Chapter 13: Temperature and Ideal
Gas
•What is Temperature?
•Temperature Scales
•Thermal Expansion
•Molecular Picture of a Gas
•The Ideal Gas Law
•Kinetic Theory of Ideal Gases
•Chemical Reaction Rates
•Collisions Between Molecules
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§13.1 Temperature
Heat is the flow of energy due to a temperature difference.
Heat always flows from objects at high temperature to
objects at low temperature.
When two objects have the same temperature, they are in
thermal equilibrium.
2
The Zeroth Law of Thermodynamics:
If two objects are each in thermal equilibrium with a
third object, then the two objects are in thermal
equilibrium with each other.
3
§13.2 Temperature Scales
Absolute or
Kelvin scale
Fahrenheit
scale
Celsius
scale
Water boils*
373.15 K
212 °F
100 °C
Water freezes*
273.15 K
32 °F
0 °C
Absolute zero
0K
-459.67 °F
-273.15°C
(*) Values given at 1 atmosphere of pressure.
4
The temperature scales are related by:
Fahrenheit/
Celsius
TF = (1.8 °F/°C )TC + 32°F
Absolute/
Celsius
T = TC + 273.15
5
Example (text problem 13.3): (a) At what temperature (if
any) does the numerical value of Celsius degrees equal the
numerical value of Fahrenheit degrees?
TF = 1.8TC + 32 = TC
TC = −40 °C
(b) At what temperature (if any) does the numerical value
of Kelvin equal the numerical value of Fahrenheit
degrees?
TF = 1.8TC + 32
= 1.8(T − 273) + 32
= 1.8(TF − 273) + 32
TF = 574 °F
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§13.3 Thermal Expansion of Solids
and Liquids
Most objects expand when their temperature increases.
7
An object’s length after its temperature has changed is
L = (1 + αΔT )L0
α is the coefficient of
thermal expansion
where ΔT=T-T0 and L0 is the length of the object at a
temperature T0.
8
How does the area of an object change when its temperature
changes?
The blue square has an area of L02.
L0
L0+ΔL
With a temperature change ΔT each
side of the square will have a length
change of ΔL = αΔTL0.
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The fractional change in area is:
new area = A = (L0 + αΔTL0 )(L0 + αΔTL0 )
= L + 2αΔTL + α ΔT L
2
0
2
0
2
2 2
0
≈ L20 + 2αΔTL20
= A0 (1 + 2αΔT )
ΔA
= 2αΔT
A0
10
The fractional change in volume due to a temperature
change is:
ΔV
= βΔT
V0
For solids β=3α
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§13.4 Molecular Picture of a Gas
The number density of particles is N/V where N is the total
number of particles contained in a volume V.
If a sample contains a single element, the number of
particles in the sample is N = M/m. N is the total mass of
the sample (M) divided by the mass per particle (m).
12
One mole of a substance contains the same number of
particles as there are atoms in 12 grams of 12C. The number
of atoms in 12 grams of 12C is Avogadro’s number.
N A = 6.022 × 10 23 mol-1
13
A carbon-12 atom by definition has a mass of exactly 12
atomic mass units (12 u).
This is the conversion factor between the atomic mass unit
and kg (1 u = 1.66×10-27 kg). NA and the mole are defined
so that a 1 gram sample of a substance with an atomic
mass of 1 u contains exactly NA particles.
14
Example (text problem 13.39): Air at room temperature and
atmospheric pressure has a mass density of 1.2 kg/m3. The
average molecular mass of air is 29.0 u. How many air
molecules are there in 1.0 cm3 of air?
total mass of air in 1.0 cm 3
number of particles =
average mass per air molecule
The total mass of air in the given volume is:
⎛ 1.2 kg ⎞⎛ 1.0 cm
⎜
m = ρV = ⎜
3 ⎟⎜
⎝ m ⎠⎝ 1
3
⎞⎛ 1 m ⎞
⎟⎟⎜
⎟
⎠⎝ 100 cm ⎠
3
= 1.2 ×10 −6 kg
15
Example continued:
total mass of air in 1.0 cm 3
number of particles =
average mass per air molecule
1.2 ×10 −6 kg
=
(29.0 u/particle) 1.66 ×10−27 kg/u
(
)
= 2.5 ×1019 particles
16
§13.5 Absolute Temperature and
the Ideal Gas Law
Experiments done on dilute gases (a gas where interactions
between molecules can be ignored) show that:
For constant pressure
V ∝T
Charles’ Law
For constant volume
P ∝T
Gay-Lussac’s Law
17
For constant
temperature
1
P∝
V
For constant pressure
and temperature
V∝N
Boyle’s Law
Avogadro’s
Law
18
Putting all of these statements together gives the ideal gas
law (microscopic form):
PV = NkT
k = 1.38×10-23 J/K is
Boltzmann’s constant
The ideal gas law can also be written as (macroscopic
form):
PV = nRT
R = NAk= 8.31 J/K/mole is the
universal gas constant and n is
the number of moles.
19
Example (text problem 13.41): A cylinder in a car engine
takes Vi = 4.50×10-2 m3 of air into the chamber at 30 °C and
at atmospheric pressure. The piston then compresses the
air to one-ninth of the original volume and to 20.0 times the
original pressure. What is the new temperature of the air?
