Gas molar specific heats
Mean kinetic energy of a gas molecule:
1 2 3
K = mv = kT
2
2
1 2
3
3
If we have n moles of gas: U = nN A ! ( mv ) = nN A kT = nRT
2
2
2
1 !U 3
Then molar specific heat at constant
Cv =
= R
volume should be:
n !T 2
What molar specific heats, Cv, do we get experimentally?
Monatomic gases: He, Ne, Ar:
3
Cv = R
2
Diatomic gas molecules: H2, O2, N2: Cv
5
= R
2
Polyatomic gas molecules: NO2, SF6, C2H5OH:
Therefore, the adiabat exponents are γ
Cv = 3 R
= Cp/Cv = (Cv+R)/Cv is 1.67, 1.4,
and 1.33 for monatomic, diatomic, and polyatomic molecules.
Gas molar specific heats
Equipartition theorem:
When a system is in
thermodynamic equilibrium the
average energy per molecule is
½·kT per each degree of
freedom.
It means that the molar specific
heat is ½·R per degree of
freedom.
Monatomic molecules only
have 3 translational degrees of
freedom.
Diatomic molecules have 3 translational plus 2 rotational – a total of 5.
Polyatomic molecules have 3 translational and 3 rotational – a total of 6.
Is this the entire story?
Not really!!!
It takes a finite temperature to “activate”
rotational degrees of freedom.
For H2, the 2 rotational degrees of
freedom get activated at ~100 K +kT in
molar specific heat at const. volume.
Below that temperature, H2 behaves as a
monatomic gas
At still higher temperatures, you activate
further degrees of freedom, which are
due to oscillations of the atoms along the
axis connecting the dumbbell: an addition
of 2 degrees of freedom and another kT
in Cv at ~1000 K.
Reversibility.
Where do we find reversible processes?
In mechanics –
• elastic collisions;
• oscillations with no friction; http://www.myphysicslab.com/pendulum1.html
• rotation of planets…
No mechanical energy is dissipated into heat-internal energy!
You can run the movie back and it will still be a plausible process.
Irreversibility.
Where do we find irreversible processes?...
Pretty much everywhere, @#%$@!..
And we are not getting any younger either!..
You can’t possibly run
that movie back…
Losing, breaking,
destroying, saying stupid
things….
Seriously.
Three common scenarios of irreversibility in thermodynamics.
1) Mixing and loosing structural order in general. Two molecularly
mixed fluids never “unmix”. http://mutuslab.cs.uwindsor.ca/schurko/animations/irreversibility/happy.htm
A broken vase never repairs itself…
2) Conversion of mechanical energy into internal energy (dissipation
into heat).
Ordered motion of an object is converted into disordered motion of its
molecules. Never coming back…
http://mutuslab.cs.uwindsor.ca/schurko/animations/secondlaw/bounce.htm
3) Heat transfer from a hotter to a cooler object – never goes in the
opposite direction.
Entropy – the story of lost opportunities...
Gas expands without doing
any mechanical work
vs.
There was an
opportunity for a
spontaneous
process – heat
flow from Th to Tc.
Heat transfer between a hot and
cold object without mechanical
work done.
What do these two processes have
in common?
Spontaneous (NOT quasi-static) expansion of a
gas and heat transfer between two objects with
different temperatures are both irreversible
processes – lost opportunities.
What kind of simple reversible processes do we have in stock?
T = const
isothermal
Thermal reservoir
with constant
temperature
No heat transfer
at all.
!U = 0
W =Q
Q=0
W = " !U
adiabatic
How do we convert internal energy or heat into work?
We build a heat engine.
"U = Q ! W
PV = nRT W = ! PdV
Isothermal engine
!U = 0
W =Q
100% of the heat
transferred to the system
is converted to work….
& V2 #
W = nRT ln$$ !!
% V1 "
In principle one can get an unlimited amount of
work…
BUT it will require an infinitely large
expansion!
What are we going to do after the gas
expands? Run it back?
Isothermal engine
W = ! PdV
& V2 #
W = nRT ln$$ !!
% V1 "
As the system expands all the heat transferred
to the system is converted to work….
W<0
W>0
As the system
contracts back,
though, the same
amount of work is
done by the
surroundings and all
the energy is
returned to the
thermal reservoir.
Adiabatic engine
W = ! PdV
Q=0
W = "!U
The positive work is now limited by the internal
energy of the insulated system.
But again, no net work is done if you go back
and forth along the same adiabat.
W<0
W>0
We need an engine working in cycles and converting heat supplied
from the outside into mechanical work, possibly, with a high efficiency…
How efficient can it be?
The isothermal engine could convert
100% heat into work, but did not work
cyclically.
Can we match this performance with an
engine operating in cycles?
Any fundamental law prohibiting it?
The second law of thermodynamics
(Kelvin-Plank statement):
It is impossible to construct a heat engine
operating in a cycle that extracts heat
from a reservoir and delivers and equal
amount of work.
It is impossible to construct a heat engine operating in a cycle that
extracts heat from a reservoir and delivers and equal amount of work
That would be an ideal heat engine…
What is a real heat engine doing?
• Works between two temperatures, a hot reservoir and a cold reservoir.
(Hot side and cold side).
• Gets some heat Qh (obtained from,
say, burning a fuel) from the hot side
• Rejects some heat Qc to the cold side.
• Works in a cycle, so that the internal
energy does not change, ΔU=0.
• Does work W
= Qh - Qc
• Has an efficiency e
= W/Qh
W Qh ! Qc
e=
=
Qh
Qh
Carnot cycle
http://www.ntu.edu.sg/home2000/S7231633I/
A-B
Carnot cycle
!U AB = 0 WAB = Qh > 0
A
B-C
Q = 0 WBC = " !U BC > 0
B
C-D
D
C
!U CD = 0 WCD = !Qc < 0
D-A
What is the total work by the gas?
Q = 0 WDA = " !U DA < 0
!U AB + !U BC + !U CD + !U DA = 0 " !U BC + !U DA = 0
WBC + WDA = "( !U BC + !U DA ) = 0
Wtotal = WAB + WCD = Qh ! Qc