What is the increase in entropy when `11.2 L of O_(2)` are mixed with `11.2 L of H_(2)` at STP?

Options

A. `0.576 J/K`
B. `5.76 J/K`
C. `7.56 J/K`
D. `2.76 J/K`

Answer

To find the increase in entropy when 11.2 L of O₂ is mixed with 11.2 L of H₂ at standard temperature and pressure (STP), we can follow these steps:

### Step 1: Determine the number of moles of each gas
At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 L. Therefore, we can calculate the number of moles of O₂ and H₂.

- For O₂:
\[ \text{Moles of } O_2 = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{mol} \]

- For H₂:
\[ \text{Moles of } H_2 = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{mol} \]

### Step 2: Calculate the total number of moles after mixing
When we mix the two gases, the total number of moles is:
\[ \text{Total moles} = 0.5 \, \text{mol (O}_2\text{)} + 0.5 \, \text{mol (H}_2\text{)} = 1 \, \text{mol} \]

### Step 3: Calculate the initial and final volumes
- Initial volume (before mixing) = Volume of O₂ + Volume of H₂ = 11.2 L + 11.2 L = 22.4 L
- Final volume (after mixing) = 22.4 L (since the gases mix and occupy the same volume)

### Step 4: Use the entropy change formula for mixing gases
The change in entropy (ΔS) for mixing two ideal gases can be calculated using the formula:
\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]
Where:
- \( n \) = total number of moles = 1 mol
- \( R \) = universal gas constant = 8.314 J/(mol·K)
- \( V_f \) = final volume = 22.4 L
- \( V_i \) = initial volume = 11.2 L

### Step 5: Substitute the values into the formula
\[ \Delta S = 1 \, \text{mol} \times 8.314 \, \text{J/(mol·K)} \times \ln\left(\frac{22.4 \, \text{L}}{11.2 \, \text{L}}\right) \]

### Step 6: Calculate the natural logarithm
\[ \frac{22.4}{11.2} = 2 \quad \Rightarrow \quad \ln(2) \approx 0.693 \]

### Step 7: Calculate the change in entropy
\[ \Delta S = 8.314 \, \text{J/(mol·K)} \times 0.693 \approx 5.76 \, \text{J/K} \]

### Final Answer
The increase in entropy when 11.2 L of O₂ is mixed with 11.2 L of H₂ at STP is approximately **5.76 J/K**.

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