One mole of an ideal gas is expanded isothermally from 1 `dm^(3)` to 10 `dm^(3)` at 300 K. `DeltaG` will be equal to

Options

A. 5.744 kJ
B. 57.44 J
C. 574.4 J
D. `-5744J`

Answer

To solve the problem of calculating the change in Gibbs free energy (\( \Delta G \)) for the isothermal expansion of one mole of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (\( n \)) = 1 mole
- Initial volume (\( V_1 \)) = 1 dm³
- Final volume (\( V_2 \)) = 10 dm³
- Temperature (\( T \)) = 300 K
- Universal gas constant (\( R \)) = 8.314 J/(mol·K)

2. **Use the Gibbs Free Energy Change Formula for Isothermal Process:**
The change in Gibbs free energy for an isothermal process can be calculated using the formula:
\[ \Delta G = \Delta G^\circ + nRT \ln\left(\frac{V_2}{V_1}\right) \]
Since we are considering the change for an ideal gas expanding isothermally, we can assume \( \Delta G^\circ = 0 \) for this calculation, so:
\[ \Delta G = nRT \ln\left(\frac{V_2}{V_1}\right) \]

3. **Calculate the Volume Ratio:**
\[ \frac{V_2}{V_1} = \frac{10 \, \text{dm}^3}{1 \, \text{dm}^3} = 10 \]

4. **Substitute the Values into the Formula:**
\[ \Delta G = 1 \, \text{mol} \times 8.314 \, \text{J/(mol·K)} \times 300 \, \text{K} \times \ln(10) \]

5. **Calculate \( \ln(10) \):**
The natural logarithm of 10 is approximately:
\[ \ln(10) \approx 2.302 \]

6. **Calculate \( \Delta G \):**
\[ \Delta G = 1 \times 8.314 \times 300 \times 2.302 \]
\[ \Delta G \approx 5744.5 \, \text{J} \]
Since the process is an expansion, \( \Delta G \) will be negative:
\[ \Delta G \approx -5744.5 \, \text{J} \]

7. **Final Result:**
Therefore, the change in Gibbs free energy (\( \Delta G \)) for the isothermal expansion is:
\[ \Delta G \approx -5744 \, \text{J} \]

### Final Answer:
\[ \Delta G = -5744 \, \text{J} \]