Equal masses of `CH_(4)` and `H_(2)` are mixed in an empty chamber. The partial pressure of hydrogen in this chamber expressed as fraction of total pressure is :

Options

A. `1//2`
B. `8//19`
C. `1//9`
D. `8/9`

Explanation

To solve the problem of finding the partial pressure of hydrogen (H₂) in a chamber where equal masses of methane (CH₄) and hydrogen (H₂) are mixed, we can follow these steps:

### Step 1: Define the mass of the gases
Let the equal mass of CH₄ and H₂ be denoted as \( W \).

### Step 2: Calculate the number of moles of each gas
- The molar mass of hydrogen (H₂) is 2 g/mol.
- The molar mass of methane (CH₄) is 16 g/mol.

Using the formula for moles:
\[ \text{Moles of H₂} = \frac{W}{2} \]
\[ \text{Moles of CH₄} = \frac{W}{16} \]

### Step 3: Calculate the total number of moles
The total number of moles in the chamber is the sum of the moles of H₂ and CH₄:
\[ \text{Total moles} = \text{Moles of H₂} + \text{Moles of CH₄} = \frac{W}{2} + \frac{W}{16} \]

To add these fractions, we need a common denominator:
\[ \frac{W}{2} = \frac{8W}{16} \]
So,
\[ \text{Total moles} = \frac{8W}{16} + \frac{W}{16} = \frac{9W}{16} \]

### Step 4: Calculate the mole fraction of hydrogen
The mole fraction of hydrogen (Xₕ) is given by:
\[ X_{H₂} = \frac{\text{Moles of H₂}}{\text{Total moles}} = \frac{\frac{W}{2}}{\frac{9W}{16}} \]
This simplifies to:
\[ X_{H₂} = \frac{W/2}{9W/16} = \frac{W \cdot 16}{2 \cdot 9W} = \frac{16}{18} = \frac{8}{9} \]

### Step 5: Relate mole fraction to partial pressure
The partial pressure of hydrogen (Pₕ) can be expressed as:
\[ P_{H₂} = X_{H₂} \cdot P_{total} \]
Thus, the fraction of the total pressure that is due to hydrogen is:
\[ \frac{P_{H₂}}{P_{total}} = X_{H₂} = \frac{8}{9} \]

### Final Answer
The partial pressure of hydrogen in the chamber expressed as a fraction of the total pressure is:
\[ \frac{8}{9} \]