A mixture of one mole of each of `O_2(g), H_2(g)`, He(g) exists in a container of volume V at temperatureT in which partial pressure of `H_2` (g) is 2atm. the total pressure in the container is:

Options

A. 6atm
B. 18atm
C. 33atm
D. 24atm

Answer

To solve the problem, we need to find the total pressure in a container that holds a mixture of gases: \(O_2\), \(H_2\), and \(He\). We know that the partial pressure of \(H_2\) is given as 2 atm. Let's break down the solution step by step.

### Step 1: Identify the number of moles of each gas
We have:
- 1 mole of \(O_2\)
- 1 mole of \(H_2\)
- 1 mole of \(He\)

Total moles of gas = 1 + 1 + 1 = 3 moles.

### Step 2: Calculate the mole fraction of \(H_2\)
The mole fraction (\(x\)) of a gas is calculated using the formula:
\[ x_{H_2} = \frac{\text{moles of } H_2}{\text{total moles of gas}} \]
Substituting the values:
\[ x_{H_2} = \frac{1}{3} \]

### Step 3: Use the relationship between partial pressure and total pressure
The partial pressure of a gas can be expressed in terms of its mole fraction and the total pressure (\(P_t\)):
\[ P_{H_2} = x_{H_2} \cdot P_t \]
We can rearrange this to find the total pressure:
\[ P_t = \frac{P_{H_2}}{x_{H_2}} \]
Substituting the known values:
\[ P_t = \frac{2 \, \text{atm}}{\frac{1}{3}} = 2 \, \text{atm} \times 3 = 6 \, \text{atm} \]

### Conclusion
The total pressure in the container is \(6 \, \text{atm}\).