In a flexible balloon 2, moles of `SO_2` having an initial volume of 1 kL at a temperature of `27.^@C` is filled . The gas is first expanded to thrice its initial volume isobarically and then further expanded adiabatically so as to attain its initial temperature . Assuming the gas to be ideal , the work done by the gas in the whole process is `[gammaSO_2=4/3,R=25/3J mol ^(-1) K^(-1)]`

Options

A. `10 kJ`
B. `35 kJ`
C. `45 kJ`
D. `15 kJ`

Answer

To solve the problem step by step, we need to calculate the work done by the gas during two processes: isobaric expansion and adiabatic expansion.

### Step 1: Initial Conditions
- Given:
- Number of moles (n) = 2 moles
- Initial Volume (V1) = 1 kL = 1000 L
- Final Volume after isobaric expansion (V2) = 3 * V1 = 3 kL = 3000 L
- Initial Temperature (T1) = 27°C = 300 K (after conversion)
- \( R = \frac{25}{3} \, \text{J/mol·K} \)
- \( \gamma = \frac{4}{3} \)

### Step 2: Work Done in Isobaric Process
In an isobaric process, the work done (W) can be calculated using the formula:
\[ W = P \Delta V \]
Where \( \Delta V = V2 - V1 \).

To find the pressure (P), we can use the ideal gas law:
\[ PV = nRT \]
From this, we can express pressure as:
\[ P = \frac{nRT}{V1} \]

Substituting the values:
\[ P = \frac{2 \cdot \frac{25}{3} \cdot 300}{1000} \]
\[ P = \frac{2 \cdot \frac{25}{3} \cdot 300}{1000} = \frac{2 \cdot 25 \cdot 300}{3 \cdot 1000} = \frac{15000}{3000} = 5 \, \text{J/L} \]

Now, we can calculate the work done:
\[ \Delta V = V2 - V1 = 3000 L - 1000 L = 2000 L \]
\[ W = P \Delta V = 5 \cdot 2000 = 10000 \, \text{J} \]

### Step 3: Work Done in Adiabatic Process
For the adiabatic process, we can use the formula for work done:
\[ W = \frac{nR}{\gamma - 1} (T1 - T2) \]

We need to find T2 after the adiabatic expansion. Using the relation for adiabatic processes:
\[ \frac{T1}{T2} = \left(\frac{V2}{V1}\right)^{\gamma - 1} \]
Substituting the values:
\[ \frac{T1}{T2} = \left(\frac{3}{1}\right)^{\frac{4}{3} - 1} = 3^{\frac{1}{3}} \]
Thus,
\[ T2 = \frac{T1}{3^{\frac{1}{3}}} = \frac{300}{3^{\frac{1}{3}}} \]

Calculating \( 3^{\frac{1}{3}} \) (approximately 1.442):
\[ T2 \approx \frac{300}{1.442} \approx 207.3 \, K \]

Now substituting into the work done formula:
\[ W = \frac{2 \cdot \frac{25}{3}}{\frac{4}{3} - 1} (300 - 207.3) \]
\[ W = \frac{2 \cdot \frac{25}{3}}{\frac{1}{3}} (92.7) \]
\[ W = 2 \cdot 25 \cdot 92.7 = 4650 \, J \]

### Step 4: Total Work Done
Now we can find the total work done in both processes:
\[ W_{total} = W_{isobaric} + W_{adiabatic} \]
\[ W_{total} = 10000 \, J + 4650 \, J = 14650 \, J \]

### Final Answer
The total work done by the gas in the whole process is approximately:
\[ W_{total} \approx 14650 \, J \]