AL Physics/Matter/Gases K.W. LAM
GASES
1. Macroscopic Description of Ideal Gases
An ideal gas is defined such that:
(1) It obeys Boyle’s law, i.e.
the pressure of a fixed mass of gas is inversely proportional to its volume if the
temperature is constant.
1 constant
P or P=
V V
∴ PV = constant
(2) PV for a fixed mass of ideal gas is directly proportional to the temperature of the gas
measured on the ideal gas scale,
i.e., PV T, where T defines the temperature on the ideal gas scale.
Hence for an ideal gas
PV
= const.
T
for n moles of gas,
PV
= nR
T
or PV = nRT
where n = number of moles of gas
R = universal gas constant, 8.31 J mol-1 K-1
This is called the equation of state for an ideal gas. (Note: state P, V, T)
N.B. R has constant value for all gases and it is not dependent on the amount (number of
moles) of gas.
Revision
Pressure Law : At constant V, P/T = constant or P T.
Charles' Law : At constant P, V/T = constant or VT.
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AL Physics/Matter/Gases K.W. LAM
Example
What is the volume of one mole of an ideal gas at s.t.p.?
Solution
Example
A cylinder contains 0.1 m3 of compressed oxygen at 5 atm pressure and –10 oC. How
much volume will the oxygen occupy if released to 1 atm pressure at 20 oC?
Solution
Note:
Actually, ideal gases do not exist in real life but real gases can be taken as ideal gases at
LOW pressure and HIGH temperature.
Virtual Lab.
Einstein's Explanation of Brownian Motion
http://comp.uark.edu/~jgeabana/mol_dyn/KinThI.html
Brownian Motion
http://www.phy.ntnu.edu.tw/~hwang/gas2D/gas2D.html
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AL Physics/Matter/Gases K.W. LAM
2. Kinetic Theory of Gases
Kinetic theory deals mainly with the study of the microscopic properties (e.g. speed, mass) of
gas molecules.
2.1 Microscopic Definition of an Ideal Gas
An ideal gas satisfies the following assumptions:
(1) The intermolecular force is negligible except during collision.
(2) The volume occupied by molecules is negligible compared to the volume
occupied by the gas.
(3) The time of contact during collisions is negligible compared with the time
between collisions.
(4) All collisions between molecules and with the wall of container are
elastic (no energy loss).
Related AL Questions: [92/IIA/2(a)], [89/IIA/1(a)]
1
2.2 Derivation of PV = Nm c2
3
Along Ox direction,
on striking wall X,
change of momentum (or impulse)
= mu – ( mu )
= 2 mu
If it starts from O and rebounds back to O again,
rate of change of momentum at X ( = force on wall X by Newton’s 2nd law)
change of momentm
=
time
2mu
=
2l / u
mu 2
=
l
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AL Physics/Matter/Gases K.W. LAM
Total force on wall X due to impacts by all N molecules
m
= ( u12 + u22 + … + uN2 )
l
F
Pressure P = on wall X of area l 2
A
m
P= 3
( u12 + u22 + … + uN2 )
l
If u 2 is the mean value of ui2
u 2 u 2 ... u 2
u2 = 1 2 N
N
N u 2 = u12 + u22 + … + uN2
Nmu 2
∴ P=
l3
Using Pythagoras’ theorem,
c2 = u 2 + v 2 + w 2
c 2 = u2 + v 2 + w 2
Since N is large and molecules move randomly u2 = v 2 = w 2
c 2 = 3 u2
1 2
u2 = c
3
Nmc 2
∴ P =
3l 3
∵ l3 = V
1
PV = Nmc 2
3
This is called the Kinetic Theory Equation
where N = total number of gas molecules
m = mass of one gas molecule
c 2 = mean square speed of gas molecules
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AL Physics/Matter/Gases K.W. LAM
The result links the macroscopic property (pressure and volume of gas) with the
microscopic property (number, mass and speed of gas molecules) of a gas.
