Calculate the maximum work obtained when 0.75 mole of an ideal gas expands isothermally and reversibly at `27^(@)C` From a volume of 15L to 25L.

Answer

To calculate the maximum work obtained when 0.75 moles of an ideal gas expands isothermally and reversibly from a volume of 15 L to 25 L at a temperature of 27°C, we can follow these steps:

### Step 1: Understand the Formula for Work Done
In an isothermal and reversible process, the work done (W) by the gas can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin
- \( V_f \) = final volume
- \( V_i \) = initial volume

### Step 2: Convert Temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin:
\[ T(K) = T(°C) + 273 \]
\[ T = 27 + 273 = 300 \, K \]

### Step 3: Identify Given Values
From the question, we have:
- \( n = 0.75 \, \text{moles} \)
- \( R = 8.314 \, \text{J/mol·K} \)
- \( V_f = 25 \, \text{L} \)
- \( V_i = 15 \, \text{L} \)

### Step 4: Substitute Values into the Formula
Now, we can substitute these values into the work done formula:
\[ W = -0.75 \times 8.314 \times 300 \times \ln\left(\frac{25}{15}\right) \]

### Step 5: Calculate the Natural Logarithm
Calculate the ratio of the volumes:
\[ \frac{V_f}{V_i} = \frac{25}{15} = \frac{5}{3} \]
Now calculate the natural logarithm:
\[ \ln\left(\frac{5}{3}\right) \approx 0.5108 \]

### Step 6: Complete the Calculation
Now plug this value back into the equation:
\[ W = -0.75 \times 8.314 \times 300 \times 0.5108 \]
\[ W \approx -0.75 \times 8.314 \times 300 \times 0.5108 \]
\[ W \approx -0.75 \times 2494.2 \times 0.5108 \]
\[ W \approx -955.7 \, \text{J} \]

### Step 7: Final Result
Thus, the maximum work obtained is:
\[ W \approx -955.7 \, \text{J} \]