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 of finding the work done and enthalpy change for the reversible expansion of an ideal gas, we can follow these steps:

### Step 1: Identify the parameters
We have:
- Number of moles (n) = 5 moles
- Initial volume (V1) = 1 L
- Final volume (V2) = 4 L
- Temperature (T) = 27°C = 300 K (convert to Kelvin by adding 273)

### Step 2: Calculate the work done (W)
For an isothermal reversible expansion of an ideal gas, the work done can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Where:
- R = 8.314 J/(mol·K) (universal gas constant)

Substituting the values:
- \( n = 5 \)
- \( R = 8.314 \, \text{J/(mol·K)} \)
- \( T = 300 \, \text{K} \)
- \( V_1 = 1 \, \text{L} \)
- \( V_2 = 4 \, \text{L} \)

Calculating \( \frac{V_2}{V_1} \):
\[ \frac{V_2}{V_1} = \frac{4}{1} = 4 \]

Now substituting into the work formula:
\[ W = -5 \times 8.314 \times 300 \times \ln(4) \]

### Step 3: Calculate \( \ln(4) \)
Using the property of logarithms:
\[ \ln(4) = 2 \ln(2) \]
And \( \ln(2) \approx 0.693 \), thus:
\[ \ln(4) \approx 2 \times 0.693 = 1.386 \]

### Step 4: Substitute \( \ln(4) \) back into the work formula
Now substituting \( \ln(4) \):
\[ W = -5 \times 8.314 \times 300 \times 1.386 \]

### Step 5: Calculate the work done
Calculating the values:
\[ W = -5 \times 8.314 \times 300 \times 1.386 \]
\[ W = -5 \times 2494.2 \approx -12471 \, \text{J} \]
Converting to kilojoules:
\[ W \approx -12.471 \, \text{kJ} \]

### Step 6: 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
Thus, the work done and enthalpy change for the process are:
- Work done (W) = -12.471 kJ
- Enthalpy change (ΔH) = 0 kJ