Related AL Questions: [97/IIB/1(d)(i)]
2.3 Order of Magnitude of c2
1
PV = Nmc 2
3
1 Nm 2
P = c
3 V
Since Nm = total mass of N gas molecules
Nm total mass
= = density of gas
V Volume
1 2
P= c
3
where = density of gas
Virtual Lab.
Molecular Model of an Ideal Gas
http://www.phy.ntnu.edu.tw/~hwang/idealGas/idealGas.html
Example
Find
(a) the mean square speed of the air molecules, density of air is 1.29 kg m-3 at s.t.p.
(b) the mean square speed of the hydrogen molecules, density of hydrogen is 0.09 kg m-3
at s.t.p.
Solution
c 2 - root-mean-square speed (same order of magnitude but greater than the mean
speed).
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AL Physics/Matter/Gases K.W. LAM
For air, c2 =
hydrogen, c2 =
2.4 Distribution of Molecular Speed
c 2 ( or crms ) root mean square speed,
it does not tell how the speeds are distributed.
It is possible that all speeds fall in a very narrow range or that the range is very wide.
The distribution of molecular speeds c can be shown as the curve below based on the
number N of molecules in each unit range of speed.
c
It is called the Maxwell distribution.
Some general features:
(1) There is small chance of very low speed (part A) and very large speed (part C)
2
(2) The most possible speed (part B) can be shown to be crms.
3
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AL Physics/Matter/Gases K.W. LAM
If temperature increases shape of curve changes from 1 to 2 to 3
c
area under the curve = total number of molecules
So the area under the curve 1, 2 and 3 are __________ since the total number of
molecules in the gas is ___________ .
Virtual Lab.
Kinetic Theory
http://comp.uark.edu/~jgeabana/mol_dyn/KinThI.html
2.5 Avogadro Constant
Number of molecules per mole is the same for all substances called the Avogadro
constant (Symbol: NA)
NA = 6.02 × 1023 mol-1
i.e., the number of molecules in 1 mole of an ideal gas
Example
Find the value of R if the volume of 1 mole of gas = 22.4 × 10-3 m3 at s.t.p.?
Solution
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AL Physics/Matter/Gases K.W. LAM
2.6 Temperature and Kinetic Theory
For 1 mole of ideal gas, (i.e., number of molecules, N = NA)
1
PV = N A mc 2
3
2 1
PV = N A ( mc 2 )
3 2
( ∵ for 1 mole, PV = (1)RT )
2 1
∴ RT = N A ( mc 2 )
3 2
1 3 R
mc 2 = T
2 2 NA
Note:
1
1. mc 2 = average K.E. of translational motion of ONE MOLECULE, EK (or
2
called the molecular kinetic energy).
3 R
And we can write EK = T
2 NA
Q: Find the internal energy U of the gas in terms of R and T.
A: U = total K.E. of the gas
1
mc 2 T. Therefore if
2. Since R and NA are constant, so for an ideal gas
2
T increases molecular speed ____________ and hence translational K.E. EK
___________ .
Related AL Questions: [95/IIB/2(a)]
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AL Physics/Matter/Gases K.W. LAM
Q: What is the total translational K.E. per mole (i.e. molecular K.E.) of an ideal gas?
Hence, find the molecular K.E. when T = 0.
Answer
1
Translational K.E. of 1 molecule mc 2
2
there are ________ molecules in 1 mole of ideal gas
∴ K.E. of ______ molecules
From the above expression, theoretically
molecular K.E. = _________ when T = 0.
R
N.B. 1. The ratio is called the Boltzmann’s constant ( Symbol: k ) and
NA
k = 1.38 × 10-23 J K 1 .
