1 litre of `N_(2) " and " 7//8` litre of `O_(2)` at the same temperature and pressure were mixed together. What is the relation between the masses of the two gases in the mixture ?

Options

A. `M(N_2)=3M(O_2)`
B. `M(N_2)=M(O_2)`
C. `M(N_2)=M(O_2)`
D. `M(N_2)=16M(O_2)`

Explanation

To solve the problem of finding the relation between the masses of nitrogen (N₂) and oxygen (O₂) when 1 liter of N₂ is mixed with 7/8 liter of O₂ at the same temperature and pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of nitrogen (N₂) = 1 liter
- Volume of oxygen (O₂) = 7/8 liter
- Both gases are at the same temperature and pressure.

2. **Use the Ideal Gas Law:**
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

3. **Relate Moles to Mass:**
The number of moles (\( n \)) can be expressed in terms of mass (\( m \)) and molar mass (\( M \)):
\[ n = \frac{m}{M} \]
Thus, we can express the mass of each gas in terms of its volume and molar mass.

4. **Calculate Molar Masses:**
- Molar mass of nitrogen (N₂) = 28 g/mol (since N has a molar mass of 14 g/mol)
- Molar mass of oxygen (O₂) = 32 g/mol (since O has a molar mass of 16 g/mol)

5. **Calculate Moles of Each Gas:**
For nitrogen:
\[ n_{N_2} = \frac{m_{N_2}}{28} \]
For oxygen:
\[ n_{O_2} = \frac{m_{O_2}}{32} \]

6. **Set Up the Equation Using the Ideal Gas Law:**
Since both gases are at the same temperature and pressure, we can set up the following relationship:
\[ \frac{P \cdot 1}{RT} = \frac{m_{N_2}}{28} \quad \text{(for N₂)} \]
\[ \frac{P \cdot \frac{7}{8}}{RT} = \frac{m_{O_2}}{32} \quad \text{(for O₂)} \]

7. **Eliminate Common Terms:**
Since \( P \), \( R \), and \( T \) are constant, we can eliminate them from both equations:
\[ 1 = \frac{m_{N_2}}{28} \quad \text{(1)} \]
\[ \frac{7}{8} = \frac{m_{O_2}}{32} \quad \text{(2)} \]

8. **Rearrange the Equations:**
From equation (1):
\[ m_{N_2} = 28 \]
From equation (2):
\[ m_{O_2} = \frac{7}{8} \times 32 = 28 \]

9. **Establish the Relation Between Masses:**
Now we can see that:
\[ m_{N_2} = m_{O_2} = 28 \text{ grams} \]

10. **Conclusion:**
The relation between the masses of nitrogen and oxygen is:
\[ m_{N_2} = m_{O_2} \]

### Final Answer:
The mass of nitrogen is equal to the mass of oxygen.

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