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I Specific heat ratio of gas mixture

  1. Nov 10, 2016 #1
    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. jcsd
  3. Nov 10, 2016 #2
    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.
     
  4. Nov 11, 2016 #3
    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.

    $$\gamma \ne \sum_i y_i \gamma_i$$
     
  5. Nov 11, 2016 #4
    So? If you know the gammas, then you know each of the specific heat values individually.
     
  6. Nov 11, 2016 #5
    How do I get the individual specific heat values from ##\gamma##? Knowing ##\gamma## only means I know the ratio of the two.
     
  7. Nov 11, 2016 #6
    The diffence between the two is equal to R
     
  8. Nov 11, 2016 #7
    You are right! Thanks. That should solve the problem.
     
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