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 energy, and (iii) the work done by the gas during this process.

Explanation

To solve the problem step by step, we will follow the instructions provided in the question.

### Given Data:
- Initial temperature, \( T_1 = 27^\circ C = 300 \, K \)
- Number of moles, \( n = 2 \, \text{moles} \)
- Initial volume, \( V_1 = V \)
- Final volume, \( V_2 = 2V \)
- For a monoatomic ideal gas, \( C_v = \frac{3}{2} R \)

### Step 1: Calculate the final temperature of the gas \( T_2 \)

For an adiabatic process, we use the relation:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
where \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \) for a monoatomic gas.

Substituting the known values:
\[ T_1 V_1^{\frac{5}{3} - 1} = T_2 V_2^{\frac{5}{3} - 1} \]
\[ 300 \cdot V^{\frac{2}{3}} = T_2 \cdot (2V)^{\frac{2}{3}} \]
\[ 300 \cdot V^{\frac{2}{3}} = T_2 \cdot 2^{\frac{2}{3}} V^{\frac{2}{3}} \]
Dividing both sides by \( V^{\frac{2}{3}} \):
\[ 300 = T_2 \cdot 2^{\frac{2}{3}} \]
Now, solving for \( T_2 \):
\[ T_2 = \frac{300}{2^{\frac{2}{3}}} \]
Calculating \( 2^{\frac{2}{3}} \approx 1.5874 \):
\[ T_2 \approx \frac{300}{1.5874} \approx 189.0 \, K \]

### Step 2: Calculate the change in internal energy \( \Delta U \)

The change in internal energy for an ideal gas is given by:
\[ \Delta U = n C_v \Delta T \]
Where \( \Delta T = T_2 - T_1 \):
\[ \Delta T = 189 - 300 = -111 \, K \]
Now substituting the values:
\[ \Delta U = n \cdot \frac{3}{2} R \cdot \Delta T \]
Substituting \( n = 2 \) and \( R = 8.314 \, J/(mol \cdot K) \):
\[ \Delta U = 2 \cdot \frac{3}{2} \cdot 8.314 \cdot (-111) \]
Calculating:
\[ \Delta U = 3 \cdot 8.314 \cdot (-111) \approx -2767 \, J \]

### Step 3: Calculate the work done by the gas \( W \)

For an adiabatic process, the first law of thermodynamics states:
\[ Q = \Delta U + W \]
Since the process is adiabatic, \( Q = 0 \):
\[ 0 = \Delta U + W \implies W = -\Delta U \]
Substituting the value of \( \Delta U \):
\[ W = -(-2767) = 2767 \, J \]

### Final Answers:
1. Final temperature \( T_2 \approx 189.0 \, K \)
2. Change in internal energy \( \Delta U \approx -2767 \, J \)
3. Work done by the gas \( W \approx 2767 \, J \)