At NTP, 8g of nitrogen occupies a volume of 22.4 litres. Calculate the speed of sound in nitrogen if `gamma` = 1.6 for nitrogen.

Answer

To calculate the speed of sound in nitrogen at NTP, we can use the formula:

\[ V = \sqrt{\frac{\gamma \cdot P}{\rho}} \]

Where:
- \( V \) is the speed of sound,
- \( \gamma \) is the adiabatic index (given as 1.6 for nitrogen),
- \( P \) is the pressure,
- \( \rho \) is the density of the gas.

### Step 1: Calculate the Density (\( \rho \))

Density (\( \rho \)) can be calculated using the formula:

\[ \rho = \frac{m}{V} \]

Where:
- \( m \) is the mass of the gas (8 g = 0.008 kg),
- \( V \) is the volume (22.4 liters = 0.0224 m³).

Substituting the values:

\[ \rho = \frac{0.008 \text{ kg}}{0.0224 \text{ m}^3} = 0.3571 \text{ kg/m}^3 \]

### Step 2: Use Standard Pressure at NTP

At Normal Temperature and Pressure (NTP), the pressure \( P \) is:

\[ P = 1 \text{ atm} = 101325 \text{ Pa} \]

### Step 3: Substitute Values into the Speed of Sound Formula

Now, substituting \( \gamma = 1.6 \), \( P = 101325 \text{ Pa} \), and \( \rho = 0.3571 \text{ kg/m}^3 \) into the speed of sound formula:

\[ V = \sqrt{\frac{1.6 \cdot 101325}{0.3571}} \]

### Step 4: Calculate the Speed of Sound

Calculating the numerator:

\[ 1.6 \cdot 101325 = 162120 \]

Now, divide by the density:

\[ \frac{162120}{0.3571} \approx 454000 \]

Now take the square root:

\[ V = \sqrt{454000} \approx 673.5 \text{ m/s} \]

### Final Answer

Thus, the speed of sound in nitrogen at NTP is approximately:

\[ V \approx 673.5 \text{ m/s} \]