Work of reversible isothermal expansion

2.6.1 Work of reversible isothermal expansion

Using the Leiden form of the virial approach in Eq. (1.10) we get

\begin{equation*} \begin{split} p & = n R T \left(\frac{1}{V} + \frac{n\,B}{V^2} + \frac{n^2\,C}{V^3} + \cdots \right) \quad \text{, i.e.}\\ w & = -\int_{V_1}^{V_2}p dV = -n R T \left[\int_{V_1}^{V_2} \frac{dV}{V} + n B \int_{V_1}^{V_2} \frac{dV}{V^2} + n^2 C \int_{V_1}^{V_2} \frac{dV}{V^3} + \cdots \right] \quad \Rightarrow \\ w & = -n R T \left[\ln \frac{V_2}{V_1} - n B \left(\frac{1}{V_2}-\frac{1}{V_1} \right) -\frac{1}{2}n^2 C \left(\frac{1}{V_2^2}-\frac{1}{V_1^2} \right) + \cdots\right]\\ \end{split} \label{eq:w_Leiden} \end{equation*}

(2.17)

The first term of $w$ represents the work of a perfect gas; for the following discussion we will neglect the third term. For an expansion $1/V_2 - 1/V_1$ is negative. We already know that $B(T)$ is positive at high $T$ and negative at low $T$, i.e. at high $T$ we find an increase of the expansion work and at low $T$ a decrease of the expansion work for real gases compared to perfect gases. Thus repulsive forces must be dominant at high $T$.
We can get a corresponding result using the vdW equation (1.12) for analyzing work contributions

\begin{equation*} \begin{split} p & = \frac{n R T}{V-n b}-\frac{n^2\, a}{V^2} \quad \Rightarrow \quad w = - n R T \int_{V_1}^{V_2} \frac{dV}{V - n b} + n^2 a \int_{V_1}^{V_2} \frac{dV}{V^2} \\ w & = - n R T \ln \frac{V_2 - n b}{V_1 - n b} - n^2 a \left( \frac{1}{V_2} - \frac{1}{V_1} \right)\\ & \approx - n R T \ln \frac{V_2}{V_1} + n^2 R T b \left( \frac{1}{V_2} - \frac{1}{V_1} \right) - n^2 a \left( \frac{1}{V_2} - \frac{1}{V_1} \right) \quad \mbox{(in linear order in $b$)}\\ & = w^0 - n^2 \left( \frac{1}{V_2} - \frac{1}{V_1} \right) \left(a - R T b\right) \quad \mbox{($w^0$ work of perfect gas)}\\ \end{split} \label{eq:w_vdW} \end{equation*}

(2.18)

So for a reversible compression we find $w \lt w^0$ for $b R T \lt a$ and $w \gt w^0$ for $b R T \gt a$, i.e. if attractive exceed repulsive forces ($a \gt b R T$) the compression of a vdW gas needs less work compared to a perfect gas. While $w^0$ scales with $n$ the interaction term in linear order scales with $n^2$. This is a must, since at least two particles are needed for any interaction and $n^2$ is the essential part for the probability of two particles to meet.

With frame

Reversible vs. irreversible processes