# Work done in reversible isothermal expansion

A cylinder fitted with a frictionless piston is filled with $$10~\mathrm{mol}$$ of gaseous carbon tetrachloride and immersed in a large bath of water maintained at $$300~\mathrm{K}$$. The volume of the gas is initially at $$V_1~\mathrm{L}$$, and the external pressure on the piston is slowly decreased until the volume reaches $$V_2~\mathrm{L}$$. Calculate the work for this process, in $$\mathrm{kJ}$$, assuming the gas can be treated as ideal.

I think I have the general idea of how to solve this problem but I'm not sure how to find the external pressure. So far, I'm using the equation

$$\int \mathrm{d}w = \int_{V_2}^{V_1}-p_{\mathrm{ext}}\,\mathrm{d}V$$

since it is pressure-volume work and have tried calculating $$p_{\mathrm{ext}}$$ using $$pV=nRT$$ and the final volume of the system, but this doesn't seem to give the correct answer. How does it have to be solved?

• If you could show us your work, then someone here might be able to point out where exactly you went wrong. Cheers! – getafix Sep 20 '16 at 2:45

I agree with getafix, if you would like an answer that is more tailored to you, you should show us exactly what you've done.

However, I am going to make a (hopefully educated) guess that what you did was to pull $p_\mathrm{ext}$ out of the integral. That is incorrect, because $p_\mathrm{ext}$ is not a constant here.

This process is known as an isothermal expansion - isothermal because the temperature remains constant throughout - and expansion because volume is increased. In thermodynamics it is very important to note which variables are held constant, because then that lets you decide which formula is appropriate to use, or how to derive such formulae).

Since the process is reversible, the external pressure must always be equal to the pressure exerted by the gas, which can be calculated via the ideal gas law $pV = nRT$. Therefore, you have (where 1 and 2 denote the initial and final state respectively)

\begin{align} w &= -\int_1^2 p\,\mathrm{d}V \\ &= -\int_1^2 \frac{nRT}{V}\,\mathrm{d}V \end{align}

and now since $T$ is a constant, you can take it out of the integral (along with $n$ and $R$ which are also constants)

\begin{align} w = &= -nRT\int_1^2 \frac{1}{V}\,\mathrm{d}V \\ &= -nRT\ln\left(\frac{V_2}{V_1}\right) \end{align}

Since you weren't given numerical values for $V_2$ and $V_1$, you can't evaluate a numerical value for $w$. The closest you can get is therefore

$$w = (-24.9~\mathrm{kJ})\ln\left(\frac{V_2}{V_1}\right)$$

• It's worth noting that carbon tetrachloride has a boiling point (at atmospheric pressure) of well over 300 K, so we can at least infer that both V1 and V2 correspond to starting and ending pressures that are significant below one atmosphere. – Curt F. Jul 11 at 22:58