Question

Calculate the work done by one mole of a van der Waals’ gas if during its isothermal
expansion its volume increases from 1m3 to 2m3 at a temperature 300K.
Take a = 1.39 × 10−6 atm m6 mol−2 and b = 39.1 × 10−6 m3 mol−1.

Answer 1

Calculate the work done by one mole of a Van der Waals' gas if during its isothermal expansion its volume increases from $1\mathrm{m}^3$ to $2\mathrm{m}^3$ at a temperature 300 K. Take $a = 1.39\cdot 10^{-6}$ atm·m$^6$·mol$^{-2}$ and $b = 39.1\cdot 10^{-6}$ m$^3$·mol$^{-1}$.

Solution: As you know, work of 1 mol of gas can be calculated as: $W = \int_{\nu_1}^{\nu_2} pd\nu$; where $\nu$ is the molar volume, $\mathrm{m}^3/\mathrm{mol}$. The Van der Waals equation of state for 1 mole of gas is: $\left(p + \frac{a}{\nu^2}\right) \cdot (\nu - b) = R_0 T$; From this equation we obtain that $p = \frac{R_0 T}{\nu - b} - \frac{a}{\nu^2}$;

Then, $W = \int_{\nu_1}^{\nu_2}\left(\frac{R_0T}{\nu - b} -\frac{a}{\nu^2}\right)d\nu = \int_{\nu_1}^{\nu_2}\frac{R_0T}{\nu - b} d\nu -\int_{\nu_1}^{\nu_2}\frac{a}{\nu^2} d\nu = R_0T\cdot \ln \frac{\nu_2 - b}{\nu_1 - b} +a\cdot \left(\frac{1}{\nu_2} -\frac{1}{\nu_1}\right);$

Where $\nu = \frac{V}{n}$; $\nu_{1} = \frac{1}{1} = 1\frac{\mathrm{m}^{3}}{\mathrm{mol}}$; $\nu_{2} = \frac{2}{1} = 2\frac{\mathrm{m}^{3}}{\mathrm{mol}}$; also, we must convert the value of coefficient $a$ into the SI units: $a = 1.39\cdot 10^{-6}\cdot 101,325\mathrm{N}\cdot \mathrm{m}^{-2}\cdot \mathrm{m}^{6}\cdot \mathrm{mol}^{-2} = 0.141\mathrm{N}\cdot \mathrm{m}^{4}\cdot \mathrm{mol}^{-2}$.

$$W = 8.314 \cdot 300 \cdot \ln \frac {2 - 39.1 \cdot 10^{-6}}{1 - 39.1 \cdot 10^{-6}} + 0.141 \cdot \left(\frac {1}{2} - \frac {1}{1}\right) = 1728.8 \mathrm{~J};$$

Answer: 1728.8 J.