Question

Two litres of an ideal gas at a pressure of $10$ atm expands isothermally to a final volume of $10$ litres against a constant external pressure of $1$ atm. What will be the work done if process is done reversibly?

Answer 1

Heat absorbed for an isothermal expansion is given as, $nRTln\left(\frac{V_2}{V_1}\right)$ where, n is the number of moles, T is the constant temperature associated with the process, $V_2$ is the final volume and $V_1$ was the initial volume.

Considering, $n=1$ applying ideal gas law, we can find the temperature,

so, $P_1{V}_1=nRT$

or, $T=\frac{P_1{V}_1}{nR}$

or, $RT=P_1{V}_1$, for $n=1$

so, $RT=10\times 2=20atm.L$

So, heat absorbed is, $1\times 10\times ln\left(\frac{10}{2}\right)=32.2atm.L$

And we know from $1^{st}$ law of thermodynamics, change in internal energy for an isothermal process is zero.

So, work done $=-$ heat absorbed $=-32.2atm.L$