Five moles of an ideal monoatomic gas with an initial temperature of `127^@C` expand and in the process absorb 1200J of heat and do 2100J of work. What is the final temperature of the gas?

Explanation

To find the final temperature of the gas, we will use the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W).

### Step-by-Step Solution:

1. **Convert Initial Temperature to Kelvin**:
The initial temperature \( T_1 \) is given as \( 127^\circ C \). To convert this to Kelvin:
\[ T_1 = 127 + 273 = 400 \, K \]

2. **Identify Given Values**:
- Number of moles, \( n = 5 \, \text{moles} \)
- Heat absorbed, \( Q = 1200 \, J \)
- Work done, \( W = 2100 \, J \)

3. **Apply the First Law of Thermodynamics**:
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Substituting the values:
\[ \Delta U = 1200 \, J - 2100 \, J = -900 \, J \]

4. **Relate Change in Internal Energy to Temperature Change**:
For an ideal monoatomic gas, the change in internal energy is given by:
\[ \Delta U = n C_v \Delta T \]
Where \( C_v \) for a monoatomic gas is \( \frac{3}{2} R \). Thus:
\[ \Delta U = n \left(\frac{3}{2} R\right) (T_f - T_1) \]

5. **Substituting Known Values**:
We know \( R = 8.31 \, J/(mol \cdot K) \). Now substituting the values:
\[ -900 \, J = 5 \left(\frac{3}{2} \times 8.31 \, J/(mol \cdot K)\right) (T_f - 400 \, K) \]

6. **Calculate \( C_v \)**:
First, calculate \( n C_v \):
\[ n C_v = 5 \times \frac{3}{2} \times 8.31 = 5 \times 12.465 = 62.325 \, J/K \]

7. **Substituting \( n C_v \) back into the equation**:
\[ -900 = 62.325 (T_f - 400) \]

8. **Solving for \( T_f \)**:
Rearranging the equation:
\[ T_f - 400 = \frac{-900}{62.325} \]
\[ T_f - 400 = -14.43 \]
\[ T_f = 400 - 14.43 = 385.57 \, K \]

9. **Final Answer**:
Rounding to the appropriate significant figures, the final temperature \( T_f \) is approximately:
\[ T_f \approx 385 \, K \]