What are the equations for $\Delta S_{mix}$, $\Delta G_{mix}$, and $\Delta H_{mix}$ for a real gas?


In one of the most well known experiments in thermodynamics, two chambers are placed beside each other with a membrane in between. The thought experiment is used to teach thermodynamic principles by calculating thermodynamic property changes after the membrane ruptures. Much like this:


However, this is mainly done for ideal gases and where typically the temperature , number of moles, and pressure are provided for each side of the chamber (such as 1 bar, 298K and 1 mol). Therefore, the thermodynamic properties after rupture can be solved easily with conventional well documented equations such as:

$\Delta S_{mix}^{Ideal Gas} = n_A R \ln(\frac{V_{total}}{V_A}) + n_B R \ln(\frac{V_{total}}{V_B})$

$\Delta H_{mix}^{Ideal Gas} = 0$

$\Delta G_{mix}^{Ideal Gas} = \Delta H_{mix}^{Ideal Gas} - T \Delta S_{mix}^{Ideal Gas}$

These are quite easy to solve when given the inital information.

However, if the gas behaves like a real gas (for instance a Van der Waals gas), how does this change?

Using the Van der Waals Equation of State for example, we know we will need the $a$ and $b$ values for the VdW Equation:

$P = \frac{nRT}{V-nb}-\frac{n^2a}{V^2}$



These $a$ and $b$ constants can be applied to a pure gas in the two initial states to solve for instance $V$ if $P$, $T$, and $n$ are given for each side. However, once the two gases mix, the $a$ and $b$ will now be different since the molecules will likely interact differently with each other than they do with themselves. So, we need tabulated values for $a_{mix}$ and $b_{mix}$ in order to solve the VdW EOS after mixing.

However, even if we do all of this, I have not found hte thermodynamic property equation for mix a real gas. Have these been figured out or is everything for these real gases experimental extrapolation?

$\Delta S_{mix}^{Real Gas}=?$

$\Delta G_{mix}^{Real Gas}=?$

$\Delta H_{mix}^{Real Gas}=?$

  • $\begingroup$ "However, once the two gases mix, the a and b will now be different since the molecules will likely interact differently with each other than they do with themselves." If the gas on both sides of the partition are the same, except for temperature, then a and b can be determined. When the 2 gases are different, then a and b will be more difficult to determine. $\endgroup$ – LDC3 Mar 30 '14 at 22:39
  • $\begingroup$ Well, if the two gases are the same, then the problem will become much simpler due to Gibbs paradox, since $\Delta S_{mix}$ will be 0 due to indistinguishability of the particles. So we are likely much more interested in the case which they are not the same. Although, I do not know if the $\Delta H_{mix}$ will be affected by expanding and mixing of a same gas. $\endgroup$ – John Mar 30 '14 at 22:42
  • $\begingroup$ You're right. I was focused on the difference in temperature. I forgot there should be different gases. $\endgroup$ – LDC3 Mar 30 '14 at 22:50

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