Theoretical Physics
Prof. Ruiz, UNC Asheville, doctorphys on YouTube
Chapter G Notes. The Ideal Gas Law and Thermodynamics
G1. The Ideal Gas Law. The following apply to the ideal gas.
-------------------------------------------------------------------------------------------------------------------Boyle's Law. Pressure of a gas is inversely related to volume at constant temperature.
P∼
1
V
and
P1 V1 = P2 V2
as constant temperature.
When temperature is used with the equations in this section,
the Kelvin scale is used, i.e., T = Celsius Temperature − 273.
Robert Boyle (1627-1691)
Images from the School of Mathematics and Statistics, Univ. of St. Andrews, Scotland
-------------------------------------------------------------------------------------------------------------------Charles's Law. Volume of a gas is proportional to temperature at constant pressure.
V ∼T
and
V1 V2
=
T1 T2
as constant pressure
and we use the absolute Kelvin temperature scale. Consider
this as a definition for the absolute temperature scale: as
you cool the gas down at constant pressure, the volume
shrinks to zero as temperature goes to zero.
Jacques Charles (1746-1823)
-------------------------------------------------------------------------------------------------------------------Gay-Lussac's Law. Pressure of a gas is proportional to temperature at constant volume.
P ∼T
and
P1 P2
=
T1 T2
at constant volume.
Think of this as an alternative definition for temperature on
the absolute scale. As you lower the pressure at constant
volume, the temperature lowers, both heading towards zero.
Joseph Gay-Lussac (1778-1850)
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We can incorporate all these laws in the form:
P1 V1 P2 V2
=
T1
T2
The chemists write the ideal gas law as
The physicists often like to write
The definition of the mole is
n=
N
NA
, where
Since PV = nRT =
chemist's constant as
N A = 6.022 x10 23 is called Avogadro's number.
NkT = nN A kT
k=
, the physicist's constant is related to the
R
NA
.
G2. "Derivation" of the Ideal Gas Law
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The ideal gas law is an excellent example of a combination of experimental results. We
will attempt to "derive" this law since our course is called "Theoretical Physics."
Sir Isaac Newton (1642-1727)
Let's see what we can derive from Newton's Second
Law
F = ma .
Bur first, let's summarize what have we accomplished so
far in our theoretical analysis of fundamental laws.
We started with the following: Newton's Second Law, Newton's Universal Law of
Gravitation, Coulomb's Law, and Special Relativity.
FG =
F = ma
GMm
r2
FE =
kQq
r2
" x = ct "
We used the first, third, and fourth to derive the Maxwell equations. From the Maxwell
equations we derived the existence of electromagnetic waves, i.e., we obtained optics.
F = ma
FE =
kQq
r2
" x = ct "
=> E&M and Optics
Now we will attempt to arrive at thermodynamics from Newton's Second Law.
F = ma
=> Thermodynamics
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We will consider a box containing a gas. Below is the basic idea of a discrete
distribution of velocities and a continuous one.
The total probability must be 1 in each case, i.e., the probability that a given gas particle
has some velocity is 1. Therefore, for the discrete case,
Ni
∑i N = 1 , which is consistent with N = ∑i N i .
For the continuous case, the total area under the curve must be 1.
∫ f (v) dv = 1 .
Classically, one integrates from zero to infinity. We are justified in doing this even
knowing relativity since our particles no way approach the speed of light. Therefore, our
function f(v) will drop to zero before we even get close to the speed of light and it is
easier to integrate to infinity when you are dealing with exponential-type functions. We
will not get into such details here anyway and at times leave the integration limits off. It
is understood that you integrate over all velocities.
We will consider a continuous distribution of velocities for our gas particles. The neat
thing about our analysis is that we will never need to worry about the exact form of the
function f(v). Watch!
Consider a box with N particles and volume V . The pressure is defined as force per
unit area and the force is the change in momentum with respect to time:
P=F/A
and
F = dp / dt .
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A slanted region is shown below where some particles are heading towards the area
shown on the floor. From the geometry at the lower right we will determine the
probability for a particle to be traveling along the slanted region at velocity v.
The volume element below is
∆A cos θ dl . We need another piece. The change in
momentum when a particle coming in along the slant with velocity v bounces off the
bottom floor. The change in momentum is
2mv cos θ
.
The following is from Derek L. Livesey, Atomic and Nuclear Physics (Waltham, MA,
Blaisdell Publishing, 1966). Let the probability distribution for the angle
∫
π /2
0
dP =
g (θ ) dθ = 1
g (θ ) dθ =
θ
be
g (θ ) .
2π r sin θ r dθ
= sin θ dθ
2π r 2
.
N
1
2mv cos θ  1
[ ∆A cos θ dl ]  f (v) dv  [sin θ dθ ] 
 ∆A
V
dt
2


dP =
N
V
dl   1


f
v
dv
cos
θ
(
)

 [ sin θ dθ ][ 2mv cos θ ]
dt   2
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dP =
Nm  dl 
2

v
f
(
v
)
dv
cos
θ sin θ dθ 
[
]

