If 1 mole of an ideal gas expands isothermally at `37^(@)C` from 15 litres to 25 litres, the maximum work obtained is `:`

Options

A. `1316.8J`
B. `6.43J`
C. `8.57J`
D. `2.92J`

Explanation

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

### Step 1: Convert Temperature to Kelvin
The given temperature is \(37^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
\[ T = 37 + 273.15 = 310.15 \, K \approx 310 \, K \]

### Step 2: Identify Given Values
From the problem, we have:
- Number of moles, \(n = 1 \, \text{mol}\)
- Initial volume, \(V_1 = 15 \, \text{L}\)
- Final volume, \(V_2 = 25 \, \text{L}\)
- Gas constant, \(R = 8.314 \, \text{J/(mol K)}\)

### Step 3: Use the Formula for Work Done in Isothermal Expansion
The work done by an ideal gas during isothermal expansion can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Since we need the maximum work, we will use the logarithmic form:
\[ W = -2.303 \, nRT \log\left(\frac{V_2}{V_1}\right) \]

### Step 4: Substitute the Values into the Formula
Now we can substitute the known values into the equation:
\[ W = -2.303 \times 1 \times 8.314 \times 310 \times \log\left(\frac{25}{15}\right) \]

### Step 5: Calculate the Logarithm
First, calculate the ratio of the volumes:
\[ \frac{V_2}{V_1} = \frac{25}{15} = \frac{5}{3} \approx 1.6667 \]
Now calculate the logarithm:
\[ \log\left(\frac{5}{3}\right) \approx 0.2218 \]

### Step 6: Substitute the Logarithm Value Back into the Equation
Now we can substitute this value back into the work formula:
\[ W = -2.303 \times 1 \times 8.314 \times 310 \times 0.2218 \]

### Step 7: Perform the Calculation
Calculating the work done:
\[ W \approx -2.303 \times 8.314 \times 310 \times 0.2218 \]
\[ W \approx -1316.8 \, \text{J} \]

### Step 8: Final Result
Thus, the maximum work obtained is approximately:
\[ W \approx -1316.8 \, \text{J} \]