At `27^@C` two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas, (ii) change in its internal enegy, and (iii) the work done by the gas during this process.

Explanation

To solve the problem step by step, we will follow the three parts of the question: finding the final temperature, change in internal energy, and the work done by the gas during the adiabatic expansion.

### Step 1: Calculate the Final Temperature of the Gas

1. **Identify Given Values:**
- Initial temperature \( T_1 = 27^\circ C = 273 + 27 = 300 \, K \)
- Initial volume \( V_1 = V \)
- Final volume \( V_2 = 2V \)
- Number of moles \( N = 2 \)
- For a monoatomic ideal gas, \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \)

2. **Use the Adiabatic Relation:**
The adiabatic condition for an ideal gas can be expressed as:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Substituting the known values:
\[ T_1 V^{\frac{5}{3} - 1} = T_2 (2V)^{\frac{5}{3} - 1} \]
Simplifying:
\[ T_1 V^{\frac{2}{3}} = T_2 (2^{\frac{2}{3}} V^{\frac{2}{3}}) \]
Dividing both sides by \( V^{\frac{2}{3}} \):
\[ T_1 = T_2 \cdot 2^{-\frac{2}{3}} \]
Rearranging gives:
\[ T_2 = T_1 \cdot 2^{\frac{2}{3}} \]

3. **Substituting the Values:**
\[ T_2 = 300 \cdot 2^{-\frac{2}{3}} = 300 \cdot \frac{1}{2^{\frac{2}{3}}} \]
Calculating \( 2^{\frac{2}{3}} \approx 1.5874 \):
\[ T_2 \approx \frac{300}{1.5874} \approx 189 \, K \]

### Step 2: Calculate Change in Internal Energy

1. **Use the Formula for Change in Internal Energy:**
\[ \Delta U = n C_v \Delta T \]
For a monoatomic gas, \( C_v = \frac{3R}{2} \).

2. **Calculate \( \Delta T \):**
\[ \Delta T = T_2 - T_1 = 189 - 300 = -111 \, K \]

3. **Substituting Values:**
\[ \Delta U = 2 \cdot \frac{3R}{2} \cdot (-111) \]
Using \( R = 8.314 \, J/(mol \cdot K) \):
\[ \Delta U = 2 \cdot \frac{3 \cdot 8.314}{2} \cdot (-111) = 3 \cdot 8.314 \cdot (-111) \]
Calculating:
\[ \Delta U \approx 3 \cdot 8.314 \cdot (-111) \approx -2767 \, J \]

### Step 3: Calculate the Work Done by the Gas

1. **Use the First Law of Thermodynamics:**
\[ Q = \Delta U + W \]
In an adiabatic process, \( Q = 0 \):
\[ 0 = \Delta U + W \Rightarrow W = -\Delta U \]

2. **Substituting the Value of \( \Delta U \):**
\[ W = -(-2767) = 2767 \, J \]

### Final Answers
1. Final Temperature \( T_2 \approx 189 \, K \)
2. Change in Internal Energy \( \Delta U \approx -2767 \, J \)
3. Work Done by the Gas \( W \approx 2767 \, J \)