Kineetic Theo
ory Of Gaas‐Pressu
ure Exerrted By A
A Gas Pressure
e exerted byy a perfect gaas: Let us consider c
a perfect gas enclosed e
in cube c
the waalls of which are perfecttly elastic. The three edges off the cube arre taken to b
be X, Y & Z axxes. Given: m = masss of each mo
olecule of th
he gas n = the ttotal number of moleculees of the gass Let C1, C
C2, C3, Cn bee the velocitiies of molecu
ules of the gaas. →
Let us consider
c
a molecule
m
havving a veloccity C .
→
Let u, v,
v w be the components
c
of this veloocity vector C along X, Y and Z axees respectiveely.
→
∴ C = u 2 + v2 + w 2
(
)
C 2 = u 2 + v 2 + w 2 → (1)
Let us first consider the collision of that moleecule with th
he walls ABC
CD and EFGH perpendicular to X axis. Thee molecule m
moves with vvelocity U alo
ong X axis , sttrikes the waall ABCD and
d since the co
ollision is perfectlyy elastic the molecule rebounds with
h same speed
d U along ‐X axis, goes an
nd strikes the
e wall EFGH, reebounds with velocity U and so on. ©Selfstu
udy.in IEMS ‐ High School T
Tutorial Class Notes Geneeral Physics Page 1 Kinetic Theory Of Gas‐Pressure Exerted By A Gas The momentum of the molecule before and after the collision with the wall ABCD are mu and –mu respectively. Therefore the change in momentum of the molecule due to collision with the wall ABCD = mu – (‐mu) = 2mu Between two successive collisions with the wall ABCD the molecule covers a distance 2l with a speed u and takes time 2l/u. Thus in every 2l/u second molecule takes 1 collision in every 1 second molecule takes 1/(2l/u) = u/2l collision. Hence the change in momentum of the molecule in 1 sec due to collision with the wall ABCD = change in momentum in 1 collision x no. of collision in 1 sec = 2mu x u/2l =mu2/l According to Newton’s second law the rate of change of momentum is impressed force. Therefore the force exerted by a molecule due to the motion along X‐axis = mu2/l Therefore the total force exerted by all the molecules due to their motion along X‐axis mu 2
l
The pressure exerted by all the gas molecules due to their motion along X axis
F 1 m
Px = = 2 ∑ u 2
A l l
m
Px = 3 ∑ u 2
l
3
But l = V = Volume of the gas
m
∴ Px = ∑ u 2 → (2)
V
=∑
©Selfstudy.in IEMS ‐ High School Tutorial Class Notes General Physics Page 2 Kinetic Theory Of Gas‐Pressure Exerted By A Gas Similarly by considering the motion of the molecules along Y and Z axes we get
m
∑ v 2 → (3)
V
m
Pz = ∑ w 2 → (4)
V
Since the pressure of the gas inside a container is same in all directions
Py =
P + P + P = P + P + P = 3P
∴
x
y
z
m
m
m
∑ u 2 + ∑ v2 + ∑ w 2
V
V
V
m
∑ u 2 + ∑ v2 + ∑ w 2
3V
m 2
u1 + u 22 + u 32 + ........ + u n2 + v12 + v 22 + v32 + ........ + v n2 + w12 + w22 + w32 + ........ + wn2
3V
m 2
u1 + v12 + w12 + u 22 + v 22 + w22 + u 32 + v32 + w32 + .......... + u n2 + v n2 + wn2
3V
m 2
c1 + c 22 + c32 + ............. + c n2
3V
mn ⎡ c12 + c 22 + c32 + ............. + c n2 ⎤
⎢
⎥ → (5)
n
3V ⎣
⎦
3P =
P=
P =
P =
P=
P=
[
]
[(
) (
[(
) (
[
) (
) (
)
(
)]
)]
]
c12 + c 22 + c32 + ............. + c n2 →
= c = Root mean square velocity → (6)
n
Putting
(6) in (5)
→
P=
1 mn c 2
→ (7)
3 V
Equation(7) gives the pressure exerted by a perfect gas.