1 3
So we can write molecular kinetic energy, EK = mc 2 = k T
2 2
3kT
Q : Show that c2 = .
m
A:
2. Recalling the equation of state
PV = nRT
N total no. of molecules
= RT (∵ no. of mole, n = )
NA Avogadro constant
R
= N( )T
NA
= NkT
So we can write the equation of state in 2 forms:
PV = nRT = NkT
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AL Physics/Matter/Gases K.W. LAM
1 2
Q : Using (1) P = c and (2) PV = NkT
3
3kT
show that c2 = .
m
A:
Related AL Questions: [96/IIB/3(a)]
2.7 Avogadro’s Law
Equal volumes of ideal gases existing under the same conditions of temperature and
pressure contain equal numbers of molecules.
Consider two ideal gases 1 and 2,
1
Gas 1: P1V1 = N1m1c12
3
1
Gas 2: P2V2 = N2m2 c 22
3
If they have same pressures and volumes, P1 = P2 and V1 = V2 ,
P1V1 = P2 V2
1 1
N1m1c12 = N2m2 c 22 …………………… (1)
3 3
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AL Physics/Matter/Gases K.W. LAM
If they have the same temperatures,
they will have the same energy.
1 1
m1c 12 = m 2 c 2 2 …………………………. (2)
2 2
∴ from (1) and (2), N1 = N2
Related AL Questions: [97/IIB/1(d)(ii)], [99/IIB/4,part of (a)(i)]
N.B.:
When two gases of different states are mixed together,
1. total number of moles of two gases is constant
2. total energy of two gases is constant.
3. Real Gases
3.1 Van der Waal’s Equation
Ideal gases : 4 assumptions are valid.
A real gas behaves like an ideal gas and obeys the equation PV = RT (for 1 mole) at
LOW pressure and HIGH temperature.
At HIGH pressure and LOW temperature, a real gas has great departure from ideal
behaviour.
Two of the 4 assumptions are not valid.
1. Volume of gas molecules may NOT be negligible.
2. Attraction force between molecules may NOT be negligible.
(A) Volume of a Real Gas
When the gas molecules approach very closely (due to high pressure or low
temperature), molecules have a particular volume because repulsive forces
occur.
The ‘free-volume’ of the space inside a container is ( V b ) where b is the
factor called the ‘co-volume’ depending on the actual volume of the molecules.
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AL Physics/Matter/Gases K.W. LAM
(B) Attractive Molecular Forces
molecules attract each other when they are close together (due to high pressure
or low temperature).
molecules at A, net force = 0. (since forces in different directions balance each
other)
molecules at B (wall), attracted by the
molecules behind it (left net force)
strike the wall with smaller force
pressure is reduced by P
1
Since P no. of molecules at the wall. P
V
1
and P no. of molecules behind wall P
V
1 a
So P or P =
V 2
V2
a
∴ Pideal = Preal +
V2
So van der Waal’s equation of state for 1 MOLE of a real gas is:
a
(P + )(V b) = RT
V2
Notes:
For n moles of gas, van der Waal’s equation of state becomes
n2a
(P + 2 )(V nb) = nRT
V
Related AL Questions: [95/IIB/2(b)]
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AL Physics/Matter/Gases K.W. LAM
3.2 Critical Point
Ideal gases:
obey PV = RT (for 1 mole)
Real gases:
obey van der Waal’s equation
a
(P + )(V -b) = RT
V2
At high temperature, isotherms are similar to that of an ideal gas
P
V
Middle isotherm at critical temperature, there is a point of inflexion C
the critical point.
Above critical temperature, a gas cannot be liquefied by increasing its pressure.
Below the critical temperature, isotherms such as EABF are obtained by using van
der Waal’s equation.
The region AB of positive slope which P increases with V is impossible.
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AL Physics/Matter/Gases K.W. LAM
The corrected graph was obtained experimentally by Andrew for carbon dioxide in
which the graph between E and F is a horizontal line.
The straight section EF represents the changing between liquid and vapour.
We can distinguish a gas and a vapour:
a gas is above its critical temperature and a vapour below.
The critical temperature of carbon dioxide is 304 K (31oC) and of oxygen 154 K
( 119oC).
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