At what temperature is the velocity of sound in nitrogen equal to its velocity in oxygen at `20^(@)` C? The atomic weights of oxygen and nitrogen are in the ratio `16 :14`

Answer

To find the temperature at which the velocity of sound in nitrogen equals the velocity of sound in oxygen at 20°C, we can follow these steps:

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

### Step 2: Identify the values for nitrogen and oxygen
For nitrogen (N₂) and oxygen (O₂):
- The atomic weights are given in the ratio \( 16:14 \). This means:
- Molar mass of O₂ = 32 g/mol (since \( 16 \times 2 = 32 \))
- Molar mass of N₂ = 28 g/mol (since \( 14 \times 2 = 28 \))

- Both gases are diatomic, so the value of \( \gamma \) for both nitrogen and oxygen is the same:
\[ \gamma = \frac{7}{5} = 1.4 \]

### Step 3: Set up the equation for the velocities
We want to find the temperature \( T \) at which the velocities of sound in nitrogen and oxygen are equal:
\[ \sqrt{\frac{\gamma R T_{N}}{M_{N}}} = \sqrt{\frac{\gamma R T_{O}}{M_{O}}} \]
Since \( \gamma \) and \( R \) are constants, we can simplify this to:
\[ \frac{T_{N}}{M_{N}} = \frac{T_{O}}{M_{O}} \]

### Step 4: Substitute known values
We know:
- \( T_{O} = 20°C = 293 K \) (converting to Kelvin)
- \( M_{O} = 32 \, \text{g/mol} \)
- \( M_{N} = 28 \, \text{g/mol} \)

Substituting these values into the equation gives:
\[ \frac{T_{N}}{28} = \frac{293}{32} \]

### Step 5: Solve for \( T_{N} \)
Rearranging the equation to solve for \( T_{N} \):
\[ T_{N} = \frac{293 \times 28}{32} \]

Calculating this:
\[ T_{N} = \frac{8204}{32} = 256.375 \, K \]

### Step 6: Conclusion
The temperature at which the velocity of sound in nitrogen equals that in oxygen at 20°C is approximately:
\[ T_{N} \approx 256.37 \, K \]