Question

In one second, 95 moles of He gas particles are hitting a wall of a cubic container of volume 1 dm\(^3\). If the average velocity component of the particles perpendicular to the wall is 1000 m s\(^{-1}\), then the pressure of the gas in the container is \( X \times 10^5 \, \text{N m}^{-2} \). The value of \( X \) is ...............

Hint:

To calculate the pressure of a gas using the kinetic theory of gases, ensure to use the correct mass for individual molecules and apply the formula \( P = \frac{1}{3} \cdot \frac{N}{V} \cdot m \cdot v^2 \).

Answer 1

The pressure exerted by a gas on the walls of a container can be calculated using the kinetic theory of gases. The formula for the pressure \( P \) is given by: \[ P = \frac{1}{3} \cdot \frac{N}{V} \cdot m \cdot v^2 \] where:
- \( N \) is the number of gas molecules,
- \( V \) is the volume of the container,
- \( m \) is the mass of a gas molecule,
- \( v \) is the average velocity of the molecules perpendicular to the wall.
Step 1: Determine the number of molecules.
The number of molecules in 95 moles of He is: \[ N = 95 \, \text{mol} \times 6.02 \times 10^{23} \, \text{mol}^{-1} = 5.71 \times 10^{25} \, \text{molecules} \] Step 2: Determine the mass of a single He atom.
The mass of one mole of He is 4 g, so the mass of one helium atom is: \[ m = \frac{4 \, \text{g}}{6.02 \times 10^{23}} = 6.64 \times 10^{-23} \, \text{g} = 6.64 \times 10^{-26} \, \text{kg} \] Step 3: Calculate the pressure.
The volume is given as 1 dm\(^3\), which is \( 1 \times 10^{-3} \, \text{m}^3 \). The average velocity \( v = 1000 \, \text{m/s} \). Substituting all the values into the pressure equation: \[ P = \frac{1}{3} \times \frac{5.71 \times 10^{25}}{1 \times 10^{-3}} \times 6.64 \times 10^{-26} \times (1000)^2 \] \[ P = \frac{1}{3} \times 5.71 \times 10^{28} \times 6.64 \times 10^{-26} \times 10^6 \] \[ P = \frac{1}{3} \times 3.79 \times 10^9 \, \text{N/m}^2 = 1.26 \times 10^9 \, \text{N/m}^2 \] Thus, the pressure is \( 1.26 \times 10^5 \, \text{N/m}^2 \).
Final Answer: \[ \boxed{1.26} \]