`4 g` hydrogen is mixed with 11.2 litre of He at (STP) in a container of volume 20 litre. If the final temperature is `300 K`, find the pressure.

Explanation

To find the pressure of the gas mixture in the container, we can follow these steps:

### Step 1: Calculate the number of moles of Hydrogen (H₂)
The molar mass of hydrogen (H₂) is 2 g/mol. Given that we have 4 g of hydrogen:

\[ \text{Number of moles of H₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ moles} \]

### Step 2: Calculate the number of moles of Helium (He)
At STP (Standard Temperature and Pressure), 22.4 liters of any gas corresponds to 1 mole. Given that we have 11.2 liters of helium:

\[ \text{Number of moles of He} = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles} \]

### Step 3: Calculate the total number of moles in the container
Now, we can find the total number of moles (n) in the container:

\[ n = n_{H2} + n_{He} = 2 \text{ moles} + 0.5 \text{ moles} = 2.5 \text{ moles} \]

### Step 4: Use the Ideal Gas Law to find the pressure
The Ideal Gas Law is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in Pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles
- \( R \) = universal gas constant = 8.314 J/(mol·K)
- \( T \) = temperature (in Kelvin)

Given:
- \( V = 20 \text{ L} = 20 \times 10^{-3} \text{ m}^3 \)
- \( T = 300 \text{ K} \)

Rearranging the Ideal Gas Law to solve for pressure \( P \):

\[ P = \frac{nRT}{V} \]

Substituting the values:

\[ P = \frac{(2.5 \text{ moles}) \times (8.314 \text{ J/(mol·K)}) \times (300 \text{ K})}{20 \times 10^{-3} \text{ m}^3} \]

### Step 5: Calculate the pressure
Calculating the numerator:

\[ 2.5 \times 8.314 \times 300 = 6235.5 \text{ J} \]

Now, substituting into the pressure equation:

\[ P = \frac{6235.5}{20 \times 10^{-3}} = \frac{6235.5}{0.02} = 311775 \text{ Pa} = 3.12 \times 10^5 \text{ Pa} \]

### Final Answer
The pressure in the container is:

\[ P = 3.12 \times 10^5 \text{ Pa} \]

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