Two moles of an ideal gas is expanded irreversibly and isothermally at `37^(@)C` until its volume is doubled and `3.41 KJ` heat is absorbed from surrounding. `DeltaS_("total")("system +surrounding")` is:

Options

A. `-0.52J//K`
B. `0.52J//K`
C. `22.52J//K`
D. 0

Explanation

To solve the problem of calculating the total change in entropy (\( \Delta S_{\text{total}} \)) for the system and surroundings during the isothermal expansion of an ideal gas, we will follow these steps:

### Step 1: Identify the given values
- Number of moles (\( n \)) = 2 moles
- Initial temperature (\( T \)) = 37°C = 310 K (converted from Celsius to Kelvin)
- Heat absorbed (\( Q \)) = 3.41 kJ = 3410 J (converted from kJ to J)
- Initial volume (\( V_1 \)) = \( V \)
- Final volume (\( V_2 \)) = \( 2V \)

### Step 2: Calculate the change in entropy of the surroundings (\( \Delta S_{\text{surr}} \))
The change in entropy of the surroundings is given by the formula:
\[ \Delta S_{\text{surr}} = -\frac{Q}{T} \]
Substituting the values:
\[ \Delta S_{\text{surr}} = -\frac{3410 \, \text{J}}{310 \, \text{K}} = -11 \, \text{J/K} \]

### Step 3: Calculate the change in entropy of the system (\( \Delta S_{\text{sys}} \))
For an isothermal process, the change in entropy of the system can be calculated using:
\[ \Delta S_{\text{sys}} = nR \ln\left(\frac{V_2}{V_1}\right) \]
Where \( R \) (the universal gas constant) = 8.314 J/(mol·K). Since \( V_2 = 2V \) and \( V_1 = V \):
\[ \Delta S_{\text{sys}} = 2 \times 8.314 \, \text{J/(mol·K)} \times \ln\left(\frac{2V}{V}\right) = 2 \times 8.314 \, \text{J/(mol·K)} \times \ln(2) \]
Calculating \( \ln(2) \):
\[ \ln(2) \approx 0.693 \]
Now substituting this value:
\[ \Delta S_{\text{sys}} = 2 \times 8.314 \times 0.693 \approx 11.526 \, \text{J/K} \]

### Step 4: Calculate the total change in entropy (\( \Delta S_{\text{total}} \))
The total change in entropy is the sum of the changes in entropy of the system and surroundings:
\[ \Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \]
Substituting the values we calculated:
\[ \Delta S_{\text{total}} = 11.526 \, \text{J/K} - 11 \, \text{J/K} = 0.526 \, \text{J/K} \]

### Final Answer
\[ \Delta S_{\text{total}} \approx 0.526 \, \text{J/K} \]