Gamma is defined as $$\frac{C_p}{C_v}$$

Now say you have a mixture of gases eg: 1 mol of monoatomic gas and 2 moles of diatomic gas, then how would I find the value of gamma of this new mixture of gases?

  • $\begingroup$ What are your thoughts on this? How is Cp of an ideal gas mixture related to the Cp's of the individual components? What is the heat of mixing of ideal gases? $\endgroup$ Apr 20, 2020 at 11:51
  • $\begingroup$ Oh so you're saying to use energy conservation whilst writing energy of gas as $U = n C_v \Delta T?$ $\endgroup$ Apr 20, 2020 at 14:10
  • $\begingroup$ No. You need to understand and quantify the mixture rules for ideal gases. Each gas species behaves as if the other gas species are not present. $\endgroup$ Apr 20, 2020 at 14:12
  • $\begingroup$ Sure, yeah the pressures add up , the volumes add up and the temperature has a funky relation. How do I apply this to find gamma? $\endgroup$ Apr 20, 2020 at 15:01
  • $\begingroup$ Have this same question. $\endgroup$
    – Kashmiri
    Jan 29, 2021 at 9:26

1 Answer 1


Once the gases are mixed and at same temperature,

$$ (n_1 + n_2) \overline{C_v} T = n_1 C_{v_1} T + n_2 C_{v_2} T$$

Cancelling temperatures,

$$(n_1+n_2)\bar{C_v}=n_1C_{v1}+n_2C_{v2}$$ where $\bar{C_v}$ is the molar average heat capacity at constant volume of the mixture. So $$\bar{C_v}=x_1C_{v1}+x_2C_{v2}$$and $$\bar{C_p}=x_1C_{v1}+x_2C_{v2}+R$$So, for the mixture, $$\gamma=\frac{\bar{C_p}}{\bar{C_v}}=1+\frac{R}{x_1C_{v1}+x_2C_{v2}}$$or $$\gamma=1+\frac{1}{\frac{x_1}{(\gamma_1-1)}+\frac{x_2}{(\gamma_2-1)}}$$ I'm sure that this can be simplified some more, but I'll leave that up to you.

  • $\begingroup$ Question: How did you know you could directly add C_vs scaled by molar oefficent ? $\endgroup$ Apr 20, 2020 at 16:43
  • $\begingroup$ Because I know that the heat of mixing of ideal gases is zero. $\endgroup$ Apr 20, 2020 at 16:52
  • 2
    $\begingroup$ That's what I said in the comment by applying energy conversation bruh $\endgroup$ Apr 20, 2020 at 18:07
  • $\begingroup$ I came back to this question but I still can't understand how you achieved the premise. The only way that equation can be stated (afaik) is if gases are mixed at same temperature, did you assume this? $\endgroup$ Oct 6, 2020 at 14:50
  • $\begingroup$ @ Buraian Of course. In a gas mixture/solution, how can they not be at the same temperature. And, of course, for ideal gasses, the enthalpy and internal energy changes on mixing are zero. So the partial molar internal energies and enthalpies of the individual components in the mixture are the same as for the pure gases at the same temperature. $\endgroup$ Oct 6, 2020 at 15:02

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