If 200mL of He at 0.66 atm and 400 mL of `O_(2)` at 0.52 atm pressure are raised in 400 mL vessel at `20^(@)C` then find the partial pressures of He and `O_(2)` ?

Options

A. 0.33 and 0.55
B. 0.33 and 0.52
C. 0.38 and 0.52
D. 0.25 and 0.45

Answer

To solve the problem of finding the partial pressures of helium (He) and oxygen (O₂) when they are combined in a 400 mL vessel, we can use Boyle's Law, which states that for a given amount of gas at constant temperature, the product of pressure and volume is constant (P₁V₁ = P₂V₂).

### Step-by-Step Solution:

**Step 1: Calculate the partial pressure of Helium (He)**

- Given:
- Initial pressure of He (P₁) = 0.66 atm
- Initial volume of He (V₁) = 200 mL
- Final volume of the combined gas (V₂) = 400 mL

Using Boyle's Law:
\[ P₁V₁ = P₂V₂ \]

Substituting the known values:
\[ 0.66 \, \text{atm} \times 200 \, \text{mL} = P₂ \times 400 \, \text{mL} \]

Now, calculate:
\[ 0.66 \times 200 = P₂ \times 400 \]
\[ 132 = P₂ \times 400 \]

Now, solve for P₂:
\[ P₂ = \frac{132}{400} \]
\[ P₂ = 0.33 \, \text{atm} \]

**Step 2: Calculate the partial pressure of Oxygen (O₂)**

- Given:
- Initial pressure of O₂ (P₁) = 0.52 atm
- Initial volume of O₂ (V₁) = 400 mL
- Final volume of the combined gas (V₂) = 400 mL

Again, using Boyle's Law:
\[ P₁V₁ = P₂V₂ \]

Substituting the known values:
\[ 0.52 \, \text{atm} \times 400 \, \text{mL} = P₂ \times 400 \, \text{mL} \]

Now, since V₁ = V₂ for O₂:
\[ 0.52 \times 400 = P₂ \times 400 \]

Dividing both sides by 400:
\[ P₂ = 0.52 \, \text{atm} \]

### Final Result:
- The partial pressure of He = 0.33 atm
- The partial pressure of O₂ = 0.52 atm