At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate
(a) final temperature of the gas
(b) change in its internal energy and
(c) the work done by the gas during the process. [ `R=8.31J//mol-K`]

Explanation

To solve the problem step by step, we will calculate the final temperature, change in internal energy, and work done by the gas during the adiabatic expansion.

### Given Data:
- Initial temperature, \( T_1 = 27^\circ C = 300 \, K \)
- Number of moles, \( n = 2 \)
- Initial volume, \( V_1 = V \)
- Final volume, \( V_2 = 2V \)
- Gas constant, \( R = 8.31 \, J/(mol \cdot K) \)
- For a monatomic ideal gas, \( C_V = \frac{3}{2}R \) and \( C_P = \frac{5}{2}R \)
- The ratio of specific heats, \( \gamma = \frac{C_P}{C_V} = \frac{5/2}{3/2} = \frac{5}{3} \)

### (a) Final Temperature of the Gas

Using the adiabatic condition:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

Substituting the known values:
\[ T_1 = 300 \, K, \quad V_1 = V, \quad V_2 = 2V, \quad \gamma = \frac{5}{3} \]

We can rewrite the equation as:
\[ 300 \cdot V^{\frac{5}{3} - 1} = T_2 \cdot (2V)^{\frac{5}{3} - 1} \]

This simplifies to:
\[ 300 \cdot V^{\frac{2}{3}} = T_2 \cdot 2^{\frac{2}{3}} \cdot V^{\frac{2}{3}} \]

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 \( 2^{\frac{2}{3}} \):
\[ 2^{\frac{2}{3}} \approx 1.5874 \]

Thus:
\[ T_2 \approx \frac{300}{1.5874} \approx 189.0 \, K \]

### (b) Change in Internal Energy

The change in internal energy \( \Delta U \) for an ideal gas is given by:
\[ \Delta U = n C_V \Delta T \]

Where:
\[ \Delta T = T_2 - T_1 = 189.0 \, K - 300 \, K = -111.0 \, K \]
\[ C_V = \frac{3}{2} R = \frac{3}{2} \cdot 8.31 \approx 12.465 \, J/(mol \cdot K) \]

Now substituting the values:
\[ \Delta U = 2 \cdot 12.465 \cdot (-111.0) \]
\[ \Delta U \approx 2 \cdot 12.465 \cdot -111.0 \approx -2767.23 \, J \]

### (c) Work Done by the Gas

Using the first law of thermodynamics:
\[ \Delta U = Q + W \]

For an adiabatic process, \( Q = 0 \):
\[ \Delta U = W \implies W = -\Delta U \]

Thus:
\[ W = -(-2767.23) = 2767.23 \, J \]

### Final Answers:
(a) Final Temperature, \( T_2 \approx 189.0 \, K \)
(b) Change in Internal Energy, \( \Delta U \approx -2767.23 \, J \)
(c) Work Done by the Gas, \( W \approx 2767.23 \, J \)