A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with `gamma=(5/3)` and some amount of gas B with `gamma=7/5` at a temperature T.
The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B if `gamma` for the gaseous mixture is `(19/13)`.

Options

A. 2
B. 12
C. 16
D. 8

Explanation

To solve the problem, we need to find the number of gram moles of gas B in a mixture of gases A and B, given their specific heat ratios (gamma values) and the gamma for the mixture.

### Step-by-Step Solution:

1. **Identify the given values:**
- Gas A: 1 gram mole, \( \gamma_A = \frac{5}{3} \)
- Gas B: \( \mu_B \) gram moles, \( \gamma_B = \frac{7}{5} \)
- Mixture gamma: \( \gamma_{mixture} = \frac{19}{13} \)

2. **Calculate \( C_V \) for gases A and B:**
- For gas A:
\[ C_{V_A} = \frac{R}{\gamma_A - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2} R \]
- For gas B:
\[ C_{V_B} = \frac{R}{\gamma_B - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5}{2} R \]

3. **Calculate \( C_V \) for the mixture:**
- The formula for \( C_V \) of the mixture is:
\[ C_{V_{mixture}} = \frac{\mu_A C_{V_A} + \mu_B C_{V_B}}{\mu_A + \mu_B} \]
- Substituting the values:
\[ C_{V_{mixture}} = \frac{1 \cdot \frac{3}{2} R + \mu_B \cdot \frac{5}{2} R}{1 + \mu_B} \]
- Simplifying:
\[ C_{V_{mixture}} = \frac{\frac{3}{2} R + \frac{5}{2} \mu_B R}{1 + \mu_B} = \frac{R\left(\frac{3}{2} + \frac{5}{2} \mu_B\right)}{1 + \mu_B} \]

4. **Set up the equation using the gamma of the mixture:**
- We know:
\[ C_{V_{mixture}} = \frac{R}{\gamma_{mixture} - 1} = \frac{R}{\frac{19}{13} - 1} = \frac{R}{\frac{6}{13}} = \frac{13}{6} R \]
- Equating the two expressions for \( C_{V_{mixture}} \):
\[ \frac{R\left(\frac{3}{2} + \frac{5}{2} \mu_B\right)}{1 + \mu_B} = \frac{13}{6} R \]
- Cancel \( R \) from both sides:
\[ \frac{3}{2} + \frac{5}{2} \mu_B = \frac{13}{6}(1 + \mu_B) \]

5. **Clear the fraction by multiplying through by 6:**
\[ 6 \left(\frac{3}{2} + \frac{5}{2} \mu_B\right) = 13(1 + \mu_B) \]
\[ 9 + 15 \mu_B = 13 + 13 \mu_B \]

6. **Rearranging the equation:**
\[ 15 \mu_B - 13 \mu_B = 13 - 9 \]
\[ 2 \mu_B = 4 \]
\[ \mu_B = 2 \]

### Final Answer:
The number of gram moles of gas B is \( \mu_B = 2 \) gram moles.

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