Let an ideal gas be enclosed in a cubical vessel of side. L. Area of cross wall be A = F. Let V, be the velocity of the molecule hitting the plane of wall ( of the vessel) YZ. After collision, it rebounds with the same speed but in the opposite direction along a line. The change in velocity of the molecule along the X direction is `v_(x) - (-v_(x)) = 2v_(x)`
The change in linear momentum imparted to the wall in the collision process = `(2mv_(x))/(Delta t)`
Pressure exerted on the wall due to 1 molecule = `("Force")/("Area")`
`therefore` pressure due to single molecule = `(2mv_(x'))/((A^(2))Deltat)"where" A = F`
Total number of molecules hiting the wall and returning
back = (Volume) (number density)
Dsitance travelled by the molecule = `(v_(x)Deltat)`
Volume covered = `(v_(x)Deltat)(A)`
Total number of molecule hitting hitting the wall on an
average `=(1)/(2)(v_(x)DeltatA)(n)`
Hence pressure exerted on the wall = p
`(p=(2mv_(x))/(ADeltat))((1)/(2)v_(x)DeltaAn)`
`i.e, " " p =mnv_(x)^(2)`
By symmetry (isotropic condition of speed) `bar(V)_(x)^(2)=bar(V)_(y)^(2)=bar(V)_(x)^(2)`
So, the mean of the squared speed along any one axis = `((1)/(3))bar(V^3)`
Thus `p=mn((1)/(3)bar(V^2))`
i.e. `p=(1)/(3)mbar(V)^(2)` where n is number density of molecules.
