Derive an expression for pressure exerted by a gas?
Answers
Answer:
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
Read more on Brainly.in - https://brainly.in/question/8560050#readmore
Answer:
The pressure exerted by the gas is because of this force exerted by gas particles per unit area on the walls of the container.