking the lo

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$\frac{4}{9}\%$

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(A) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^\gamma = k$,where $\gamma = C_P/C_V = 3/2$.Taking the logarithm on both sides: $\ln P + \gamma \ln V = \ln k$.Differentiating both sides: $\frac{\Delta P}{P} + \gamma \frac{\Delta V}{V} = 0$.Therefore,$\frac{\Delta V}{V} = -\frac{1}{\gamma} \frac{\Delta P}{P}$.Given $\frac{\Delta P}{P} = \frac{2}{3}\% = \frac{2}{300}$ and $\gamma = 3/2$,we have:$\frac{\Delta V}{V} = -\frac{1}{3/2} 	imes \frac{2}{3}\% = -\frac{2}{3} 	imes \frac{2}{3}\% = -\frac{4}{9}\%$.The negative sign indicates a decrease in volume.Thus,the percentage decrease in volume is $\frac{4}{9}\%$.

If the pressure in an adiabatic process increases by $\frac{2}{3}\%$,then the percentage decrease in volume will be ....... . Assume $C_P/C_V = 3/2$.

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