What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of 10 L to 100 L at 300 K ?

Options

A. `"19.14 J K"^(-1)`
B. `"109.12 J K"^(-1)`
C. `"29.12 J K"^(-1)`
D. `"10 J K"^(-1)`

Explanation

To find the entropy change when 1 mole of oxygen gas expands isothermally and reversibly from an initial volume of 10 L to 100 L at a temperature of 300 K, we can use the formula for entropy change (ΔS) for an ideal gas during an isothermal process:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 1 mole
- Initial volume (V1) = 10 L
- Final volume (V2) = 100 L
- Temperature (T) = 300 K
- Gas constant (R) = 8.314 J/(mol·K)

2. **Use the Entropy Change Formula:**
The formula for the change in entropy (ΔS) for an isothermal expansion of an ideal gas is given by:
\[ \Delta S = nR \ln\left(\frac{V2}{V1}\right) \]
Since we are using logarithm base 10, we will convert it using:
\[ \Delta S = 2.303 \cdot nR \log\left(\frac{V2}{V1}\right) \]

3. **Calculate the Volume Ratio:**
\[ \frac{V2}{V1} = \frac{100 \, \text{L}}{10 \, \text{L}} = 10 \]

4. **Calculate the Logarithm:**
\[ \log\left(10\right) = 1 \]

5. **Plug in the Values:**
Now substituting the values into the entropy change formula:
\[ \Delta S = 2.303 \cdot (1 \, \text{mol}) \cdot (8.314 \, \text{J/(mol·K)}) \cdot 1 \]

6. **Perform the Calculation:**
\[ \Delta S = 2.303 \cdot 8.314 \approx 19.14 \, \text{J/K} \]

7. **Final Result:**
Therefore, the entropy change when 1 mole of oxygen gas expands isothermally and reversibly from 10 L to 100 L at 300 K is approximately:
\[ \Delta S \approx 19.14 \, \text{J/K} \]