Question

At constant temperature, the volume of a gas is to be decreased by 4%. The pressure must be increased by

• A. 4%
• B. 4.16%
• C. 8%
• D. 3.86%

Answer 1

Answer: (B)

At constant temperature, $p_1{V}_1=p_2{V}_2$
$\Rightarrow \frac{p_1}{p_2}=\frac{V_2}{V_1}$
Hence fractional change in volume
$\Rightarrow \frac{V_1-V_2}{V_1}=\frac{4}{100}=\frac{1}{25}$
$\Rightarrow 1-\frac{V_2}{V_1}=\frac{1}{25}$
$\Rightarrow \frac{V_2}{V_1}=\frac{24}{25}$
$\Rightarrow \frac{p_1}{p_2}=\frac{V_2}{V_1}=\frac{24}{25}$
$\Rightarrow \frac{p_2-p_{!}}{p_1}=\frac{24}{25}-1=\frac{1}{24}$
Percentage increase in pressure = $\frac{100}{24}=$ 4.16%.