Question

In a period of 1.00 s, 5.00×1023 nitrogen molecules strike a wall with an area of 8.00cm2. Assume the molecules move with a speed of 300 m/s and strike the wall head-on in elastic collisions. What is the pressure exerted on the wall? Note: The mass of one N2 molecule is 4.65×10−26kg.

Answer 1

A wall with an area of 8 cm² is struck by 5.00 x 10²³ nitrogen molecules over the course of 1 s. Assume that the molecules are moving at a speed of 300 m/s and colliding with the wall in an elastic manner. The pressure acting on the wall is  17.43 kPa.

Mass in motion is a definition of momentum. Mass and velocity are multiplied to determine momentum. That is, p = mv where p,m, and v denotes momentum, mass, and velocity.

Given the change in the time period, Δt is 1 second, the number of nitrogen molecules is 5.00×10²³ molecules, the change in velocity Δv is 300 m/s, and the area is 8×10⁻⁴ m².

The momentum of one molecule is given as 2mΔv. Then, the momentum of N molecules is 2NmΔv.

The force exerted on the wall by nitrogen molecule is written as,

$\begin{aligned}\text{Force F}&=\frac{\Delta\text {momentum}}{\Delta t}\\&=\frac{2Nm\Delta v}{\Delta t}\end{aligned}$

Then, the pressure exerted by nitrogen molecule on the wall is,

$\begin{aligned}P&=\frac{F}{A}\\&=\frac{2Nm\Delta v}{\Delta t\times A}\\&=\frac{2\times 5\times10^{23}\times4.65\times10^{-26}\times300}{8\times 10^{-4}\times 1.00}\\&=\mathrm{17.43\;kPa}\end{aligned}$

Therefore, the required answer is 17.43 kPa.

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