A gaseous mixture enclosed in a vessel of volume `V` consists of one mole of gas `A` with `gamma = (C_(P))/(C_(V)) = (5)/(3)`an another gas `B` with `gamma = (7)/(5)` at a certain temperature `T`. The gram molecular weights of the gases `A` and `B` are `4` and `32` respectively. The gases `A` and `B` do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation `PV^(19//13)` = constant , in adiabatic process. Find the number of moles of the gas `B` in the gaseous mixture.

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the problem, we need to find the number of moles of gas B in a gaseous mixture consisting of one mole of gas A and an unknown number of moles of gas B. The mixture follows an adiabatic process described by the equation \( PV^{\frac{19}{13}} = \text{constant} \).

### Step-by-Step Solution:

1. **Identify Given Data:**
- Moles of gas A, \( N_A = 1 \)
- \( \gamma_A = \frac{C_P}{C_V} = \frac{5}{3} \)
- \( \gamma_B = \frac{7}{5} \)
- Volume of the vessel, \( V \) (not needed explicitly)
- We need to find \( N_B \) (number of moles of gas B).

2. **Understand the Relationship for Adiabatic Processes:**
- For an adiabatic process, we have the relation \( PV^\gamma = \text{constant} \).
- Given that \( PV^{\frac{19}{13}} = \text{constant} \), we can equate \( \gamma_{mixture} = \frac{19}{13} \).

3. **Use the Formula for Mixture of Gases:**
- The formula relating the number of moles and specific heats for the mixture is:
\[ \frac{N_A + N_B}{\gamma_{mixture} - 1} = \frac{N_A}{\gamma_A - 1} + \frac{N_B}{\gamma_B - 1} \]
- Substituting the known values:
\[ \frac{1 + N_B}{\frac{19}{13} - 1} = \frac{1}{\frac{5}{3} - 1} + \frac{N_B}{\frac{7}{5} - 1} \]

4. **Calculate the Denominators:**
- Calculate \( \gamma_{mixture} - 1 = \frac{19}{13} - 1 = \frac{19 - 13}{13} = \frac{6}{13} \)
- Calculate \( \gamma_A - 1 = \frac{5}{3} - 1 = \frac{5 - 3}{3} = \frac{2}{3} \)
- Calculate \( \gamma_B - 1 = \frac{7}{5} - 1 = \frac{7 - 5}{5} = \frac{2}{5} \)

5. **Substitute Values into the Equation:**
\[ \frac{1 + N_B}{\frac{6}{13}} = \frac{1}{\frac{2}{3}} + \frac{N_B}{\frac{2}{5}} \]
- This simplifies to:
\[ \frac{13(1 + N_B)}{6} = \frac{3}{2} + \frac{5N_B}{2} \]

6. **Clear the Fraction:**
- Multiply through by 6 to eliminate the fraction:
\[ 13(1 + N_B) = 9 + 15N_B \]
- Expanding gives:
\[ 13 + 13N_B = 9 + 15N_B \]

7. **Rearranging the Equation:**
- Rearranging terms gives:
\[ 13 - 9 = 15N_B - 13N_B \]
- This simplifies to:
\[ 4 = 2N_B \]
- Therefore:
\[ N_B = 2 \]

### Final Answer:
The number of moles of gas B in the gaseous mixture is \( N_B = 2 \).