If 6.00 moles of a monatomic ideal gas at a temperature of 235 L are expanded isothermally from a volume of 1.26 L to a volume of 4.08 L. How would you calculate the work (W) done by the gas?

Dec 28, 2016

You would do it like this:

Explanation:

There is no change in the internal energy of an ideal gas during an isothermal process - the temperature is constant.

The 1st Law of Thermodynamics tells us that all the heat added to the system is used to do work.

At a constant temperature the work integral is:

$\textsf{W = {\int}_{{V}_{i}}^{{V}_{f}} P . \mathrm{dV}}$

$\textsf{P V = n R T}$

$\therefore$$\textsf{P = \frac{n R T}{V}}$

So the integral can now be written:

$\textsf{W = n R T {\int}_{{V}_{i}}^{{V}_{f}} \frac{\mathrm{dV}}{V}}$

Integrating and putting in the limits gives:

$\textsf{W = n R T \ln \left[{V}_{f} / {V}_{i}\right]}$

Putting in the numbers (I assume you mean 235 K):

$\textsf{W = 6.00 \times 8.31 \times 235 \times \ln \left[\frac{4.08}{1.26}\right]}$

I can use litres instead of $\textsf{{m}^{3}}$ as this is a ratio.

$\textsf{W = 6.00 \times 8.31 \times 235 \times 1.175 = 1.38 \times {10}^{4} \textcolor{w h i t e}{x} J}$