Question

In an adiabatic process, the pressure is increased by $\frac{2}{3}$. If $\gamma =\frac{3}{2}$, then find the decreases in volume (approximately) ?

Answer 1

Let $P$ and $V$ be the initial pressure and volume respectively.

For an adiabatic process,

$P{V}^{\gamma }=\text{constant}$

Therefore,

$\frac{P_i}{P_f}={\left(\frac{V_f}{V_i}\right)}^{\gamma }$

Given that:-

$P_i=P$

$P_f=P+\frac{2}{3}P=\frac{5}{3}P$

$V_i=V$

$V_f=?$

$\gamma =\frac{3}{2}$

$\frac{P}{\left(\frac{5}{3}P\right)}={\left(\frac{V_f}{V}\right)}^{\frac{3}{2}}$

$\Rightarrow {\left(\frac{3}{5}\right)}^{\frac{2}{3}}=\frac{V_f}{V}$

$\Rightarrow V_f=0.84V$

Therefore,

Decrease in volume $=V_i-V_f=V-0.84V=0.16V$

Hence the volume is decreased by $0.16$ times.