1.

The mass of hydrogen molecule is `3.23 xx 10^(23)` hydrogen molecules strike `2 cm^(2)` of a wall per second at an angle of `45^(@)` with the normal when moving with a speed of `10^(5) cm s^(-1)`, what pressure do they exert on the wall ? Assume collision to be elasitc.

Answer» Velocity of each molecule along the normal `= v cos theta` where `theta` is the angle made by the velocity of the molecule with the normal.
Therefore, change in velocity
`= v cos theta - (- v cos theta ) = 2 v cos theta`
Change in momentum `= m xx 2v cos theta = 2 m v cos theta`
There, rate of change of momentum `= 2 m v cos theta xx N`
where N is the number, of molecules striking the wall each second.
Force exerted by molecules `= 2 m v N cos theta`
`Pressure = ("Force")/("Area")`
`= (2 m v N cos theta)/(A) = (2 xx 3.32 xx 10^(-27) xx cos 45^(@))/(2 xx 10^(-4))`
`= 2.35 xx 10^(3) N m^(-2)`


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