Question

For adiabatic expansion of a monoatomic perfect gas, the volume increases by $2.4\%$. What is the percentage decreases in pressure ?

• A. $2.4\%$
• B. $4.0\%$
• C. $4.8\%$
• D. $7.1\%$

Answer 1

The correct option is B $4.0\%$

For an adiabatic process: $P{V}^{\gamma }=constant$
Differentiating on both sides: $V^{\gamma }\delta P+\gamma P{V}^{\gamma -1}\delta V=0$
Dividing by: $P{V}^{\gamma }=constant$
$\frac{\delta P}{P}+\gamma \frac{\delta V}{V}=0$

Percentage increase in volume is 2.4%
i.e. $\frac{\delta V}{V}=\frac{2.4}{100}$ and $\gamma =\frac{5}{3}$

i.e. $\frac{\delta P}{P}=-\gamma \frac{\delta V}{V}=-\frac{5}{3}\times \frac{2.4}{100}=-\frac{4}{100}$

Hence, decrease in pressure is $4\%$.