Question

Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2.

Answer 1

γ satisfies the equation 1 − 21−γ = (γ − 1) ln2 of two samples A and B of same gas with equal volumes and pressures

Explanation:

Given  Data

Initial gas pressure = $P_1$

Initial gas volume = $V_1$

Final gas pressure =   $P_2$

Final gas  volume = $V_2$

Given for each case $V_2 = 2 V_1$

Step 1:

In a method of isothermal expansion;

$\text { work done }=n R T_{1} \ln \left(\frac{V_{2}}{V_{1}}\right)$

Work done adiabatically,

$W=\frac{\mathrm{P}_{1} \mathrm{V}_{1}-\mathrm{P}_{2} \mathrm{V}_{2}}{\gamma-1}$

In both cases, the same work is done.

$n R T_{1} \ln \left(\frac{V_{2}}{V_{1}}\right)=\frac{P_{1} V_{1}-P_{2} V_{2}}{\gamma-1} \ldots(1)$

Step 2:

In case of adiabatic process,

$\mathrm{P}_{2}=\mathrm{P}_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\mathrm{P}_{1}\left(\frac{1}{2}\right)^{\gamma}$

From equation (1),

$n R T_{1} \ln (2)=\frac{P_{1} V_{1}\left(1-\frac{1}{2^{V}} \times 2\right)}{\gamma-1}$

and   $n R T_{1}=\mathrm{P}_{1} \mathrm{V}_{1}$

$\ln 2=\frac{\left(1-\frac{1}{2^{y}} \times 2\right)}{\gamma-1}$

(or) $(\gamma-1) \ln 2=1-2^{1-\gamma}$

Therefore it is proved that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2 of two samples A and B f same gas with equal volumes and pressure, where A is expanded isothermally and B is expanded adiabatically to double their volume with same work done for two two cases.