Question

The
pressure exerted by the gas is because of this
force exerted by gas particles per unit area
on the walls of the container.
.....Explain this line pls​

Answer 1

Answer:

because the particles of in air have lot of space between them and r moving randomly at high speed

the particles of gas collide with each other and hits the walls of container and hence exert a force on the walls of container

Explanation:

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Answer 2

Explanation:The molecules of a gas are in a state of random motion. They continuously collide against the walls of the container. During each collision, momentum is transfered to the walls of the container. The pressure exerted by the gas is due to the continuous  collision  of  the  molecules against the walls of the container. Due to this  continuous  collision, the  walls experience a continuous force which is equal to the total momentum imparted to the walls per second. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. Consider a cubic container of side containing n molecules of perfect gas moving with velocities C1, C2, C3  ... Cn as shown in figure. A molecule moving with a velocity C1, will have velocities u1, v1 and w1 as components along the x, y and z axes respectively. Similarly u2, v2 and w2 are the velocity components of the second molecule and so on. Let a molecule P as shown in figure having velocity C  collide against the wall marked I (BCFG) perpendicular to the x-axis. Only the x-component of the velocity of the molecule is relevant for the wall I. Hence momentum of the molecule before collision is mu1 where m is the mass of the molecule. Since the collision is elastic, the molecule will rebound with the velocity u1 in the opposite direction. Hence momentum of the molecule after collision is –mu1.

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

So,

 

where C is called the root mean square (RMS) velocity, which is defined as the square root of the mean value of the squares of velocities of individual molecules.

That is, C = √[C12+C23+ …...+Cn2/n]