Specific heat ratio of gas mixture

AI Thread Summary
The discussion centers on calculating the specific heat ratio, ##\gamma##, for a gas mixture in multi-fluid hydrodynamic modeling. The user presents a formula derived from the thermal energy density equations, expressing ##\gamma## in terms of the number densities and specific heat ratios of individual species. A clarification is made that the specific heat ratio cannot be simply averaged from the individual species' ##\gamma_i## values. Instead, the correct approach involves using the heat capacities at constant pressure and volume, which are weighted by mole fractions. The conversation concludes with an acknowledgment that understanding the relationship between the specific heat values and the ideal gas constant R helps solve the user's problem.
colinjohnstoe
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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?
 
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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.
 
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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.

$$\gamma \ne \sum_i y_i \gamma_i$$
 
colinjohnstoe said:
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.

$$\gamma \ne \sum_i y_i \gamma_i$$
So? If you know the gammas, then you know each of the specific heat values individually.
 
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Chestermiller said:
So? If you know the gammas, then you know each of the specific heat values individually.

How do I get the individual specific heat values from ##\gamma##? Knowing ##\gamma## only means I know the ratio of the two.
 
colinjohnstoe said:
How do I get the individual specific heat values from ##\gamma##? Knowing ##\gamma## only means I know the ratio of the two.
The diffence between the two is equal to R
 
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Chestermiller said:
The diffence between the two is equal to R

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