Question

If by keeping the temperature constant, the pressure of a gas is decreased by $20\%$, then the percentage change in volume of the gas will be

• A. + $20\%$
• B. - $20\%$
• C. + $25\%$
• D. - $25\%$

Answer 1

The correct option is C + $25\%$

According to Boyle's law, we know that
$PV=\text{constant}$ at constant temperature
$\therefore$ we can write that,
$P_1{V}_1=P_2{V}_2..............\left(1\right)$
Given,
$P$ is decreased by $20\%$
Thus,
Initial pressure $\left(P_1\right)=P$
Final pressure $\left(P_2\right)=P-\frac{20}{100}P$
Initial volume $\left(V_1\right)=V$
Final volume $\left(V_2\right)=V^{\prime }$
From $\left(1\right)$, we can write that
$PV=\frac{80}{100}P{V}^{\prime }$
$\Rightarrow V=\frac{80}{100}{V}^{\prime }.......\left(2\right)$
Percentage change in volume is given by,
$\frac{\text{Final volume}\left(V^{\prime }\right)-\text{Initial volume}\left(V\right)}{\text{Initial volume}\left(V\right)}\times 100$
Using $\left(2\right)$ in the above formula, we get
$\frac{V^{\prime }-V}{V}\times 100=\frac{100-80}{80}\times 100=+25\%$
Thus, option (c) is the correct answer.