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 solve the problem of finding the temperature at which the velocity of sound in oxygen (O₂) is the same as that in nitrogen (N₂) at 15°C, we can follow these steps:

### Step 1: Convert the temperature of nitrogen from Celsius to Kelvin
The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]

Given that the temperature of nitrogen is 15°C:
\[ T(N₂) = 15 + 273 = 288 \, K \]

### Step 2: Write the formula for the velocity of sound in a gas
The velocity of sound in a gas can be expressed as:
\[ v = \sqrt{\frac{\gamma R T}{M}} \]
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.

### Step 3: Identify the values for nitrogen and oxygen
For nitrogen (N₂):
- Molar mass \( M(N₂) = 28 \, g/mol \)
- \( \gamma(N₂) \) for diatomic gases is approximately \( 1.4 \)

For oxygen (O₂):
- Molar mass \( M(O₂) = 32 \, g/mol \)
- \( \gamma(O₂) \) for diatomic gases is also approximately \( 1.4 \)

### Step 4: Write the equation for the velocities of sound in both gases
For nitrogen:
\[ v(N₂) = \sqrt{\frac{\gamma(N₂) R T(N₂)}{M(N₂)}} = \sqrt{\frac{1.4 R \cdot 288}{28}} \]

For oxygen:
\[ v(O₂) = \sqrt{\frac{\gamma(O₂) R T(O₂)}{M(O₂)}} = \sqrt{\frac{1.4 R T(O₂)}{32}} \]

### Step 5: Set the velocities equal to each other
Since we want the velocities to be equal:
\[ \sqrt{\frac{1.4 R \cdot 288}{28}} = \sqrt{\frac{1.4 R T(O₂)}{32}} \]

### Step 6: Square both sides to eliminate the square root
\[ \frac{1.4 R \cdot 288}{28} = \frac{1.4 R T(O₂)}{32} \]

### Step 7: Cancel out common terms
Since \( 1.4 R \) appears on both sides, we can cancel it:
\[ \frac{288}{28} = \frac{T(O₂)}{32} \]

### Step 8: Solve for \( T(O₂) \)
Cross-multiplying gives:
\[ 288 \cdot 32 = 28 \cdot T(O₂) \]
\[ T(O₂) = \frac{288 \cdot 32}{28} \]

### Step 9: Calculate \( T(O₂) \)
Calculating the right-hand side:
\[ T(O₂) = \frac{9216}{28} = 329 \, K \]

### Step 10: Convert Kelvin back to Celsius
To convert back to Celsius:
\[ T(°C) = T(K) - 273 \]
\[ T(O₂) = 329 - 273 = 56 \, °C \]

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

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