Question

In a reversible adiabatic expansion of air (assume it a mixture of $N_2$ and $O_2$), the volume increases by $25\%$, then the percentage change in pressure is:
(Take ${0.8}^{1.4}=0.73$)

• A. decreases by $35\%$
• B. increases by $40\%$
• C. increases by $27\%$
• D. decreases by $27\%$

Answer 1

The correct option is D decreases by $27\%$

For a reversible adiabatic expansion,
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
According to the question, after the expansion, the volume increases by $25\%$.
Thus,
$P{V}^{\gamma }=P_1(\frac{125}{100}V{)}^{\gamma }$

$\therefore \frac{P_1}{P}\times {\left(\frac{125}{100}\right)}^{\gamma }=1$
or $\frac{P_1}{P}={\left[\frac{100}{125}\right]}^{\frac{7}{5}}$

As $N_2$ and $O_2$ are diatomic, so $(\gamma \text{for}{N}_2\text{or}{O}_2=7/5)$

$\therefore \frac{P_1}{P}=(0.8{)}^{1.4}$
So we found that
$P_1=0.73P$
$\therefore 27\%$ decrease in pressure.