mn
= ρ = density of the gas
mn = mass of the gas ∴
V
1 →
∴ P = ρ c 2 → (8)
3
©Selfstudy.in IEMS ‐ High School Tutorial Class Notes General Physics Page 3 Kinetic Theory Of Gas‐Pressure Exerted By A Gas Deductions from Kinetic Theory: (1) Boyle’s law from kinetic theory : From kinetic theory pressure exerted by a perfect gas is 1 mn 2
C
3 V
2
1
PV = 3 mnC → (1)
For a given mass of gas mn = constant
P=
We also know that the r.m.s velocity of the molecules of a given as depends only on temperature
i.e. C ∝ T
If the temperature of the gas is kept constant then C is also constant.
From equation(1) for a given mass of a gas at constnat temperature the R.M.S velocity is constant
i.e.PV = constant ( Boyles law )
©Selfstudy.in IEMS ‐ High School Tutorial Class Notes General Physics Page 4 Kinetic Theory Of Gas‐Pressure Exerted By A Gas (2) Kinetic interpretation of temperature: From kinetic theory pressure exerted by a perfect gas PV = 1 mnC → (1)
2
3
Fromequationof state
PV = αRT → (2)
Massof the gas
mn
Whereα = no.of moleculesof the gas = Molecularweight = M
Fromequation(1) and (2)
2
1
mnC = αRT
3
1 mnC 2 = mn RT → (3)
3
M
3R
C=
T → (4)
M
Sincefor a givengas M and R are constant
C∝ T
Thusfor a givengas the R.M.Svelocityis proportional to thesquareroot of absolutetemperature of the gas
From equation(3)
2 mn
1
mnC =
RT
3
M
M = m × N whereN = Avogradro's no.
1
mn
2
mnC =
RT
3
m× N
R
K = = Boltzman' s constant
N
2
1
3
2 mnC = 2 KT → (5)
Mean kineticenergyper moleculeof the gas = 3 KT
2
If T = 0, Mean K.E per moleculeof the gas = 0 thus absolutezero can also be definedas the temperature
at which the mean kineticenergyper moleculeof the gas becomeszero.
©Selfstudy.in IEMS ‐ High School Tutorial Class Notes General Physics Page 5 Kinetic Theory Of Gas‐Pressure Exerted By A Gas (3) Avogadro’s hypothesis from kinetic theory: let us consider two different gases say Gas1 and Gas2 m
1 and m 2 = mass of each molecule of gas1 and gas2
n1 and n 2 = total no of molecules of gas1 and gas2
−
−
c1 and c 2 = R.M.S velocity of molecules of gas1 and gas2
P1 and P2 = Pressure of gas1 and gas2
V1 and V2 = Volume of gas1 and gas2
T1 and T2 = Temperature of gas1 and gas2
Using the kinetic theory the pressure exerted by gas1:
1 m1 n1 −2
P1 =
c1
3 V1
−
1
2
P1 V1 = m1 n1 c1 → (1)
3
The pressure exerted by gas 2
−
1
2
P2 V2 = m2 n2 c 2 → (2)
3
If P1 = P2 and V1 = V2 then P1 V1 = P2 V2
From equation(1) and (2) we get
−
−
1
1
2
2
m1 n1 c1 = m2 n2 c 2
3
3
−
−
m1 n1 c1 = m2 n2 c 2 → (3)
2
2
Usingkineticinterpretation of gas1
−
K.Eper moleculeof thegas1= 1 m c 2 = 3 KT → (4)
Themean
1 1
1
2
2
−
3
1
2
= m2 c2 = KT2 → (5)
2
2
−
−
1
1
2
2
If T1 = T2 thenfromequation(4) and(5) m1 c1 = m2 c2
2
2
Dividingequation(3) by equation(6)
ThemeanK.Eper moleculeof gas2
−
2
1 1 1
−
2
1 1
−
2
mn c
mn c
= 2 2 2−
1
1
2
mc
m2 c2
2
2
∴n1 = n2
Thusat equaltemperature and pressureequalvolumeof twodifferentgasescontainequalnumberof molecules
- Avogradro's hypothesis.
©Selfstudy.in IEMS ‐ High School Tutorial Class Notes General Physics Page 6