# I Specific heat ratio of gas mixture

1. Nov 10, 2016

### colinjohnstoe

I am doing some multi-fluid hydrodynamic modelling and I have a quick question. I think I know the answer, but I am not convinced. One of the things that I need to know is the specific heat ratio, $\gamma$, for the gas and my question is, how does one calculate this from the values of each species in the mixture.

At a given point, I know for each species the specific heat ratio, $\gamma_i$, and number density, $n_i$. The thermal energy density is

$$\epsilon = \frac{p}{\gamma-1},$$

where $p$ is the thermal pressure. The thermal energy density is equal to the sums of the values for each individual species

$$\epsilon = \sum_i \epsilon_i = \sum_i \frac{p_i}{\gamma_i - 1}.$$

Inserting $p = n k_B T$ (where $n = \sum_i n_i$) and $p_i = n_i k_B T$ (therefore assuming all species have the same temperature) gives

$$\frac{n k_B T}{\gamma - 1} = \sum_i \frac{n_i k_B T}{\gamma_i - 1},$$

which can be rearranged to give

$$\gamma = \frac{n}{\sum_i \left( \frac{n_i}{\gamma_i - 1} \right) } + 1.$$

This $\gamma$ is the value that I want. Am I correct here or have I made a mistake somewhere?

2. Nov 10, 2016

### Staff: Mentor

For an ideal gas mixture, the heat capacity of the mixture at constant pressure or at constant volume is a weighted average of the corresponding heat capacities of the pure gases, weighted in proportion to their mole fractions:

$$C_{mixture}=\sum_{i=1}^n{y_iC_i}$$
where n is the number of gases present.

3. Nov 11, 2016

### colinjohnstoe

Thanks for the response. It doesn't really answer the question though. I am asking about the specific heat ratio $\gamma=C_\mathrm{P}/C_\mathrm{V}$. If we use the weighted sums, we get

$$\gamma=\frac{C_\mathrm{P}}{C_\mathrm{V}} = \frac{\sum_i y_i C_{\mathrm{P},i}}{ \sum_i y_i C_{\mathrm{V},i} }$$

For this quantity, you can't simply take the weighted averages of $\gamma$ for each species.

$$\gamma \ne \sum_i y_i \gamma_i$$

4. Nov 11, 2016

### Staff: Mentor

So? If you know the gammas, then you know each of the specific heat values individually.

5. Nov 11, 2016

### colinjohnstoe

How do I get the individual specific heat values from $\gamma$? Knowing $\gamma$ only means I know the ratio of the two.

6. Nov 11, 2016

### Staff: Mentor

The diffence between the two is equal to R

7. Nov 11, 2016

### colinjohnstoe

You are right! Thanks. That should solve the problem.