The velocity of sound in air is `332 ms^(-1)` at `0^@ C` .At what temperature will the velocity become 1/2 time that at `0^@C`?

Answer

To solve the problem of finding the temperature at which the velocity of sound in air becomes half of its value at \(0^\circ C\), we can follow these steps:

### Step 1: Understand the relationship between velocity and temperature
The velocity of sound in air can be expressed using the formula:
\[ V = \sqrt{\frac{\gamma R T}{M}} \]
where:
- \(V\) is the velocity of sound,
- \(\gamma\) is the adiabatic index (which is constant for air),
- \(R\) is the universal gas constant,
- \(T\) is the absolute temperature in Kelvin,
- \(M\) is the molar mass of air.

### Step 2: Determine the initial conditions
At \(0^\circ C\) (which is \(273 K\)), the velocity of sound is given as:
\[ V_0 = 332 \, \text{m/s} \]

### Step 3: Calculate the target velocity
We want to find the temperature at which the velocity of sound becomes half of \(332 \, \text{m/s}\):
\[ V = \frac{332}{2} = 166 \, \text{m/s} \]

### Step 4: Set up the equation for the new velocity
Using the formula for the velocity of sound, we can set up the equation for the new velocity:
\[ 166 = \sqrt{\frac{\gamma R T}{M}} \]

### Step 5: Relate the two velocities
From the initial condition, we also have:
\[ 332 = \sqrt{\frac{\gamma R \cdot 273}{M}} \]

### Step 6: Square both equations to eliminate the square root
Squaring both equations gives us:
\[ 166^2 = \frac{\gamma R T}{M} \quad \text{(1)} \]
\[ 332^2 = \frac{\gamma R \cdot 273}{M} \quad \text{(2)} \]

### Step 7: Divide equation (1) by equation (2)
Dividing equation (1) by equation (2) allows us to eliminate \(\frac{\gamma R}{M}\):
\[ \frac{166^2}{332^2} = \frac{T}{273} \]

### Step 8: Solve for \(T\)
Rearranging gives:
\[ T = 273 \cdot \frac{166^2}{332^2} \]

Calculating \(166^2\) and \(332^2\):
\[ 166^2 = 27556, \quad 332^2 = 110224 \]
Thus:
\[ T = 273 \cdot \frac{27556}{110224} \]

### Step 9: Calculate the temperature
Calculating the fraction:
\[ \frac{27556}{110224} \approx 0.25 \]
Then:
\[ T \approx 273 \cdot 0.25 = 68.25 \, \text{K} \]

### Step 10: Convert to Celsius
To convert Kelvin to Celsius:
\[ T_{Celsius} = T_{Kelvin} - 273 = 68.25 - 273 \approx -204.75^\circ C \]

### Final Answer
The temperature at which the velocity of sound becomes half of that at \(0^\circ C\) is approximately:
\[ -204.75^\circ C \]