Question

derive the expression of pressure exerted by the gas on the wall of the container​

Answer 1

Change in the momentum of the molecule

= Final momentum - Initial momentum

= –mu1 – mu1 = –2mu1

During each successive collision on face I the molecule must travel a distance 2l from face I to face II and back to face I.

Time taken between two successive collisions is = 2l / u1

∴ Rate of change of momentum = Change in the momentum/Time taken

= – 2mu1/(2l/u1) = -2mu12/2l = – mu12/l

(i.e) Force exerted on the molecule = – mu12/l

Components of Velocity

∴ According to Newton’s third law of motion, the force exerted by the molecule,

= – (– mu12)/l = mu12/l

Force exerted by all the n molecules is

Fx = mu12/l + mu22/l + …...+mun2/l

Pressure exerted by the molecules,

Px = Fx/A

= 1/l2 (mu12/l + mu22/l + …...+mun2/l)

= m/l3 (u12+ u22 + …...+un2)

Similarly, pressure exerted by the molecules along Y and Z axes are,

Py = m/l3 (v12+ v22 + …...+vn2)

Pz = m/l3 (ω12+ ω22 + …...+ωn2)

Since the gas exerts the same pressure on all the walls of the container

Px = Py = Pz = P

P = [Px+Py+Pz]/3

P = [(1/3) (m/l3)] [(u12+ u22 + …...+un2) + (v12+ v22 + …...+vn2) + (ω12+ ω22 + …...+ωn2)]

P = [(1/3) (m/l3)] [(u12 + v12 +ω12) + (u22 + v22 +ω22) +.............+ (un2 + vn2 +ωn2)  

P = [(1/3) (m/l3)] [C12+C23+ …...+Cn2]

Here, C12 = (u12 + v12 +ω12)

P = [(1/3) (mn/l3)] [C12+C23+ …...+Cn2/n]

P = (1/3) (mn/V) C2