One mole of air `(C_(V) = 5R//2)` is confined at atmospheric pressure in a cylinder with a piston at `0^(@)C`. The initial volume occupied by gas is`V`. After the equivalent of `13200 J` of heat is transferred to it, the volume of gas `V` is nearly `(1 atm = 10^(5) N//m^(3))`

Options

A. `37 L`
B. `22 L`
C. `60 L`
D. `30 L`

Explanation

To solve the problem step by step, we will use the concepts of thermodynamics, specifically the ideal gas law and the relationship between heat, temperature, and volume changes in gases.

### Step 1: Identify the given values
- Number of moles of air, \( n = 1 \) mole
- Heat added, \( \Delta Q = 13200 \) J
- Specific heat at constant volume, \( C_V = \frac{5R}{2} \)
- Initial temperature, \( T_i = 0^\circ C = 273 \, \text{K} \)
- Initial pressure, \( P = 1 \, \text{atm} = 10^5 \, \text{N/m}^2 \)

### Step 2: Calculate \( C_P \) using the relation \( C_P - C_V = R \)
Using Meyer’s equation:
\[ C_P = C_V + R = \frac{5R}{2} + R = \frac{5R}{2} + \frac{2R}{2} = \frac{7R}{2} \]

### Step 3: Calculate the initial volume using the ideal gas law
The ideal gas law is given by:
\[ PV = nRT \]
Substituting the known values:
\[ V_i = \frac{nRT_i}{P} = \frac{1 \times R \times 273}{10^5} \]
Using \( R = 8.3 \, \text{J/(mol K)} \):
\[ V_i = \frac{1 \times 8.3 \times 273}{10^5} \approx 0.0227 \, \text{m}^3 = 22.7 \, \text{liters} \]

### Step 4: Relate heat added to change in volume
At constant pressure, the heat added can be expressed as:
\[ \Delta Q = n C_P \Delta T \]
Rearranging gives:
\[ \Delta T = \frac{\Delta Q}{n C_P} \]
Substituting \( C_P = \frac{7R}{2} \):
\[ \Delta T = \frac{\Delta Q}{n \cdot \frac{7R}{2}} = \frac{2 \Delta Q}{7R} \]

### Step 5: Substitute \( \Delta Q \) and \( R \) to find \( \Delta T \)
\[ \Delta T = \frac{2 \times 13200}{7 \times 8.3} \approx \frac{26400}{58.1} \approx 454.3 \, \text{K} \]

### Step 6: Calculate the final temperature
\[ T_f = T_i + \Delta T = 273 + 454.3 \approx 727.3 \, \text{K} \]

### Step 7: Calculate the final volume using the ideal gas law again
Using the ideal gas law for the final state:
\[ V_f = \frac{nRT_f}{P} = \frac{1 \times 8.3 \times 727.3}{10^5} \]
Calculating:
\[ V_f \approx \frac{6048.19}{10^5} \approx 0.0604 \, \text{m}^3 = 60.4 \, \text{liters} \]

### Final Answer
The final volume of the gas is approximately **60.4 liters**.