If the density of a gas at NTP is `1.3mg//cm^(3)` and velocity of sound in it is 330 m/s. The number of degree of freedom of gas molecule is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the number of degrees of freedom of a gas molecule given its density at NTP and the velocity of sound in it, we can follow these steps:

### Step 1: Understand the relationship between density, pressure, and the velocity of sound.
The velocity of sound in a gas can be expressed as:
\[ v = \sqrt{\frac{\gamma P}{\rho}} \]
where \( v \) is the velocity of sound, \( \gamma \) is the adiabatic index, \( P \) is the pressure, and \( \rho \) is the density of the gas.

### Step 2: Convert the given density to appropriate units.
The density is given as \( 1.3 \, \text{mg/cm}^3 \). We need to convert this to \( \text{kg/m}^3 \):
\[ 1.3 \, \text{mg/cm}^3 = 1.3 \times 10^{-3} \, \text{g/cm}^3 = 1.3 \times 10^{-3} \times 1000 \, \text{kg/m}^3 = 1.3 \, \text{kg/m}^3 \]

### Step 3: Use the standard pressure at NTP.
At Normal Temperature and Pressure (NTP), the pressure \( P \) is approximately:
\[ P = 1.01 \times 10^5 \, \text{Pa} \]

### Step 4: Substitute the values into the velocity of sound equation.
We have:
\[ v = 330 \, \text{m/s} \]
Substituting the values into the equation:
\[ 330 = \sqrt{\frac{\gamma \cdot 1.01 \times 10^5}{1.3}} \]

### Step 5: Square both sides to eliminate the square root.
\[ 330^2 = \frac{\gamma \cdot 1.01 \times 10^5}{1.3} \]
Calculating \( 330^2 \):
\[ 108900 = \frac{\gamma \cdot 1.01 \times 10^5}{1.3} \]

### Step 6: Solve for \( \gamma \).
Rearranging gives:
\[ \gamma = \frac{108900 \cdot 1.3}{1.01 \times 10^5} \]
Calculating \( \gamma \):
\[ \gamma \approx 1.40 \]

### Step 7: Relate \( \gamma \) to degrees of freedom.
The relationship between \( \gamma \) and the degrees of freedom \( f \) is given by:
\[ \gamma = 1 + \frac{2}{f} \]
Substituting \( \gamma = 1.40 \):
\[ 1.40 = 1 + \frac{2}{f} \]
Subtracting 1 from both sides:
\[ 0.40 = \frac{2}{f} \]

### Step 8: Solve for \( f \).
Rearranging gives:
\[ f = \frac{2}{0.40} = 5 \]

### Conclusion
The number of degrees of freedom of the gas molecule is \( f = 5 \). This indicates that the gas is likely a diatomic gas.

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