`5mol` of an ideal gas at `27^(@)C` expands isothermally and reversibly from a volume of `6L` to `60L`. The work done in `kJ` is

Options

A. `-14.7`
B. `-28.72`
C. `+28.72`
D. `-56.72`

Explanation

To solve the problem of calculating the work done by an ideal gas during isothermal and reversible expansion, we can follow these steps:

### Step 1: Understand the formula for work done
The work done (W) during isothermal and reversible expansion of an ideal gas is given by the formula:
\[ W = -2.303 \, nRT \log\left(\frac{V_2}{V_1}\right) \]
where:
- \( n \) = number of moles of the gas,
- \( R \) = universal gas constant (8.314 J/(mol·K)),
- \( T \) = absolute temperature in Kelvin,
- \( V_1 \) = initial volume,
- \( V_2 \) = final volume.

### Step 2: Convert the temperature to Kelvin
Given the temperature is \( 27^\circ C \):
\[ T = 27 + 273 = 300 \, K \]

### Step 3: Identify the values
From the problem statement:
- \( n = 5 \, \text{mol} \)
- \( R = 8.314 \, \text{J/(mol·K)} \)
- \( V_1 = 6 \, L \)
- \( V_2 = 60 \, L \)

### Step 4: Substitute values into the formula
Now, we can substitute the known values into the work done formula:
\[ W = -2.303 \times 5 \, \text{mol} \times 8.314 \, \text{J/(mol·K)} \times 300 \, K \times \log\left(\frac{60}{6}\right) \]

### Step 5: Calculate the logarithm
Calculate the logarithm:
\[ \log\left(\frac{60}{6}\right) = \log(10) = 1 \]

### Step 6: Substitute and calculate
Now substitute the logarithm back into the equation:
\[ W = -2.303 \times 5 \times 8.314 \times 300 \times 1 \]
\[ W = -2.303 \times 5 \times 8.314 \times 300 \]

Calculating this step-by-step:
1. Calculate \( 5 \times 8.314 = 41.57 \)
2. Calculate \( 41.57 \times 300 = 12471 \)
3. Finally, calculate \( -2.303 \times 12471 \approx -28720.713 \, J \)

### Step 7: Convert Joules to Kilojoules
To convert Joules to Kilojoules:
\[ W = \frac{-28720.713 \, J}{1000} \approx -28.72 \, kJ \]

### Final Answer
The work done during the isothermal and reversible expansion is approximately:
\[ W \approx -28.72 \, kJ \]