how much work in kJ `mol^(-1)` unit is done by reversible and isothermal expansion of 1.2 mole of an ideal gas to 10 times of its original volume at `27^(@)C`?

Options

A. `4.191`
B. `6.892`
C. `-6.892`
D. `-4.191`

Explanation

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

### Step 1: Understand the formula for work done
The work done (W) during the isothermal expansion of an ideal gas can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Where:
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (8.314 J/mol·K or 0.008314 kJ/mol·K)
- \( T \) = absolute temperature in Kelvin
- \( V_1 \) = initial volume
- \( V_2 \) = final volume

### Step 2: Convert temperature to Kelvin
The given temperature is 27°C. To convert this to Kelvin:
\[ T(K) = 27 + 273.15 = 300.15 \approx 300 \, \text{K} \]

### Step 3: Identify the volumes
The problem states that the gas expands to 10 times its original volume. Let’s denote the initial volume as \( V_1 = V \) and the final volume as:
\[ V_2 = 10V \]

### Step 4: Substitute values into the formula
Now, substituting the values into the work done formula:
- \( n = 1.2 \, \text{moles} \)
- \( R = 0.008314 \, \text{kJ/mol·K} \)
- \( T = 300 \, \text{K} \)
- The ratio \( \frac{V_2}{V_1} = \frac{10V}{V} = 10 \)

Thus, the work done becomes:
\[ W = -1.2 \times 0.008314 \times 300 \times \ln(10) \]

### Step 5: Calculate \( \ln(10) \)
Using the natural logarithm:
\[ \ln(10) \approx 2.303 \]

### Step 6: Substitute and calculate
Now substituting \( \ln(10) \):
\[ W = -1.2 \times 0.008314 \times 300 \times 2.303 \]

### Step 7: Perform the calculation
Calculating step by step:
1. Calculate \( 1.2 \times 0.008314 \):
\[ 1.2 \times 0.008314 \approx 0.0099768 \]
2. Calculate \( 0.0099768 \times 300 \):
\[ 0.0099768 \times 300 \approx 2.99204 \]
3. Finally, calculate \( 2.99204 \times 2.303 \):
\[ 2.99204 \times 2.303 \approx 6.892 \]

Thus, the work done is:
\[ W \approx -6.892 \, \text{kJ/mol} \]

### Final Answer
The work done by the reversible and isothermal expansion of 1.2 moles of an ideal gas to 10 times its original volume at 27°C is approximately:
\[ W \approx -6.892 \, \text{kJ/mol} \]

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