The ratio of densities of nitrogen and oxygen is 14 : 16. The temperature at which the speed of sound in nitrogen will be same as that in oxygen at `55^@C` is

Options

A. `35^@C`
B. `48^@C`
C. `65^@C`
D. `14^@C`

Answer

To solve the problem, we need to find the temperature at which the speed of sound in nitrogen will be the same as that in oxygen at 55°C. We know the ratio of the densities of nitrogen and oxygen is given as 14:16.

### Step-by-Step Solution:

1. **Understand the relationship between density and molecular mass**:
The ratio of densities of two gases is equal to the ratio of their molecular masses. Therefore, we can express this as:
\[ \frac{\text{Density of Nitrogen}}{\text{Density of Oxygen}} = \frac{\text{Molecular Mass of Nitrogen}}{\text{Molecular Mass of Oxygen}} = \frac{14}{16} \]

2. **Use the formula for the speed of sound in a gas**:
The speed of sound \( v \) in a gas is given by the formula:
\[ v = \sqrt{\frac{\gamma R T}{M}} \]
where \( \gamma \) is the adiabatic index, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molecular mass of the gas.

3. **Set up the equation for both gases**:
Let \( T_{N2} \) be the temperature of nitrogen and \( T_{O2} \) be the temperature of oxygen (which is 55°C or 328 K). We can write the equations for the speed of sound in nitrogen and oxygen:
\[ v_{N2} = \sqrt{\frac{\gamma R T_{N2}}{M_{N2}}} \]
\[ v_{O2} = \sqrt{\frac{\gamma R T_{O2}}{M_{O2}}} \]

4. **Set the speeds equal to each other**:
Since we want the speeds to be equal:
\[ \sqrt{\frac{\gamma R T_{N2}}{M_{N2}}} = \sqrt{\frac{\gamma R T_{O2}}{M_{O2}}} \]
Squaring both sides and canceling out common terms gives:
\[ \frac{T_{N2}}{M_{N2}} = \frac{T_{O2}}{M_{O2}} \]

5. **Substituting known values**:
Rearranging the equation gives:
\[ T_{N2} = T_{O2} \cdot \frac{M_{N2}}{M_{O2}} \]
We know \( T_{O2} = 55°C = 328 K \) and the ratio of molecular masses from the density ratio:
\[ \frac{M_{N2}}{M_{O2}} = \frac{14}{16} = \frac{7}{8} \]
Now substituting these values:
\[ T_{N2} = 328 K \cdot \frac{7}{8} \]

6. **Calculating \( T_{N2} \)**:
\[ T_{N2} = 328 K \cdot 0.875 = 287 K \]

7. **Convert Kelvin to Celsius**:
To convert Kelvin back to Celsius:
\[ T_{N2} = 287 K - 273 = 14°C \]

### Final Answer:
The temperature at which the speed of sound in nitrogen will be the same as that in oxygen at 55°C is **14°C**.