At STP a contains has 1 mole of `H_(2)` 2 mole `Ne,` 3 mole `O_(2)` and 4 mole `N_(2)`. Without changing total pressure if 2 mole of `O_(2)` is removed, the partial pressure of `O_(2)` will be decreased by

Options

A. `26 %`
B. `40%`
C. `58.33%`
D. `66.66%`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the total number of moles of gas in the mixture.
The mixture contains:
- 1 mole of H₂
- 2 moles of Ne
- 3 moles of O₂
- 4 moles of N₂

Total moles (n_total) = 1 + 2 + 3 + 4 = 10 moles

### Step 2: Calculate the initial partial pressure of O₂.
The partial pressure of a gas can be calculated using the formula:
\[ P_{O_2} = \chi_{O_2} \times P_{total} \]
where \( \chi_{O_2} \) is the mole fraction of O₂.

Mole fraction of O₂ (χ_O₂):
\[ \chi_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{3}{10} \]

Thus, the initial partial pressure of O₂:
\[ P_{O_2(initial)} = \frac{3}{10} \times P_{total} \]

### Step 3: Determine the new number of moles of O₂ after removing 2 moles.
After removing 2 moles of O₂:
Remaining moles of O₂ = 3 - 2 = 1 mole

### Step 4: Calculate the new total number of moles after removal.
New total number of moles (n_new_total):
\[ n_{new\_total} = n_{total} - 2 = 10 - 2 = 8 \text{ moles} \]

### Step 5: Calculate the new partial pressure of O₂.
New mole fraction of O₂ (χ'_O₂):
\[ \chi'_{O_2} = \frac{n'_{O_2}}{n_{new\_total}} = \frac{1}{8} \]

Thus, the new partial pressure of O₂:
\[ P_{O_2(new)} = \chi'_{O_2} \times P_{total} = \frac{1}{8} \times P_{total} \]

### Step 6: Calculate the decrease in the partial pressure of O₂.
The decrease in partial pressure of O₂ can be calculated as:
\[ \Delta P_{O_2} = P_{O_2(initial)} - P_{O_2(new)} \]

Substituting the values:
\[ \Delta P_{O_2} = \left(\frac{3}{10} P_{total}\right) - \left(\frac{1}{8} P_{total}\right) \]

To perform this subtraction, we need a common denominator:
The least common multiple of 10 and 8 is 40.

Converting both fractions:
\[ P_{O_2(initial)} = \frac{3}{10} P_{total} = \frac{12}{40} P_{total} \]
\[ P_{O_2(new)} = \frac{1}{8} P_{total} = \frac{5}{40} P_{total} \]

Now, substituting back:
\[ \Delta P_{O_2} = \left(\frac{12}{40} P_{total}\right) - \left(\frac{5}{40} P_{total}\right) = \frac{7}{40} P_{total} \]

### Step 7: Calculate the percentage decrease in partial pressure of O₂.
The percentage decrease is given by:
\[ \text{Percentage Decrease} = \left(\frac{\Delta P_{O_2}}{P_{O_2(initial)}}\right) \times 100 \]

Substituting the values:
\[ \text{Percentage Decrease} = \left(\frac{\frac{7}{40} P_{total}}{\frac{12}{40} P_{total}}\right) \times 100 \]
\[ = \left(\frac{7}{12}\right) \times 100 \approx 58.33\% \]

### Final Answer:
The partial pressure of O₂ will decrease by approximately **58.33%**.

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