Find the ratio of number of moles of a monoatomic and a diatomic gas whose mixture has a value of adiabatic exponent `gammaa`=3/2.

Answer

To find the ratio of the number of moles of a monoatomic gas (n1) and a diatomic gas (n2) in a mixture with an adiabatic exponent (γ) of 3/2, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Values of γ for Each Gas:**
- For a monoatomic gas, the adiabatic exponent (γ1) is given by:
\[ \gamma_1 = \frac{5}{3} \]
- For a diatomic gas, the adiabatic exponent (γ2) is given by:
\[ \gamma_2 = \frac{7}{5} \]

2. **Write the Expression for the Mixture's Adiabatic Exponent:**
- The adiabatic exponent for the mixture (γ) can be expressed as:
\[ \frac{n_1}{n_1 + n_2} \cdot \gamma_1 + \frac{n_2}{n_1 + n_2} \cdot \gamma_2 = \gamma \]
- Substituting the known values:
\[ \frac{n_1}{n_1 + n_2} \cdot \frac{5}{3} + \frac{n_2}{n_1 + n_2} \cdot \frac{7}{5} = \frac{3}{2} \]

3. **Multiply Through by (n1 + n2) to Eliminate the Denominator:**
- This gives:
\[ n_1 \cdot \frac{5}{3} + n_2 \cdot \frac{7}{5} = \frac{3}{2} (n_1 + n_2) \]

4. **Rearrange the Equation:**
- Expanding the right side:
\[ n_1 \cdot \frac{5}{3} + n_2 \cdot \frac{7}{5} = \frac{3}{2} n_1 + \frac{3}{2} n_2 \]
- Rearranging gives:
\[ n_1 \cdot \frac{5}{3} - \frac{3}{2} n_1 + n_2 \cdot \frac{7}{5} - \frac{3}{2} n_2 = 0 \]

5. **Combine Like Terms:**
- Factor out n1 and n2:
\[ n_1 \left(\frac{5}{3} - \frac{3}{2}\right) + n_2 \left(\frac{7}{5} - \frac{3}{2}\right) = 0 \]

6. **Calculate the Coefficients:**
- For n1:
\[ \frac{5}{3} - \frac{3}{2} = \frac{10}{6} - \frac{9}{6} = \frac{1}{6} \]
- For n2:
\[ \frac{7}{5} - \frac{3}{2} = \frac{14}{10} - \frac{15}{10} = -\frac{1}{10} \]

7. **Set Up the Ratio:**
- The equation becomes:
\[ n_1 \cdot \frac{1}{6} = n_2 \cdot \frac{1}{10} \]
- Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{1/10}{1/6} = \frac{6}{10} = \frac{3}{5} \]

8. **Final Ratio:**
- Thus, the ratio of the number of moles of monoatomic gas to diatomic gas is:
\[ \frac{n_1}{n_2} = \frac{3}{5} \]