Here, Vf = Vi/9, Pf = 20.0Pi, and Ti = 30 °C = 303 K.
PiVi = NkTi
Pf V f = NkT f
The ideal gas law holds
for each set of parameters
(before compression and
after compression).
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Example continued:
Take the ratio:
Pf V f
PiVi
=
NkT f
NkTi
=
Tf
Ti
⎛ Pf
The final temperature is T f = ⎜⎜
⎝ Pi
⎞⎛ V f ⎞
⎟⎟⎜⎜ ⎟⎟Ti
⎠⎝ Vi ⎠
⎛ Vi ⎞
⎛ 20.0 Pi ⎞⎜ 9 ⎟
⎟⎟⎜
= ⎜⎜
⎟(303 K ) = 673 K
⎝ Pi ⎠⎜ Vi ⎟
⎝
⎠
The final temperature is 673 K = 400 °C.
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§13.6 Kinetic Theory of the Ideal
Gas
An ideal gas is a dilute gas where the particles act as point
particles with no interactions except for elastic collisions.
22
Gas particles have random motions. Each time a particle
collides with the walls of its container there is a force exerted
on the wall. The force per unit area on the wall is equal to
the pressure in the gas.
The pressure will depend on:
•The number of gas particles
•Frequency of collisions with the walls
•Amount of momentum transferred during each collision
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The pressure in the gas is
2N
P=
K tr
3V
Where <Ktr> is the average translational kinetic energy of
the gas particles; it depends on the temperature of the gas.
K tr
3
= kT
2
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The average kinetic energy also depends on the rms speed
of the gas
K tr
1
1 2
2
= m v = mvrms
2
2
where the rms speed is
3
1 2
kT = mvrms
2
2
3kT
vrms =
m
K tr =
25
The distribution of speeds in a gas is given by the MaxwellBoltzmann Distribution.
26
Example (text problem 13.60): What is the temperature of
an ideal gas whose molecules have an average
translational kinetic energy of 3.20×10-20 J?
3
K tr = kT
2
2 K tr
T=
= 1550 K
3k
27
Example (text problem 13.70): What are the rms speeds of
helium atoms, and nitrogen, hydrogen, and oxygen
molecules at 25 °C?
vrms =
3kT
m
On the Kelvin scale T = 25 °C = 298 K.
Element Mass (kg)
rms speed (m/s)
He
6.64×10-27
1360
H2
3.32 ×10-27
1930
N2
4.64 ×10-26
515
O2
5.32 ×10-26
482
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§13.7 Temperature and Reaction
Rates
For a chemical reaction to proceed, the reactants must
have a minimum amount of kinetic energy called activation
energy (Ea).
29
If
3
Ea >> kT
2
then only molecules in the high speed tail of MaxwellBoltzmann distribution can react. When this is the situation,
the reaction rates are an exponential function of T.
reaction rates ∝ e
−⎛⎜
⎝
Ea
⎞
kT ⎟⎠
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Example (text problem 13.76): The reaction rate for the
hydrolysis of benzoyl-l-arginine amide by trypsin at 10.0 °C
is 1.878 times faster than at 5.0 °C. Assuming that the
reaction rate is exponential, what is the activation energy?
r1 ∝ e
r2 ∝ e
−⎛⎜
⎝
Ea
−⎛⎜
⎝
Ea
⎞
kT 1 ⎟⎠
⎞
kT 2 ⎟⎠
where T1 = 10.0 °C = 283 K and T2
= 5 °C = 278 K; and r1 = 2.878 r2.
⎛ Ea
Ea ⎞
r1
⎟⎟
The ratio of the reaction rates is
= exp⎜⎜ −
+
r2
⎝ kT1 kT2 ⎠
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Example continued:
Solving for the activation energy gives:
⎛ r1 ⎞
k ln⎜⎜ ⎟⎟
r2 ⎠
⎝
Ea =
⎛1 1⎞
⎜⎜ − ⎟⎟
⎝ T2 T1 ⎠
(
1.38 × 10
=
)
J/K ln (1.878)
= 1.37 × 10 −19 J
1 ⎞
⎛ 1
−
⎜
⎟
278
K
283
K
⎠
⎝
− 23
32
§13.8 Collisions Between Gas
Molecules
On average, a gas particle will be able to travel a distance
Λ=
1
2πd 2 ( N / V )
before colliding with another particle. This is the mean free
path. The quantity πd2 is the cross-sectional area of the
particle.
33
After a collision, the molecules involved will have their
direction of travel changed. Successive collisions
produce a random walk trajectory.
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Substances will move from areas of high concentration to
areas of lower concentration. This process is called
diffusion.
In a time t, the rms displacement in one direction is:
xrms = 2 Dt
D is the diffusion constant
(see table 13.3).
35
Example (text problem 13.81): Estimate the time it takes a
sucrose molecule to move 5.00 mm in one direction by
diffusion in water. Assume there is no current in the water.
xrms = 2 Dt
Solve for t
(
(
−3
)
2
x
5.00 × 10 m
=
= 25000 s
t=
−10
2
2D 2 5.0 × 10 m /s
2
rms
)
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Summary
•Definition of Temperature
•Temperature Scales (Celsius, Fahrenheit, Absolute)
•Thermal Expansion
•Origin of Pressure in a Gas
•Ideal Gas Law
•Exponential Reaction Rates
•Mean Free Path
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