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Find the pressure exerted by 6xx10^(23) ...

Find the pressure exerted by `6xx10^(23)` hydrogen molecules which will strike per second a wall of area `10^(-4) km^(2)` at `60^(@)` with normal.The mass ofhydrogen molecules and speed are `3.32xx10^(-27)` kg and `10^(-27)` kg and `10^(3) m//s` respectively.

A

`19.92xx10^(3)N//m^(2)`

B

`18.2xx10^(3)N//m^(2)`

C

`1.992xx10^(3)N//m^(2)`

D

`0.1992xx10^(3)N//m^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Change in momentum `=`2mv cos `theta`
Force, F= 2mv cos `thetaxxn`
`p=F/A=(2mv n cos theta)/(A)`
`=(2xx3.32xx106(-27)xx10^(3)xx(1)/(2)xx6xx10^(23))/(10^(-4))`
`=19.92xx10^(3)N//m^(2)`
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