Two diatomic gases are mixed in mole ratio 1:2. Find the value of adiabatic exponent for this mixture of gases.

Answer

To find the adiabatic exponent (γ) for the mixture of two diatomic gases in a mole ratio of 1:2, follow these steps:

### Step 1: Understand the Adiabatic Exponent
The adiabatic exponent (γ) is defined as the ratio of the molar specific heat at constant pressure (Cp) to the molar specific heat at constant volume (Cv):
\[ \gamma = \frac{C_p}{C_v} \]

### Step 2: Determine the Specific Heats for Diatomic Gases
For diatomic gases, the degrees of freedom (F) is 5. The formulas for Cp and Cv are:
\[ C_p = \frac{F}{2}R + R = \frac{5}{2}R + R = \frac{7}{2}R \]
\[ C_v = \frac{F}{2}R = \frac{5}{2}R \]

### Step 3: Set Up the Mixture Formula
For a mixture of gases, the effective Cp and Cv can be calculated using the mole fractions of the gases. Let:
- n1 = moles of gas 1
- n2 = moles of gas 2

Given the mole ratio of gas 1 to gas 2 is 1:2, we can set:
\[ n_1 = 1 \quad \text{and} \quad n_2 = 2 \]

### Step 4: Calculate the Mixture's Cp and Cv
Using the mole fractions:
\[ \text{Mole fraction of gas 1} = \frac{n_1}{n_1 + n_2} = \frac{1}{1 + 2} = \frac{1}{3} \]
\[ \text{Mole fraction of gas 2} = \frac{n_2}{n_1 + n_2} = \frac{2}{1 + 2} = \frac{2}{3} \]

Now, we can calculate the effective Cp and Cv for the mixture:
\[ C_{p,\text{mixture}} = \left(\frac{n_1}{n_1 + n_2} C_{p1}\right) + \left(\frac{n_2}{n_1 + n_2} C_{p2}\right) \]
\[ C_{v,\text{mixture}} = \left(\frac{n_1}{n_1 + n_2} C_{v1}\right) + \left(\frac{n_2}{n_1 + n_2} C_{v2}\right) \]

Substituting the values:
\[ C_{p,\text{mixture}} = \left(\frac{1}{3} \cdot \frac{7}{2}R\right) + \left(\frac{2}{3} \cdot \frac{7}{2}R\right) = \frac{7}{6}R + \frac{14}{6}R = \frac{21}{6}R = \frac{7}{2}R \]
\[ C_{v,\text{mixture}} = \left(\frac{1}{3} \cdot \frac{5}{2}R\right) + \left(\frac{2}{3} \cdot \frac{5}{2}R\right) = \frac{5}{6}R + \frac{10}{6}R = \frac{15}{6}R = \frac{5}{2}R \]

### Step 5: Calculate the Adiabatic Exponent for the Mixture
Now, substituting the values of Cp and Cv into the formula for γ:
\[ \gamma_{\text{mixture}} = \frac{C_{p,\text{mixture}}}{C_{v,\text{mixture}}} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} = 1.4 \]

### Final Answer
The value of the adiabatic exponent for the mixture of the two diatomic gases is:
\[ \gamma = 1.4 \]

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