In an adiabatic process, the pressure of a monoatomic ideal gas increases by 0.5%. The volume will decrease by___%

In an adiabatic process ${\mathrm{PV}}^{\mathrm{\gamma }}=\text{ constant }\left(\mathrm{k}\right)$.

Taking logarithm,

                     $\log \mathrm{P}+\mathrm{\gamma log}\mathrm{V}=\log \mathrm{k}$

                    Differentiating,

$\begin{array}{l} \frac{\mathrm{\Delta P}}{\mathrm{P}}+\mathrm{\gamma }\frac{\mathrm{\Delta V}}{\mathrm{V}}=0\\ \Rightarrow \frac{\mathrm{\Delta V}}{\mathrm{V}}=-\frac{1}{\mathrm{\gamma }}\frac{\mathrm{\Delta P}}{\mathrm{P}}\end{array}$

For a monoatomic gas, $\mathrm{\gamma }=\frac{5}{3}$. Therefore $\frac{\mathrm{\Delta V}}{\mathrm{V}}=-\frac{3}{5}\times 0.5\mathrm{\%}=-0.3\mathrm{\%}$

The negative sign shows that the volume decreases.