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PHYSICAL CHEMISTRY

Proving the kinetic model presure and volume of a gas could sometimes seem to be a menace

PV=mnc²/3 (The velocity c1 of the molecule is solved in to 3 components x ,y and z

So that c1= c1x + c1y + c1z ie a cube with opposite phase [A'] )

Thats why am going to share some light on it today. OK let's start

Firstly a particle in motion moves with a momentum (M=mc) mc1x.......i

Where (m) is the mass of the molecule and (c) the velocity. if the molecule bounces back from the wall of its container it goes against gravity with the momentum=~~mc1x.......ii~~

~~In combination we derive the total collision as~~

~~mc1x~~(-mc1x)= 2mc1x

and when we apply this for a number of molecules we have (n'=number of moles so we now have )..... f= 2mn'c1x

During the course of collision the molecule travelled a distance of (L) and another distance of L when it bounced back so total distance =2L

The rate of change of momentum per second will be

f= 2mn'c1x/2L = mn'c1x/L........iii

Thesame quantity of momentum is experienced in the opposite phase so equ......iii becomes

2mn'c/L

Pressure(p) = F/A

F= force ,A= Area

Substituting f for F we have

P = 2mn'c²/L A

Where Area of a cube= 6 (l × l) and c² = root mean velocity

P= 2mn'c²/6(l × l× l) .......... P= 2mn'c²/6 l³

Volume (V)= l³

P= 2mn'c²/6V ie cross multiplying

PV= 2mn'c²/6 ie PV = mn'c²/3

Proving the kinetic model presure and volume of a gas could sometimes seem to be a menace

PV=mnc²/3 (The velocity c1 of the molecule is solved in to 3 components x ,y and z

So that c1= c1x + c1y + c1z ie a cube with opposite phase [A'] )

Thats why am going to share some light on it today. OK let's start

Firstly a particle in motion moves with a momentum (M=mc) mc1x.......i

Where (m) is the mass of the molecule and (c) the velocity. if the molecule bounces back from the wall of its container it goes against gravity with the momentum=

and when we apply this for a number of molecules we have (n'=number of moles so we now have )..... f= 2mn'c1x

During the course of collision the molecule travelled a distance of (L) and another distance of L when it bounced back so total distance =2L

The rate of change of momentum per second will be

f= 2mn'c1x/2L = mn'c1x/L........iii

Thesame quantity of momentum is experienced in the opposite phase so equ......iii becomes

2mn'c/L

Pressure(p) = F/A

F= force ,A= Area

Substituting f for F we have

P = 2mn'c²/L A

Where Area of a cube= 6 (l × l) and c² = root mean velocity

P= 2mn'c²/6(l × l× l) .......... P= 2mn'c²/6 l³

Volume (V)= l³

P= 2mn'c²/6V ie cross multiplying

PV= 2mn'c²/6 ie PV = mn'c²/3

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