14 g of `N_(2)` and 36 g of ozone are at the same pressure and temperature. Their volumes will be related as

Options

A. `2V_(N_(2))=3V_(O_(3))`
B. `3V_(N_(2))=2V_(O_(3))`
C. `3V_(N_(2))=4V_(O_(3))`
D. `4V_(N_(2))=3V_(O_(3))`

Explanation

To find the relationship between the volumes of 14 g of nitrogen gas (N₂) and 36 g of ozone (O₃) at the same pressure and temperature, we can use the ideal gas law and the concept of moles. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of N₂
The molecular weight of nitrogen (N₂) is 28 g/mol. To find the number of moles (n) of N₂, we use the formula:

\[ n = \frac{\text{mass}}{\text{molecular weight}} \]

For N₂:

\[ n_{N_2} = \frac{14 \, \text{g}}{28 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Calculate the number of moles of O₃
The molecular weight of ozone (O₃) is 48 g/mol. Similarly, we calculate the number of moles of O₃:

\[ n_{O_3} = \frac{36 \, \text{g}}{48 \, \text{g/mol}} = 0.75 \, \text{mol} \]

### Step 3: Relate the volumes using the ideal gas law
According to the ideal gas law, at constant temperature and pressure, the volume (V) of a gas is directly proportional to the number of moles (n):

\[ V \propto n \]

Thus, the volume ratio of N₂ to O₃ can be expressed as:

\[ \frac{V_{N_2}}{V_{O_3}} = \frac{n_{N_2}}{n_{O_3}} \]

### Step 4: Substitute the values of moles
Now substituting the values we calculated:

\[ \frac{V_{N_2}}{V_{O_3}} = \frac{0.5 \, \text{mol}}{0.75 \, \text{mol}} = \frac{2}{3} \]

### Step 5: Conclusion
Thus, the volumes of N₂ and O₃ are related as:

\[ V_{N_2} : V_{O_3} = 2 : 3 \]

### Final Answer
The volumes of 14 g of N₂ and 36 g of O₃ at the same pressure and temperature will be in the ratio of 2:3.

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