Question

In a period of `1.00 s, 5 xx 10^(23)` nitrogen molecules strike a wall with an area of `8.00 cm^(2)`. 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 wait?
Note : The mass of one `N_(2)` molecules is `4.65 xx 10^(-26) kg`
A. `17.4 kPa`
B. `24.5 kPa`
C. `36.2 kPa`
D. `8.24 kPa`

Answer 1

Correct Answer - A
To find the pressure exerted by the nytrogen molecules, we first calculate the average force exerted by the molecules

`bar(F) = Nm_(0)(Delta v)/(Delta t)`
Here `N=5.00xx10^(23), m_(0)=4.65 xx 10^(-26)kg`
`Delta v = [300-(-300)]=2xx300m//s`
`=((500xx10^(-23))[(4.65xx10^(-26)kg)2(300m//s)])/(1.00s)`
`=14.0N`
The pressure is then
`P = (barF)/(A) = (14.0N)/(8.00xx10^(-4)m^(2)) = 17.4k Pa`.