The molecule of a gas room in a state of random motion. They repetitively collide versus the walls of the container. During each collision, inert is transfered come the walls of the container.

Pressure exerted by a gas

The molecule of a gas space in astate of arbitrarily motion. They repeatedly collide against the walls of thecontainer. During each collision, inert is transfered to the walls of thecontainer. The press exerted by the gas is because of the continuous collisionof the molecules versus the walls of the container.Due to this constant collision,the walls endure a continuousforce which is equal to the complete momentum imparted to the walls per second.The pressure experienced per unit area that the walls of the container determinesthe push exerted by the gas.

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Consider a cubic container ofside together containing n molecule of perfect gas moving with velocities C1, C2, C3... Cn (Fig.). A molecule moving with a velocityC1, will have actually velocities u1, v1 and w1 as materials along the x, y and also z axesrespectively. Similarly u2, v2 and w2 are the velocity contents of the secondmolecule and also so on. Let a molecule p (Fig.) having velocity C1 collide against thewall significant I (BCFG) perpendicular to the x-axis. Just the x-component the thevelocity that the molecule is pertinent for the wall I. Therefore momentum the themolecule prior to collision is mu1 whereby m is the mass of the molecule. Since thecollision is elastic, the molecule will certainly rebound v the velocity u1 in theopposite direction. For this reason momentum the the molecule after ~ collision is ?mu1.

Change in the momentum of themolecule

= final momentum - Initialmomentum

= ?mu1 ? mu1 = ?2mu1

During each successive collisionon challenge I the molecule have to travel a street 2l from face I to confront II andback to face I.

Time taken between two successivecollisions is = 2l/u1


∴ price of change of momentum = adjust in the inert / Timetaken

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

= (-2mu12) / (2l)

= -mu12 / l

(i.e) force exerted onthe molecule = -mu12/ l

∴ according to Newton?s 3rd law that motion, the pressure exerted bythe molecule

= -(-mu12/ l) = mu12 / l

Force exerted by allthe n molecules is

Fx= mu12/ together + mu22 / together + ?? + mun2 / l

Pressure exerted bythe molecules


=1/l2 ( /l+ /l +?..+ /l )

=m/l3(u12+ u22 + ?.+ un2 )

Similarly, pressureexerted by the molecules along Y and also Z axes are

Py = m/l3(v12+v22+ ???+ vn2)

Px = m/l3(w12+w 22+ ???+ w n2)

Since the gas exertsthe same press on every the walls of the container

Px = Py= Pz = P

P = /3

P = 1/3 . M/l3.

where C12= (u12 + v12 + w12)

P = 1/3 . Mn/V . C2

where C is dubbed theroot mean square (RMS) velocity, i m sorry is defined as the square source of themean value of the squares the velocities of individual molecules.

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(i.e.) C = root< (C12+ C22 +?.+ Cn2 ) / n >

Relation between the pressure exerted through a gas and the mean

kinetic energy oftranslation per unit volume the the gas

Pressure exerted byunit volume the a gas, p =mnC2 / 3

P = 1/3 . ρC2 (∵ mn = mass per unitvolume that the gas ; mn = ρ, density of the gas)

Mean kinetic energy oftranslation per unit volume the the gas

E =1/2 . ρC2

P/E = 2/3

P=2/3 . E

Average kinetic energy per molecule that the gas

Let us think about onemole that gas of massive M and also volume V

P = 1/3 . ρC2

P = 1/3 . M/V . C2

PV = 1/3 . M . C2

From gas equation


∴ RT =1/3 . M C2

3/2 . RT = ? . M C2

(i.e) average kineticenergy the one mole of the gas is equal to 3/2 . RT

Since one mole of thegas contains N number of atoms where N is the Avogadro number

we have M = Nm

(1/2 ) . MNC2= 3/2 . RT

? mC2 = 3/2. R/N . T

=3/2 kT where k =R/N, isthe Boltzmann continuous Its value is 1.38 ? 10-23 J K-1

∴ median kinetic power per molecule the the gas is same to 3/2kT

Hence, that is clearthat the temperature the a gas is the measure of the median translational kineticenergy every molecule the the gas.