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. 12.87 L-atm
B. 6.43 L-atm
C. 8.57 L-atm
D. 2.92 L-atm

Answer

To solve the problem of calculating the maximum work done by 1 mole of an ideal gas during isothermal expansion from 15 liters to 25 liters at a temperature of 37°C, we can follow these steps:

### Step 1: Convert Temperature to Kelvin
The temperature given is in Celsius. To convert it to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
For this problem:
\[ T = 37 + 273 = 310 \, K \]

**Hint**: Always convert Celsius to Kelvin when using gas laws and thermodynamic equations.

### Step 2: Identify the Variables
We need to identify the variables for the work done formula:
- \( n = 1 \) mole (number of moles)
- \( R = 8.314 \, J/(mol \cdot K) \) (universal gas constant)
- \( V_1 = 15 \, L \) (initial volume)
- \( V_2 = 25 \, L \) (final volume)

**Hint**: Make sure to note down all the values you will need for calculations.

### Step 3: Use the Work Done Formula for Isothermal Expansion
The work done \( W \) during isothermal expansion is given by the formula:
\[ W = -2.303 \, nRT \log \left( \frac{V_2}{V_1} \right) \]
Substituting the known values:
\[ W = -2.303 \times 1 \times 8.314 \times 310 \log \left( \frac{25}{15} \right) \]

**Hint**: Pay attention to the logarithm base; here, it is base 10 (logarithm with base 10).

### Step 4: Calculate the Logarithm
First, calculate the ratio of volumes:
\[ \frac{V_2}{V_1} = \frac{25}{15} = 1.6667 \]
Now calculate the logarithm:
\[ \log(1.6667) \approx 0.2224 \]

**Hint**: Use a calculator for logarithmic calculations to ensure accuracy.

### Step 5: Substitute and Calculate Work Done
Now substitute the logarithm back into the work done formula:
\[ W = -2.303 \times 1 \times 8.314 \times 310 \times 0.2224 \]
Calculating this gives:
\[ W \approx -1316.8 \, J \]

**Hint**: Keep track of the negative sign, which indicates work done by the system.

### Step 6: Convert Work Done to Liter-Atmospheric
To express the work done in liter-atmospheric, we convert joules to liter-atmospheric using the conversion factor \( 1 \, L \cdot atm = 101.325 \, J \):
\[ W_{L \cdot atm} = \frac{-1316.8 \, J}{101.325 \, J/(L \cdot atm)} \approx -12.99 \, L \cdot atm \]

**Hint**: Remember that we are interested in the magnitude of work done, so we can ignore the negative sign.

### Final Answer
The maximum work obtained is approximately:
\[ \boxed{12.99 \, L \cdot atm} \]