In the isothermal reversible compression of `52.0m` mol of a perfect gas at `260 K`, the volume of the gas is reduced to one`-`third of its initial value. Calculate `w` of this process.

Options

A. 0
B. `+123` J
C. `-123` J
D. `+246` J

Explanation

To solve the problem of calculating the work done (W) during the isothermal reversible compression of a perfect gas, we can follow these steps:

### Step 1: Understand the Given Data
- Number of moles (n) = 52.0 mmol = 52.0 x 10^-3 mol = 0.052 mol
- Temperature (T) = 260 K
- The volume of the gas is reduced to one-third of its initial value, meaning V2 = (1/3)V1.

### Step 2: Identify the Formula for Work Done
For an isothermal reversible process, the work done (W) can be calculated using the formula:
\[ W = -nRT \ln \left( \frac{V_2}{V_1} \right) \]
Since V2 = (1/3)V1, we can substitute this into the equation:
\[ W = -nRT \ln \left( \frac{1}{3} \right) \]

### Step 3: Substitute the Known Values
Now, we substitute the values of n, R, and T into the equation:
- R (gas constant) = 8.314 J/(mol·K)
- n = 0.052 mol
- T = 260 K

So, we have:
\[ W = - (0.052) (8.314) (260) \ln \left( \frac{1}{3} \right) \]

### Step 4: Calculate the Natural Logarithm
Calculate \( \ln \left( \frac{1}{3} \right) \):
\[ \ln \left( \frac{1}{3} \right) \approx -1.0986 \]

### Step 5: Substitute the Logarithm Value
Now substitute this value back into the equation:
\[ W = - (0.052) (8.314) (260) (-1.0986) \]

### Step 6: Perform the Calculation
Now calculate the work done:
1. Calculate \( 0.052 \times 8.314 \):
\[ 0.052 \times 8.314 \approx 0.432 \]
2. Now multiply by 260:
\[ 0.432 \times 260 \approx 112.32 \]
3. Finally, multiply by -1.0986:
\[ 112.32 \times 1.0986 \approx 123.0 \]

### Step 7: Final Result
Thus, the work done (W) during the isothermal reversible compression is approximately:
\[ W \approx 123 \text{ Joules} \]

### Conclusion
The work done during the process is \( +123 \text{ Joules} \).