How to compute this change of Gibbs energy for a Van der Waals gas?


Solution 1

I think shifting the variable of integration should work: \begin{align} \int_{p_1}^{p_2}dp~V & =\int_\text{state 1}^\text{state 2} d(pV)-\int_{V_1}^{V_2} dV~p \\ & = p_2V_2-p_1V_1-\int_{V_1}^{V_2} dV~\left[ \frac{nRT}{V-bn}-a\frac{n^2}{V^2} \right] \end{align}

Solution 2

This problem contains a functional form that is very difficult to work with. A numeric integration is one approach that will give an approximate (but good) answer. To do this, the following procedure can be used:

  1. Start at pressure $P_1$, and establish a "small" value for $dP$.
  2. Use a trial-and-error method to calculate $V$.
  3. Multiply $V$ by $dP$ and keep track of the sum of $V(dP)$.
  4. Increment pressure by $dP$.
  5. Repeat steps 2-4 until $P_2$ is reached.

The size of $dP$ is arbitrary. To ensure that your value of $dP$ is appropriate, it would be helpful to calculate the integral with a value of $dP$, then divide $dP$ by $2$ and repeat the calculation. If both calculations give approximately the same answer, you are done.


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Updated on December 26, 2020


  • Nikolaj-K
    Nikolaj-K over 2 years

    The change of Gibbs energy at constant temperature and species numbers, $\Delta G$, is given by an integral $\int_{p_1}^{p_2}V\,{\mathrm d}p$. For the ideal gas law $$p\,V=n\,RT,$$ this comes down to $$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$ That logarithm is at fault for a lot of the formulas in chemistry.

    I find I have a surprisingly hard time computing $\Delta G$ for gas governed by the equation of state $$\left(p + a\frac{n^2}{V^2}\right)\,(V - b\,n) = n\, R T,$$ where $a\ne 0,b$ are small constants. What is $\Delta G$, at least in low orders in $a,b$?

    One might be able to compute ΔG by an integral in which not V is the integrand.

    Edit 19.8.15: My questions are mostly motivated by the desire to understand the functional dependencies of the chemical potential $\mu(T)$, that is essentially given by the Gibbs energy. For the ideal gas and any constant $c$, we see that a state change from e.g. the pressure $c\,p_1$ to another pressure $c\,p_2$ doesn't actually affect the Gibbs energy. The constant factors out in $\frac{1}{p}\,{\mathrm d}p$, resp. $\ln\frac{p_2}{p_1}$. However, this is a mere feature of the gas law with $V\propto \frac{1}{p}$, i.e. it likely comes from the ideal gas law being a model of particles without interaction with each other.

    • Gert
      Gert almost 8 years
      The change in Gibbs Free Energy for which transformative process? Going from which initial state to which final state? $\Delta G$ is a state function.
    • Nikolaj-K
      Nikolaj-K almost 8 years
      @Gert: I'm not sure which aspect of your question isn't answered by the first sentence.