6mole of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litre at 27° C .The maximum work is done

Options

A. 47kJ
B. 100kJ
C. 0
D. 34.46kJ

Answer

To find the maximum work done during the isothermal and reversible expansion of an ideal gas, we can use the formula:

\[ W = -2.303 \, nRT \, \log \left( \frac{V_2}{V_1} \right) \]

### Step 1: Identify the given values
- Number of moles, \( n = 6 \, \text{moles} \)
- Initial volume, \( V_1 = 1 \, \text{L} \)
- Final volume, \( V_2 = 10 \, \text{L} \)
- Temperature, \( T = 27^\circ \text{C} \)

### Step 2: Convert temperature to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 3: Calculate the gas constant \( R \)
The value of the gas constant \( R \) is:
\[ R = 8.314 \, \text{J/(mol·K)} \]

### Step 4: Substitute values into the work formula
Now we can substitute the values into the work formula:
\[ W = -2.303 \times n \times R \times T \times \log \left( \frac{V_2}{V_1} \right) \]
Substituting the known values:
\[ W = -2.303 \times 6 \times 8.314 \times 300 \times \log \left( \frac{10}{1} \right) \]

### Step 5: Calculate the logarithm
Since \( \log(10) = 1 \):
\[ W = -2.303 \times 6 \times 8.314 \times 300 \times 1 \]

### Step 6: Perform the multiplication
Calculating the expression:
\[ W = -2.303 \times 6 \times 8.314 \times 300 \]
Calculating step by step:
1. \( 2.303 \times 6 = 13.818 \)
2. \( 13.818 \times 8.314 = 114.536 \)
3. \( 114.536 \times 300 = 34360.8 \)

Thus,
\[ W = -34360.8 \, \text{J} \]

### Step 7: Convert to kilojoules
To convert joules to kilojoules, divide by 1000:
\[ W = -34.36 \, \text{kJ} \]

### Final Answer
The maximum work done is:
\[ \boxed{34.36 \, \text{kJ}} \]