One gram of hydrogen and 112 g of nitrogen are enclosed in two separate containers each of volume 5 L and at `27^@C` . If the pressure of the hydrogen is 1 atm, then the pressure of nitrogen would be

Options

A. 16 atm
B. 12 atm
C. 8 atm
D. 4 atm

Answer

To solve the problem, we will use the Ideal Gas Law and the relationship between pressure and the number of moles when volume and temperature are constant.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of hydrogen (H₂) = 1 g
- Molar mass of hydrogen (H₂) = 2 g/mol
- Mass of nitrogen (N₂) = 112 g
- Molar mass of nitrogen (N₂) = 28 g/mol
- Volume of each container = 5 L
- Temperature = 27°C (which is 300 K when converted to Kelvin)
- Pressure of hydrogen (P_H₂) = 1 atm

2. **Calculate the Number of Moles:**
- For hydrogen:
\[ n_{H_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ mol} \]
- For nitrogen:
\[ n_{N_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{112 \text{ g}}{28 \text{ g/mol}} = 4 \text{ mol} \]

3. **Use the Relationship Between Pressures and Moles:**
- Since the volume and temperature are constant, we can use the relation:
\[ \frac{P_{H_2}}{P_{N_2}} = \frac{n_{H_2}}{n_{N_2}} \]
- Plugging in the values:
\[ \frac{1 \text{ atm}}{P_{N_2}} = \frac{0.5 \text{ mol}}{4 \text{ mol}} \]

4. **Cross Multiply to Solve for \( P_{N_2} \):**
\[ P_{N_2} \cdot 0.5 = 1 \cdot 4 \]
\[ P_{N_2} \cdot 0.5 = 4 \]
\[ P_{N_2} = \frac{4}{0.5} = 8 \text{ atm} \]

5. **Conclusion:**
- The pressure of nitrogen (P_N₂) is 8 atm.

### Final Answer:
The pressure of nitrogen would be 8 atm.

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