Question

Two samples A and B of same gas have same initial pressure and volume. A undergoes an Isothermal expansion and B is subjected to adiabatic expansion. If the final volume in both cases is double the initial volume and work done in both the cases are same, then prove that
$2^{1-\gamma }-1=ln2,\gamma =\frac{C_p}{C_{\gamma }}$

Answer 1

Let the initial pressure, $p_i=p_1$
Initial volume, $V_i=V_1$
final volume, $V_f=V_2$
$=2V_1$
Work done in isothermal process,
$W_{iso}=2.303RTlog\left(\frac{V_2}{V_1}\right)$
Since $PV=RT$ (1 mol of gas),
$W_{iso}=P_1{V}_{1}l{n}_2$ ..........(1)
work done in adiabatic process,
$W_{ad}=\frac{1}{\gamma -1}[P_1{V}_1-P_2{V}_2]$
$=\frac{P_1{V}_1}{\gamma -1}[1-\frac{P_2{V}_2}{P_1{V}_1}]$
For adiabatic process,
$P_1{V}_1^{\gamma }=P_1{V}_2^{\gamma }=K\left(constant\right)$
$\therefore \frac{P_2}{P_1}=\frac{V_1^{\gamma }}{V_2^{\gamma }}$
$\therefore W_{ad}=\frac{P_1{V}_1}{\gamma -1}[1-\frac{V_2^{1-\gamma }}{V_1^{1-\gamma }}]$
$W_{ad}=\frac{P_1{V}_1}{\gamma -1}[1-2^{1-\gamma }]$ .............(2)
since work done is same for both processes,
$W_{iso}=W_{ad}$
$\therefore P_1{V}_{1}l2=\frac{P_1{V}_1}{\gamma -1}(1-2^{1-\gamma })$ .........(3)
Re-arranging and simplifying (3),
$2^{1-\gamma }-1=(1-\gamma )ln2.$