7 moles of a tetra-atomic non-linear gas 'A' at 10 atm and T K are mixed with 6 moles of another gas B at `(T)/(3)K` and 5 atm in a closed, rigid vessel without energy transfer with surroundings. If final temperature of mixture was `(5T)/(6) K,` then gas B is ? (Assuming all modes of energy are active)

Options

A. monoatomic
B. diatomic
C. triatomic
D. tetra atomic

Explanation

To solve the problem step by step, we will follow the principles of thermodynamics and the specific heat capacities of gases.

### Step 1: Identify the given data
- Moles of gas A, \( n_A = 7 \)
- Pressure of gas A, \( P_A = 10 \) atm
- Temperature of gas A, \( T_A = T \) K
- Moles of gas B, \( n_B = 6 \)
- Pressure of gas B, \( P_B = 5 \) atm
- Temperature of gas B, \( T_B = \frac{T}{3} \) K
- Final temperature of the mixture, \( T_f = \frac{5T}{6} \) K

### Step 2: Determine the heat lost by gas A
The heat lost by gas A can be expressed using the formula:
\[ Q_A = n_A C_{vA} (T_A - T_f) \]
Where \( C_{vA} \) is the molar heat capacity at constant volume for gas A.

### Step 3: Calculate \( C_{vA} \) for gas A
Gas A is a tetra-atomic non-linear gas. For such a gas, the degrees of freedom are:
- Translational: 3
- Rotational: 3
- Vibrational: 6 (since it is tetra-atomic)

Thus, the molar heat capacity at constant volume \( C_{vA} \) is given by:
\[ C_{vA} = \frac{f}{2} R = \frac{(3 + 3 + 6)}{2} R = 9R \]

### Step 4: Calculate the heat gained by gas B
The heat gained by gas B can be expressed as:
\[ Q_B = n_B C_{vB} (T_f - T_B) \]
Where \( C_{vB} \) is the molar heat capacity at constant volume for gas B.

### Step 5: Set the heat lost by gas A equal to the heat gained by gas B
Since there is no heat transfer with the surroundings, we have:
\[ Q_A = Q_B \]
Substituting the expressions for \( Q_A \) and \( Q_B \):
\[ n_A C_{vA} (T_A - T_f) = n_B C_{vB} (T_f - T_B) \]
Substituting the known values:
\[ 7 \times 9R \left( T - \frac{5T}{6} \right) = 6 C_{vB} \left( \frac{5T}{6} - \frac{T}{3} \right) \]

### Step 6: Simplify the equation
Calculating \( T - \frac{5T}{6} = \frac{T}{6} \) and \( \frac{5T}{6} - \frac{T}{3} = \frac{5T}{6} - \frac{2T}{6} = \frac{3T}{6} = \frac{T}{2} \):
\[ 7 \times 9R \times \frac{T}{6} = 6 C_{vB} \times \frac{T}{2} \]
Cancelling \( T \) from both sides (assuming \( T \neq 0 \)):
\[ 63R = 6 C_{vB} \times 1 \]
\[ C_{vB} = \frac{63R}{6} = \frac{21R}{2} \]

### Step 7: Identify the type of gas B
The value of \( C_{vB} = \frac{21R}{2} \) corresponds to a diatomic gas, which has a \( C_v \) value of \( \frac{5R}{2} \) for translational and rotational modes, plus additional vibrational modes.

### Conclusion
Thus, gas B is a **diatomic gas**.

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