V  dt 
dP =
Nm 2
v f (v ) dv  cos 2 θ sin θ dθ 
V
π /2
Nm ∞ 2
P=
v f (v ) dv ∫ cos 2 θ sin θ dθ
∫
0
V 0
π /2
Nm 2  cos 3 θ 
P=
v −

V
3

0
P=
P=
Nm 2 
1 
v  −0 − ( − ) 
V
3 

N
1
mv 2
V
3
and
PV =
N
mv 2
3
We can define temperature by the total kinetic energy of the gas particles. This total
energy is equal to the average kinetic energy the particles times the number of particles,
mv 2
KE = N
2
.
Since we want the assignment
PV =
N
mv 2 = NkT
3
, we find
1 2 3
mv = kT
2
2
In summary, we have
PV = nRT
and
PV = NkT
with
1 2 3
mv = kT
2
2
.
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.
G3. The First Law of Thermodynamics and Specific Heats
We will phrase the first law of thermodynamics in terms of our ideal gas. The first law of
thermodynamics is the law of conservation of energy: The change in energy of a system
is equal to the heat that flows into the gas minus the work that is done by the gas.
∆U = ∆Q − ∆W
We can therefore also write the first law as
∆U = ∆Q − ∆W
Here is an application using the definition of the specific heat at constant volume.
cV ≡
∆Q = ∆U + P ∆V
U =N
cV ≡
1 ∆Q
n ∆T V
and
∆QV = ∆U
1 2 3
3
mv = NkT = nRT
2
2
2
1 ∆Q
1 ∆Q 1 3 nR∆T 3
=
=
= R
n ∆T V n ∆T n 2 ∆T
2
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Here is an application using the definition of the specific heat at constant pressure.
cP ≡
1 ∆Q
n ∆T
P
3
nR∆T + P∆V = ncV ∆T + P∆V
2
Since we want constant pressure, it is better to have a ∆P in the above. We note
∆Q = ∆U + P∆V =
PV = nRT
Then
∆ ( PV ) = nR∆T = P∆V + V ∆P .
and
P∆V = nR∆T − V ∆P
and
∆Q = ncV ∆T + nR∆T − V ∆P = n(cV + R ) ∆T − V ∆P .
cP ≡
1 ∆Q
n ∆T
= cV + R
∆P = 0 '
since
P
.
Summary
PV = nRT
PV = NkT
and
U =N
1 ∆Q
n ∆T V
,
with
.
1 2 3
3
mv = NkT = nRT
2
2
2
∆U = ∆Q − P ∆V
cV ≡
1 2 3
mv = kT
2
2
cP ≡
For an ideal gas we have
∆Q = ∆U + P ∆V
and
1 ∆Q
n ∆T
, and
cP = cV + R
P
cV =
3R
2
and
cP =
5R
2 .
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G4. Four Thermodynamic Processes
1. Isometric, Isochoric (constant volume):
∆V = 0 .
W = ∫ P dV = 0
2. Isobaric (constant pressure):
∆P = 0 .
V2
V2
V1
V1
W = ∫ P dV = P ∫ dV = P (V2 − V1 )
∆T = 0 .
3. Isothermal (constant temperature):
V2
V2
V1
V1
W = ∫ P dV = ∫
V2 1
nRT
dV = nRT ∫
dV
V
1
V
V
V
W = nRT ln V V2 = nRT ln
1
V2
V1
4. Adiabatic (no heat flow): ∆Q = 0 .Note that we can't say constant heat since heat is
not a "regular" variable like P, V, and T. Heat exchange, like work, depends on a path
we take in the PV plane. Heat and work are not intrinsic properties of the gas. But
energy is. From the earlier section, write
∆Q = ∆U + P∆V = ncV ∆T + P∆V
. Remember U =
3
nRT
2
.
∆Q = ncV ∆T + nR∆T − V ∆P = n(cV + R ) ∆T − V ∆P
These equations are
∆Q = ncV ∆T + P∆V
For
∆Q = 0
and
∆Q = ncP ∆T − V ∆P .
we can write
ncV ∆T = − P∆V
and
ncP ∆T = V ∆P .
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ncP ∆T
V ∆P
=−
ncV ∆T
P ∆V
γ=
We define
cP
cV
. Then, γ = −
cP
V dP
=−
cV
P dV
leads to
.
V dP
P dV
dP
dV
= −γ
P
V
∫
P2
P1
V2 1
1
dP = − γ ∫
dV
V1 V
P
P
V
ln P P2 = −γ ln V V2
1
ln
1
P2
V
= −γ ln 2
P1
V1
V 
P
ln 2 = ln  2 
P1
 V1 
P2  V2 
= 
P1  V1 
−γ
P1 V1γ = P2 V2γ
and
, i.e.,
−γ
P2  V1 
= 
P1  V2 
γ
PV γ = const
PG1 (Practice Problem). Show that the work done by a gas in an adiabatic expansion
from volume V1 to V2 is
V2− γ +1 − V1− γ +1  P2 V2 − P1 V1
W = const 
=
1
−
γ
1− γ


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