Question

Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$. It expands isothermally to volume $4V_1$. After this, the gas expands adiabatically and its volume becomes $32V_1$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{iso}$ and $W_{adia}$, respectively. If the ratio $\raisebox{1ex}{$W_{iso}$}\!\left/ \!\raisebox{-1ex}{$W_{adia}=f\ln 2$}\right.$, then $f$ is ________.

Answer 1

It is given that the initial pressure and volume of one $\mathrm{mole}$ helium gas enclosed in a container is $P_1$ and $V_1$respectively.

After the isothermal expansion of helium, the volume of gas becomes $4V_1$ and after adiabatic expansion volume becomes $32V_1$.

Now, we have to determine $f$ if $\raisebox{1ex}{$W_{iso}$}\!\left/ \!\raisebox{-1ex}{$W_{adia}=f\ln 2$}\right.$.

For this let us determine the work done in isothermal expansion: $W_{iso}=nR{T}_1\left(\ln \frac{V_2}{V_1}\right)$

$$\begin{gathered} W_{iso}=nR{T}_1\left(\ln \frac{4V_1}{V_1}\right) \\ {W}_{iso}=nR{T}_1\left(\ln 4\right) \\ {W}_{iso}=2nR{T}_1\ln 2\_\_\_\_\_\_\_\_\_\left(1\right) \end{gathered}$$

Now, for work done in adiabatic expansion let us consider, $V_1=4V_1$ and $V_2=32V_1$, $\gamma =2f-1$

Let us first determine $T_2$ to calculate $∆T$ when $\gamma =\frac{5}{3}$ : $T_1{V_1}^{\gamma -1}=T_2{V_2}^{\gamma -1}$

$$\begin{gathered} T_1{\left(4V\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3-1$}\right.}=T_2{\left(32V\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3-1$}\right.} \\ {T}_1{\left(4\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=T_2{\left(32\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ {T}_2=\frac{T_1}{4} \end{gathered}$$

So, work done in adiabatic expansion is: $W_{adia}=\frac{nR∆T}{\gamma -1}$

$$\begin{gathered} W_{adia}=\frac{nR\left({\displaystyle T_1-T_2}\right)}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.-1}} \\ {W}_{adia=\frac{nR(T_1-{\displaystyle \raisebox{1ex}{$T_1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.})}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ {W}_{adia}=\frac{9}{8}nR{T}_1\_\_\_\_\_\_\_\_\_\left(2\right) \end{gathered}$$

Determining $f$ by dividing equation $\left(1\right)$ by $\left(2\right)$: $\frac{W_{iso}}{W_{adia}}=\frac{2nR{T}_1\ln 2}{{\displaystyle \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}nR{T}_1}$

$\frac{W_{iso}}{W_{adia}}=\frac{16}{9}\ln 2$

Now from the given equation $\raisebox{1ex}{$W_{iso}$}\!\left/ \!\raisebox{-1ex}{$W_{adia}=f\ln 2$}\right.$ and the equation determined above we can say that $f=\frac{16}{9}=1.77$.