# How do you calculate the heat capacity ratio for a multi-compound gas?

Let's say I have a heat capacity ratio of 1.4 N2 and a heat capacity ratio of 1.2 for O2 (just random placeholder values no significance), and I wanted to calculate the heat capacity ratio of their mixture, similar to air let's say 80% N2, 20% O2 (again placeholder values).

How would I go about calculating the heat capacity ratio of air in total? I have seen people say that I should find the molar fractions of N2 and O2, and then multiply the fractions by the heat capacity ratio.

BUT, I have also seen a solution manual from Thermodynamics: An Engineering Approach, instead use the mass fractions of N2 and O2, and then multiply the fractions by the specific heat capacity at a constant pressure. Then once you find the specific heat capacity at a constant pressure of the mixture, you then use Cv=Cp - R, to find the specific heat capacity at a constant volume, and then do Cp/Cv to get the heat capacity ratio of the mixture.

Can anyone confirm the correct method? And unfortunately I have checked, you do not get the same answer.

• Placeholders are the best expressed by variables instead of random values. // For mass fractions and specific heat, Cp=Cv+R is not applicable .Check dimensions of the formula. Commented Aug 1, 2022 at 5:50
• I would use 78% N2, 21% O2 and 1% Ar. It's more complicated, but besides more closely approximating real air it introduces a component that actually has a different ratio when pure. Commented Aug 1, 2022 at 9:51
• Could you please clarify how this is different from your previous question? I can see the text is different, but this seems to be the same underlying question, to the point where you've just used the same title, verbatim. In other words: why shouldn't this extra text just been edited into your previous question? Commented Aug 1, 2022 at 13:40

First off, in reality both nitrogen and oxygen have the same ratio, $$7/5=1.4$$. A more realistic example would be to include the argon component in the air, as argon has a ratio of $$5/3\approx1.67$$.

When you calculate the heat capacity, which fractions you use depends on the basis for the heat capacity. If you are calculating on a mass basis, you use mass fractions; if you are calculating on a molar basis, you use mole fractions.

The ideal gas relation $$C_p-C_v=R$$ applies on a molar basis, so for calculating the heat capacity ratio using a molar basis is preferred. Therefore use mole fractions in your mixing rule.

We can develop a formula for the overall heat capacity ratio given the molecule fractions and individual ratio of each component. For a pure gaseous compound we have

$$C_p-C_v=R$$

$$C_p/C_v=\gamma$$

and from these we easily obtain

$$C_v=\dfrac{R}{\gamma-1}$$

$$C_p=\dfrac{R\gamma}{\gamma-1}$$

Therefore for a multicomponent gas with mole fractions $$x_i$$ we render

$$\bar{\gamma}=\dfrac{\Sigma\dfrac{x_i\gamma_i}{\gamma_i-1}}{\Sigma\dfrac{x_i}{\gamma_i-1}}.$$

We now subtract $$1$$ to get

$$\bar{\gamma}-1=\dfrac{\Sigma x_i}{\Sigma\dfrac{x_i}{\gamma_i-1}}$$

and thus a weighted harmonic mean formulation:

$$\dfrac{1}{\bar{\gamma}-1}={\Sigma\dfrac{x_i}{\gamma_i-1}}.$$

• I think I should have clarified, the Cp - Cv = R relation they use is for Cp being mass specific heat capacity, [kJ/(kg x K)], and so is Cv. And R isn't the universal gas constant but the specific gas constant so Rs, which is the universal R divided by the molecular weight (kg/kmol)of the gas. So Rs = R/molecular weight. Which pops out units of [kJ/(kg)(K) ] which matches the Cp and Cv. I think you can do it your way as well, just wanted to clarify the books method. I will actually just link it here : sfu.ca/~mbahrami/ENSC%20461/Suggested%20Problems/Chapter%2013/… Commented Aug 1, 2022 at 20:22
• @onetwo what are you talking about? Commented Aug 1, 2022 at 20:29
• I mean when you mention "The ideal gas relation Cp−Cv=R applies on a molar basis" however they (the link) use it on a mass basis. In essence I just want to verify that the way they reach their heat capacity ratio is reasonable. That is the main point of my entire question. Sorry for the confusion Commented Aug 1, 2022 at 20:31

The partial molar enthalpy and partial molar internal energy of a component in an ideal gas mixture is identical to the corresponding properties of the pure component at the same temperature and pressure equal to its partial pressure in the mixture. From this, it follows that $$C_{p,mixture}=\sum{y_i}C_{p,i}$$$$C_{v,mixture}=\sum{y_i}C_{v,i}$$where the C's are the molar heat capacities of the pure components, and y's are the mole fractions in the gas.