At STP, a container has 1 mole of Ar, 2 mole of `CO_(2)`, 3 moles of `O_(2)` and 4 moles of `N_(2)`. Without changing the total pressure if one mole of `O_(2)` is removed, the partial pressure of `O_(2)`

Options

A. is changed by about 26%
B. is halved
C. is unchanged
D. changed by 33%

Explanation

To solve the problem step by step, we need to calculate the partial pressure of oxygen before and after removing one mole of it, while keeping the total pressure constant.

### Step 1: Calculate the initial total number of moles in the container.
- We have:
- Moles of Ar = 1
- Moles of CO₂ = 2
- Moles of O₂ = 3
- Moles of N₂ = 4

**Total moles = 1 + 2 + 3 + 4 = 10 moles**

### Step 2: Calculate the initial partial pressure of O₂.
- The formula for partial pressure of a gas is given by:
\[ P_{A} = \left( \frac{n_{A}}{n_{total}} \right) \times P_{t} \]
where \( n_{A} \) is the number of moles of gas A, \( n_{total} \) is the total number of moles, and \( P_{t} \) is the total pressure.

- For O₂:
- \( n_{O₂} = 3 \)
- \( n_{total} = 10 \)

**Initial partial pressure of O₂:**
\[ P_{O₂, initial} = \left( \frac{3}{10} \right) \times P_{t} \]

### Step 3: Remove one mole of O₂ and calculate the new total number of moles.
- After removing 1 mole of O₂:
- Remaining moles of O₂ = 3 - 1 = 2
- New total moles = 10 - 1 = 9

### Step 4: Calculate the new partial pressure of O₂.
- Using the same formula for partial pressure:
- For O₂ after removal:
- \( n_{O₂} = 2 \)
- \( n_{total} = 9 \)

**New partial pressure of O₂:**
\[ P_{O₂, final} = \left( \frac{2}{9} \right) \times P_{t} \]

### Step 5: Calculate the change in partial pressure of O₂.
- Change in partial pressure:
\[ \Delta P_{O₂} = P_{O₂, final} - P_{O₂, initial} \]
Substituting the values:
\[ \Delta P_{O₂} = \left( \frac{2}{9} P_{t} \right) - \left( \frac{3}{10} P_{t} \right) \]

### Step 6: Find a common denominator and simplify.
- The common denominator for 9 and 10 is 90:
\[ \Delta P_{O₂} = \left( \frac{20}{90} P_{t} \right) - \left( \frac{27}{90} P_{t} \right) = \left( \frac{-7}{90} P_{t} \right) \]

### Step 7: Calculate the percentage change in partial pressure of O₂.
- The formula for percentage change is:
\[ \text{Percentage Change} = \left( \frac{\Delta P_{O₂}}{P_{O₂, initial}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Change} = \left( \frac{-7/90 P_{t}}{3/10 P_{t}} \right) \times 100 \]
Cancelling \( P_{t} \):
\[ \text{Percentage Change} = \left( \frac{-7/90}{3/10} \right) \times 100 = \left( \frac{-7 \times 10}{90 \times 3} \right) \times 100 = \left( \frac{-70}{270} \right) \times 100 \]
Calculating:
\[ \text{Percentage Change} = -25.9\% \]

### Conclusion:
The partial pressure of O₂ decreases by approximately 26%.

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