One mole of an ideal monoatomic gas is initially at 300K. Find the final temperature if 200J of heat are added as follows:
(a) at constant volume (b) at constant pressure.

Explanation

To solve the problem, we will use the formulas for heat transfer in an ideal monoatomic gas at constant volume and constant pressure.

### Given:
- Number of moles (n) = 1 mole
- Initial temperature (T_initial) = 300 K
- Heat added (Q) = 200 J
- For a monoatomic ideal gas:
- \( C_v = \frac{3}{2} R \)
- \( C_p = \frac{5}{2} R \)
- Universal gas constant \( R = 8.31 \, \text{J/(mol K)} \)

### (a) Heat added at constant volume

1. **Formula for heat at constant volume**:
\[ Q = n C_v \Delta T \]
where \( \Delta T = T_{final} - T_{initial} \).

2. **Substituting values**:
\[ 200 = 1 \cdot \left(\frac{3}{2} \cdot R\right) \cdot (T_{final} - 300) \]
\[ 200 = 1 \cdot \left(\frac{3}{2} \cdot 8.31\right) \cdot (T_{final} - 300) \]

3. **Calculate \( C_v \)**:
\[ C_v = \frac{3}{2} \cdot 8.31 = 12.465 \, \text{J/K} \]

4. **Substituting \( C_v \) back into the equation**:
\[ 200 = 12.465 \cdot (T_{final} - 300) \]

5. **Solving for \( T_{final} \)**:
\[ T_{final} - 300 = \frac{200}{12.465} \]
\[ T_{final} - 300 \approx 16.04 \]
\[ T_{final} \approx 316.04 \, \text{K} \]

### (b) Heat added at constant pressure

1. **Formula for heat at constant pressure**:
\[ Q = n C_p \Delta T \]

2. **Substituting values**:
\[ 200 = 1 \cdot \left(\frac{5}{2} \cdot R\right) \cdot (T_{final} - 300) \]
\[ 200 = 1 \cdot \left(\frac{5}{2} \cdot 8.31\right) \cdot (T_{final} - 300) \]

3. **Calculate \( C_p \)**:
\[ C_p = \frac{5}{2} \cdot 8.31 = 20.775 \, \text{J/K} \]

4. **Substituting \( C_p \) back into the equation**:
\[ 200 = 20.775 \cdot (T_{final} - 300) \]

5. **Solving for \( T_{final} \)**:
\[ T_{final} - 300 = \frac{200}{20.775} \]
\[ T_{final} - 300 \approx 9.63 \]
\[ T_{final} \approx 309.63 \, \text{K} \]

### Final Answers:
- (a) Final temperature at constant volume: **316 K**
- (b) Final temperature at constant pressure: **310 K**