if quantity of heat 1163.4 J supplied to one mole of nitrogen gas , at room temperture at constant pressure , then the rise in temperature is

Options

A. 54 K
B. 28 K
C. 65 K
D. 40 K

Explanation

To solve the problem of finding the rise in temperature when 1163.4 J of heat is supplied to one mole of nitrogen gas at constant pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Quantity of heat (Q) = 1163.4 J
- Number of moles (n) = 1 mole
- For nitrogen gas at constant pressure, we need to use the specific heat capacity at constant pressure (C_P).

2. **Determine the Value of C_P:**
- For a diatomic gas like nitrogen (N₂), the molar specific heat capacity at constant pressure (C_P) can be calculated using the formula:
\[ C_P = \frac{7}{2} R \]
- Where R is the universal gas constant, approximately \( R = 8.31 \, \text{J/(mol·K)} \).

3. **Calculate C_P:**
\[ C_P = \frac{7}{2} \times 8.31 = 29.1 \, \text{J/(mol·K)} \]

4. **Use the Formula for Heat Transfer:**
- The relationship between heat supplied, the number of moles, and the change in temperature is given by:
\[ Q = n \cdot C_P \cdot \Delta T \]
- Rearranging this formula to find the change in temperature (\(\Delta T\)):
\[ \Delta T = \frac{Q}{n \cdot C_P} \]

5. **Substituting the Values:**
\[ \Delta T = \frac{1163.4 \, \text{J}}{1 \cdot 29.1 \, \text{J/(mol·K)}} \]

6. **Calculate \(\Delta T\):**
\[ \Delta T = \frac{1163.4}{29.1} \approx 40.02 \, \text{K} \]

7. **Conclusion:**
- The rise in temperature of the nitrogen gas is approximately \( 40 \, \text{K} \).

### Final Answer:
The rise in temperature is approximately \( 40 \, \text{K} \).

---