`1` mol of an idel gas is allowed to expand isothermally at `27^(@)C` till its volume is tripped. If the expansion is carried out reversibly then select the correct option(s) : `(R = (25)/(3)(J)/("molK"),log3=0.48)`

Options

A. `q _(sys) = 2763 J//mol`
B. `DeltaS_(sys.) = 7.21 J//K` mol
C. `DeltaS_(universe) = 0`
D. `DeltaS_(surr.) = -9.21 J//K` mol

Explanation

To solve the problem step by step, we will calculate the various thermodynamic quantities involved in the isothermal expansion of the ideal gas.

### Step 1: Understand the Given Data
- Number of moles (n) = 1 mol
- Initial temperature (T) = 27°C = 300 K (since K = °C + 273)
- Volume is tripled, so if the initial volume is V, the final volume is 3V.
- Given R = 25/3 J/(mol·K)
- Given log(3) = 0.48

### Step 2: Calculate the Change in Entropy of the System (ΔS_system)
The formula for the change in entropy for an isothermal process is:
\[ \Delta S_{system} = nR \ln\left(\frac{V_{final}}{V_{initial}}\right) \]
Substituting the values:
\[ \Delta S_{system} = 1 \cdot \frac{25}{3} \cdot 2.303 \cdot \log(3) \]
Calculating:
\[ \Delta S_{system} = 1 \cdot \frac{25}{3} \cdot 2.303 \cdot 0.48 \]
\[ \Delta S_{system} = \frac{25 \cdot 2.303 \cdot 0.48}{3} \]
\[ \Delta S_{system} \approx 9.21 \, \text{J/(K·mol)} \]

### Step 3: Calculate the Change in Entropy of the Surroundings (ΔS_surrounding)
For a reversible process, the total change in entropy of the universe is zero:
\[ \Delta S_{universe} = \Delta S_{system} + \Delta S_{surrounding} = 0 \]
Thus,
\[ \Delta S_{surrounding} = -\Delta S_{system} = -9.21 \, \text{J/(K·mol)} \]

### Step 4: Calculate the Heat Transfer (Q_system)
For an isothermal process, the heat transfer can be calculated using:
\[ Q_{system} = T \cdot \Delta S_{system} \]
Substituting the values:
\[ Q_{system} = 300 \cdot 9.21 \]
Calculating:
\[ Q_{system} = 2763 \, \text{J/mol} \]

### Step 5: Summary of Results
- \(\Delta S_{system} \approx 9.21 \, \text{J/(K·mol)}\)
- \(\Delta S_{surrounding} \approx -9.21 \, \text{J/(K·mol)}\)
- \(Q_{system} \approx 2763 \, \text{J/mol}\)
- \(\Delta S_{universe} = 0\)

### Conclusion
The correct options based on the calculations are:
- Option A: \(Q_{system} = 2763 \, \text{J/mol}\)
- Option C: \(\Delta S_{system} = 9.21 \, \text{J/(K·mol)}\)
- Option D: \(\Delta S_{surrounding} = -9.21 \, \text{J/(K·mol)}\)