Question

The mass of a hydrogen molecule is $3.32 \times 10^{-27} \, kg$. If $10^{23}$ hydrogen molecules strike, per second,a fixed wall of area $ 2 \, cm^2$ at an angle of $45^{\circ} $ to the normal, and rebound elastically with a speed of $10^3 \, m/s$ , then the pressure on the wall is nearly :

• A. $2.35 \times 10^3 \, N/m^2$
• B. $4.70 \times 10^3 \, N/m^2$
• C. $2.35 \times 10^2 \, N/m^2$
• D. $4.70 \times 10^2 \, N/m^2$

Answer 1

The Correct Option is A

To calculate the pressure exerted by hydrogen molecules on the wall, we apply the concept of momentum change. When molecules collide with the wall elastically and change direction, their momentum changes, applying a force on the wall. The pressure is then the force per unit area.

1. Calculate the change in momentum for one molecule:
   • The molecule strikes the wall at an angle of \(45^{\circ}\) to the normal. Its velocity vector can be decomposed into components:
   • Initial velocity components:
     • Normal component: \(v_n = v \cos 45^{\circ}\)
     • Lateral component: \(v_t = v \sin 45^{\circ}\)
   • Rebounding with the same speed elastically, the normal component reverses direction:
     • Change in normal component of velocity: \(\Delta v_n = -v_n - v_n = -2v_n\)
   • Change in momentum for one molecule:
     • \(\Delta p = m \cdot (-2v_n) = m \cdot (-2v \cos 45^{\circ})\)
     • \(= m \cdot (-2 \times 10^3 \times \frac{1}{\sqrt{2}}) = -m \cdot 2 \times 10^3 \times \frac{1}{\sqrt{2}} \, kg \cdot m/s\)
2. Total change in momentum per second for \(10^{23}\) molecules:
   • \(\Delta P = 10^{23} \times \Delta p = 10^{23} \times \left(-3.32 \times 10^{-27} \times 2 \times 10^3 \times \frac{1}{\sqrt{2}}\right)\)
   • Compute the force:
     • \(F = \Delta P = 10^{23} \times \left(-3.32 \times 10^{-27} \times 2 \times 10^3 \times \frac{1}{\sqrt{2}}\right) \, N\)
3. Calculate the pressure:
   • Area of the wall: \(A = 2 \, cm^2 = 2 \times 10^{-4} \, m^2\)
   • Pressure: \(P = \frac{F}{A}\)
   • Substitute the values:
     • \(Pressure = \frac{10^{23} \times (-3.32 \times 10^{-27} \times 2 \times 10^3 \times \frac{1}{\sqrt{2}})}{2 \times 10^{-4}}\)
     • Simplifying gives approximately: \(2.35 \times 10^{3} \, N/m^2\)

Therefore, the pressure exerted on the wall by the hydrogen molecules is approximately \(2.35 \times 10^{3} \, N/m^2\). Hence, the correct answer is $2.35 \times 10^3 \, N/m^2$.