Question

A given mass of a gas is compressed isothermally until its presssure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately :

• A. 1.2
• B. 1.41
• C. 1.67
• D. 1.83

Answer 1

The correct option is B 1.41

let Initial value of volume,pressure and temperature of gas be $V_0$, $P_0$, $T_0$

In step one, gas is compressed isothermally, so temperature will be constant

After first step :

Pressure = $2P_0$; Temperature = $T_0$

from $PV=nRT$ Volume = $V_0/2$;

In 2nd step, gas is allowd to expand adiabatically to restore its original volume
After 2nd step:
Pressure = $0.75P_0$, Volume = $V_0$,
from the adiabatic process relation between P and V - $P{V}^{\gamma }=constant$
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
$(\frac{V_1}{V_2}{)}^{\gamma }=\frac{P_2}{P_1}$

$V_1=V_0/2$, $P_1=2P_0$,

$V_2=V_0$ and $P_2=0.75P_0$

so $(\frac{V_0/2}{V_0}{)}^{\gamma }=\frac{0.75P_0}{2P_0}$
$(\frac{1}{2}{)}^{\gamma }=\frac{3}{8}$
on solving this:
$\gamma =1.41$