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)`.

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 of the mixture.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- For gas A:
- Number of moles (n1) = 1 mole
- Gamma (γ_A) = 5/3
- For gas B:
- Gamma (γ_B) = 7/5
- For the mixture:
- Gamma (γ_mixture) = 19/13

2. **Calculate Cv for Each Gas:**
- Using the formula \( C_v = \frac{R}{\gamma - 1} \):
- For gas A:
\[ C_{vA} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2} \]
- For gas B:
\[ C_{vB} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2} \]

3. **Calculate Cv for the Mixture:**
- For the mixture:
\[ C_{v\_mixture} = \frac{R}{\frac{19}{13} - 1} = \frac{R}{\frac{6}{13}} = \frac{13R}{6} \]

4. **Apply the Energy Conservation Equation:**
- The change in internal energy of the mixture is equal to the sum of the internal energies of gases A and B:
\[ (n_1 + n_2) C_{v\_mixture} = n_1 C_{vA} + n_2 C_{vB} \]
- Substitute the known values:
\[ (1 + n_2) \frac{13R}{6} = 1 \cdot \frac{3R}{2} + n_2 \cdot \frac{5R}{2} \]

5. **Simplify the Equation:**
- Cancel R from both sides:
\[ (1 + n_2) \frac{13}{6} = \frac{3}{2} + n_2 \cdot \frac{5}{2} \]
- Multiply through by 6 to eliminate the fraction:
\[ 13(1 + n_2) = 9 + 15n_2 \]
- Distributing gives:
\[ 13 + 13n_2 = 9 + 15n_2 \]

6. **Rearranging the Equation:**
- Move all terms involving \( n_2 \) to one side:
\[ 13 - 9 = 15n_2 - 13n_2 \]
- This simplifies to:
\[ 4 = 2n_2 \]

7. **Solve for \( n_2 \):**
\[ n_2 = \frac{4}{2} = 2 \]

### Final Answer:
The number of gram moles of gas B is **2 moles**.