Consider one mole of helium gas enclosed in a container at initial pressure $P_{1}$ and volume $V_{1}$ . It expands isothermally to volume $4 V_{1}$ . After this, the gas expands adiabatically and its volume becomes $32 V_{1}$ . The work done by the gas during isothermal and adiabatic expansion processes are $W_{i s o}$ and $W_{a d i a}$ , respectively. If the ratio $\frac{W_{i s o}}{W_{a d i a} = f \ln 2}$ , then $f$ is ________.

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Solution

It is given that for $1 m o l e$ helium gas, $P_{1}$ is initial pressure and $V_{1}$ is initial volume.
Also, when the gas expand isothermally the volume becomes $4 V_{1}$ and when gas expand adibatically the volume becomes $32 V_{1}$ .
Now, let us consider $T_{1}$ as $T$ and find work done in isothermal expansion is: $W_{i s o} = n R T_{1} (\ln \frac{V_{2}}{V_{1}}) W_{i s o} = n R T (\ln \frac{4 V_{1}}{V_{1}}) W_{i s o} = n R T (\ln 4) W_{i s o} = 2 n R T (\ln 2) . . . . . . . . (1)$
For work done in adiabatic expansion: $V_{1} = 4 V_{1}$ and $V_{2} = 32 V_{1}$ , $γ = 2 f - 1$ .
For adiabatic process: $T_{1} {V_{1}}^{γ - 1} = T_{2} {V_{2}}^{γ - 1} T {(4 V)}^{\frac{2}{3}} = T_{2} {(32 V)}^{\frac{2}{3}} T {(4)}^{\frac{2}{3}} = T_{2} 8 T_{2} = \frac{T}{4}$
Work done in adiabatic process is: $W_{a d i a} = \frac{n R (T_{1} - T_{2})}{γ - 1} W_{a d i a} = \frac{n R (T - \frac{T}{4})}{\frac{5}{3} - 1} = \frac{9}{8} n R T . . . . . . . (2)$
Now determining the ratio of work done in isothermal expansion and in adiabatic expansion by dividing eq 1 and 2: $\frac{W_{i s o}}{W_{a d i a}} = \frac{2 n R T (\ln 2)}{\frac{9}{8} n R T} \frac{W_{i s o}}{W_{a d i a}} = \frac{2 \ln 2 \times 8}{9} = \frac{16}{9} \ln 2$
As given that, $\frac{W_{i s o}}{W_{a d i a}} = f \ln 2$ and as calculated $\frac{W_{i s o}}{W_{a d i a}} = \frac{16}{9} \ln 2$
Hence, $f = \frac{16}{9} = 1.77$

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