The temperature at which the velocity of sound in oxygen will be same as that of nitrogen at `15^(@) C` is

Options

A. `112^(@) C`
B. `72^(@) C`
C. `56^(@) C`
D. `17 ^(@) C`

Explanation

To find the temperature at which the velocity of sound in oxygen (O₂) will be the same as that in nitrogen (N₂) at 15°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Velocity of Sound:**
The velocity of sound in a gas is given by the formula:
\[ v = \sqrt{\frac{\gamma RT}{M}} \]
where:
- \( v \) = velocity of sound
- \( \gamma \) = adiabatic index (ratio of specific heats)
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin
- \( M \) = molar mass of the gas

2. **Set Up the Equation:**
We need to equate the velocities of sound in both gases:
\[ \sqrt{\frac{\gamma_{O_2} R T_{O_2}}{M_{O_2}}} = \sqrt{\frac{\gamma_{N_2} R T_{N_2}}{M_{N_2}}} \]

3. **Cancel Common Terms:**
Since both gases are diatomic, their \(\gamma\) values are the same, and \(R\) is a constant, we can cancel these terms:
\[ \frac{T_{O_2}}{M_{O_2}} = \frac{T_{N_2}}{M_{N_2}} \]

4. **Substitute Known Values:**
- Molar mass of \(O_2\) (\(M_{O_2}\)) = 32 g/mol
- Molar mass of \(N_2\) (\(M_{N_2}\)) = 28 g/mol
- Temperature of \(N_2\) (\(T_{N_2}\)) = 15°C = 15 + 273 = 288 K

Plugging these values into the equation gives:
\[ \frac{T_{O_2}}{32} = \frac{288}{28} \]

5. **Solve for \(T_{O_2}\):**
Rearranging the equation to solve for \(T_{O_2}\):
\[ T_{O_2} = \frac{32}{28} \times 288 \]

6. **Calculate \(T_{O_2}\):**
\[ T_{O_2} = \frac{32 \times 288}{28} = \frac{9216}{28} \approx 329.14 \text{ K} \]

7. **Convert to Celsius:**
To convert from Kelvin to Celsius:
\[ T_{O_2} = 329.14 - 273 \approx 56.14°C \]

### Final Answer:
The temperature at which the velocity of sound in oxygen will be the same as that in nitrogen at 15°C is approximately **56°C**.