To solve the problem step by step, we will analyze the two processes: isobaric and adiabatic expansion of the gas.
### Step 1: Initial Conditions
We have:
- Number of moles, \( n = 2 \) moles
- Initial volume, \( V_1 = 1 \, \text{kL} = 1000 \, \text{L} \)
- Initial temperature, \( T_1 = 27^\circ C = 300 \, \text{K} \)
### Step 2: Isobaric Expansion
In the isobaric process, the gas expands to three times its initial volume:
- Final volume after isobaric expansion, \( V_2 = 3V_1 = 3 \times 1000 \, \text{L} = 3000 \, \text{L} \)
Using the relationship between volume and temperature in an isobaric process:
\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]
We can find \( T_2 \):
\[
T_2 = \frac{V_2}{V_1} \times T_1 = 3 \times 300 \, \text{K} = 900 \, \text{K}
\]
### Step 3: Work Done in Isobaric Process
The work done by the gas during isobaric expansion is given by:
\[
W_1 = nR(T_2 - T_1)
\]
Where \( R = 8.31 \, \text{J/(mol K)} \):
\[
W_1 = 2 \, \text{moles} \times 8.31 \, \text{J/(mol K)} \times (900 \, \text{K} - 300 \, \text{K})
\]
\[
W_1 = 2 \times 8.31 \times 600 = 9972 \, \text{J}
\]
### Step 4: Adiabatic Expansion
In the adiabatic process, we need to find the final temperature \( T_3 \) after expanding back to the initial temperature \( T_1 \) while the volume changes from \( V_2 \) to \( V_3 \).
Using the relation for adiabatic processes:
\[
T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}
\]
Where \( \gamma = \frac{4}{3} \):
\[
T_3 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1}
\]
Substituting the values:
\[
T_3 = 300 \left( \frac{1}{3} \right)^{\frac{4}{3} - 1} = 300 \left( \frac{1}{3} \right)^{\frac{1}{3}} \approx 207.3 \, \text{K}
\]
### Step 5: Work Done in Adiabatic Process
The work done during the adiabatic process is given by:
\[
W_2 = \frac{nR(T_1 - T_3)}{\gamma - 1}
\]
Substituting the values:
\[
W_2 = \frac{2 \times 8.31 \times (300 - 207.3)}{\frac{4}{3} - 1}
\]
Calculating:
\[
W_2 = \frac{2 \times 8.31 \times 92.7}{\frac{1}{3}} = 6 \times 8.31 \times 92.7 \approx 4622 \, \text{J}
\]
### Step 6: Total Work Done
The total work done by the gas in the entire process is:
\[
W_{\text{total}} = W_1 + W_2 = 9972 \, \text{J} + 4622 \, \text{J} = 14594 \, \text{J} \approx 14.6 \, \text{kJ}
\]
### Final Answer
Thus, the total work done by the gas in the whole process is approximately:
\[
\boxed{15 \, \text{kJ}}
\]
