Five moles of an ideal gas undergoes reversible expansion from 1 L to 4 L at `27^oC`. The work done and enthalpy change for the process respectively are

Options

A. `- 12.5 kJ and - 18.2 kJ`
B. `- 17.3 kJ and 0 kJ`
C. `+ 17.3 kJ and +22.5 kJ`
D. `+ 21.5 kJ and 0 kJ`

Answer

To solve the problem, we need to calculate the work done and the enthalpy change for the reversible isothermal expansion of an ideal gas. Here are the steps to find the solution:

### Step 1: Understand the Given Data
- Number of moles (n) = 5 moles
- Initial volume (V1) = 1 L
- Final volume (V2) = 4 L
- Temperature (T) = 27°C = 27 + 273 = 300 K
- Gas constant (R) = 8.314 J/(mol·K)

### Step 2: Calculate the Work Done (W)
The formula for work done during a reversible isothermal expansion of an ideal gas is given by:

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

Substituting the values into the equation:

\[ W = -5 \times 8.314 \times 300 \times \ln\left(\frac{4}{1}\right) \]

Calculating \(\ln(4)\):

\[ \ln(4) = 1.386 \]

Now substituting this value into the work done equation:

\[ W = -5 \times 8.314 \times 300 \times 1.386 \]

Calculating the work done:

\[ W = -5 \times 8.314 \times 300 \times 1.386 = -1732.32 \text{ J} \]

Converting to kilojoules:

\[ W = -1.73232 \text{ kJ} \approx -1.73 \text{ kJ} \]

### Step 3: Calculate the Enthalpy Change (ΔH)
For an isothermal process involving an ideal gas, the change in enthalpy (ΔH) is given by:

\[ \Delta H = 0 \]

### Final Answer
- Work done (W) = -1.73 kJ
- Enthalpy change (ΔH) = 0 kJ

### Summary
The work done during the reversible expansion is approximately -1.73 kJ, and the enthalpy change is 0 kJ.