Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
10.1.1 State the equation of state for an ideal
gas.
10.1.2 Describe the difference between an ideal
gas and a real gas.
10.1.3 Describe the concept of the absolute zero
of temperature and the Kelvin scale of
temperature.
10.1.4 Solve problems using the equation of state
of an ideal gas.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
State the equation of state for an ideal gas.
PV = nRT
R = 8.31 J/mol·K
(universal gas constant)
equation of
state of an
ideal gas
We will explain what an ideal gas is soon.
The variables P (pressure), V (volume), n (number
of moles), and T (absolute temperature) are all
called the state variables.
Don’t forget the following:
P = F/A
pressure
NA = 6.021023 molecules/mole
Avagadro’s
number
FYI
Topic 10 is an extension of Topic 3. Review it!
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Describe the difference between an ideal gas and
a real gas.
Recall the properties of an ideal gas.
The molecules that make up an ideal gas…
…are identical perfect spheres.
…are perfectly elastic – they don’t lose any
kinetic energy during their collisions with
each other, or the walls of their container.
…have no intermolecular forces – their
potential energy does not change.
…are so small that their volume is much smaller
than the volume of their container.
FYI
This is the kinetic model of an ideal gas.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Describe the difference between an ideal gas and
a real gas.
For a real gas…
The molecules that make up a real gas…
…are not identical perfect spheres.
…are not perfectly elastic.
…have intermolecular forces.
…are relatively large.
FYI
Ideal gases cannot be liquefied, whereas real
gases can.
Real gases have intermolecular forces.
Real gases are often polyatomic (N2, O2, H2O, NH4,
etc.) and thus not spherical.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Describe the concept of the absolute zero of
temperature and the Kelvin scale of temperature.
When gas is heated in an enclosed
space its pressure increases.
The following experiment plots
pressure vs. temperature in Celsius.
We can extrapolate the graph.
Now we repeat using different gases.
p
10 20
0
-300
-200 -100
0
100
200
300
T (°C)
30
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Describe the concept of the absolute zero of
temperature and the Kelvin scale of temperature.
The lowest pressure that can exist is zero.
Perhaps surprisingly, the temperature at which
any gas attains a pressure of zero is the same.
The Celsius temperature at which the pressure is
zero (for all gases) is -273 °C.
p
-273.15 °C
-300
-200 -100
0
100
200
300
T (°C)
Topic 10: Thermal physics
10.1 Thermodynamics
-273°C
-300
-200 -100
0
100
200
300
T (°C)
273 K
KELVIN SCALE
-273.15 °C
Absolute
zero
CELSIUS SCALE
Gas Laws
Describe the concept of the absolute zero of
temperature and the Kelvin scale of temperature.
Because the lowest pressure that can exist is
zero, this temperature is the lowest temperature
that can exist, and it is called absolute zero.
A new temperature scale that has
0° C
absolute zero as its lowest value
is called the Kelvin temperature
scale.
p
0K
Topic 10: Thermal physics
10.1 Thermodynamics
-273°C
KELVIN SCALE
0° C
Absolute
zero
FYI
Note that there is no degree
symbol on Kelvin temperatures.
EXAMPLE: Convert 100°C to Kelvin,
and 100 K to C°.
SOLUTION:
T(K) = T(°C) + 273
T = 100 + 273 = 373 K
 100 = T(°C) + 273
T = -173°C
CELSIUS SCALE
Gas Laws
Describe the concept of the absolute zero of
temperature and the Kelvin scale of temperature.
The simple relationship between the Kelvin and
Celsius scales is given here:
T(K) = T(°C) + 273
relation between Kelvin and Celsius
0K
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
An ideal gas satisfies the criteria of the
kinetic theory of gases…
(1) All of its molecules are perfect identical
spheres.
(2) All of its collisions are perfectly
elastic.
(3) It has no intermolecular forces.
(4) Its molecules are extremely small compared
to the container.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
For an ideal gas use PV = nRT.
We want n, the number of moles.
P = 20106 Pa, V = 2.0 10-2 m3.
From T(K) = T(°C) + 273
T(K) = 17 + 273 = 290 K.
Then n = PV/(RT)
n = (20106)(210-2)/[(8.31)(290)]
n = 170 mol.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
Use NA = 6.021023 atoms/mol.
N = 170 mol 
6.021023 atoms
mol
N = 1.01026 atoms.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
V = 2.010-2 m3.
N = 1.01026 atoms.
V/atom = 2.010-2 m3 / 1.01026 atoms.
V/atom = 2.010-28 m3/atom.
Assuming a cubic volume X3…
x = (2.010-28)1/3 = 5.810-10 m.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
Under high pressure or low volume real gases’
intermolecular forces come into play.
Under low pressure or large volume real gases’
obey the equation of state.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
Fixed mass and constant volume means n and V
are constant. Thus
PV = nRT  P = (nR/V)T  P = (CONST)T. (LINEAR)
Since the t axis is in ºC, but T is in Kelvin,
the horizontal intercept must be NEGATIVE…
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
Temperature is a measure of the EK of the gas.
Reducing the EK reduces the frequency of collisions.
For perfectly elastic collisions (as in an ideal
gas) contact time is zero regardless of EK.
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
Use PV = nRT.
PV = nRT  P = nRT/V  P = (CONST)1/V.
Thus P  1/V. (Inversely proportional.)
PV = nRT  V = (nR/P)T  V = (CONST)T.
Thus V  T. (Directly proportional.)
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
Use PV = nRT.
From T(K) = T(°C) + 273
Ti = 30 + 273 = 303 K.
Tf = 330 + 273 = 603 K.
Vi = Vf.
PiVi
PfVf
PiVi
Pf
Pf
= nRTi, PfVf = nRTf.
nRTf
= nRT
i
= PiTf/Ti
= (6)(603)/303 = 12
Topic 10: Thermal physics
10.1 Thermodynamics
Gas Laws
Solve problems using the equation of state of an
ideal gas.
P = 0 at
absolute
zero.
From PV = nRT:
P = (1R/V)T: