Today’s Lecture
Specific Heats
Cyclic Processes
Second Law of Thermodynamics
Carnot Cycle
Four Processes Between Two Isotherms
How do you order them in terms of heat transferred to the gas?
Things to remember:
U depends on T only!
W = ∫ PdV
Q = ΔU + W
The largest work done
by the gas will require
the most heat transfer!
What is the heat transferred to the gas in process 3?
Q = ΔU + W
W = ∫ PdV
ΔU = nCv ΔT
Summary of Temperature Changes
Process 1 – Adiabatic
Q = 0 => ΔU = -W => T1 < Ta
Process 2 – Isochoric
W = 0 => Q = nCvΔT => T2 < Ta
Process 3 – Isobaric
P3 = Pa => Q = nCpΔT => T3 > Ta
Process 4 – Isothermal
T4 = Ta => ΔU = 0 => T4 = Ta
Adiabatic Processes
Q=0
ΔU = −W
We can also find the work done by an
ideal gas from:
For an adiabatic expansion PVγ = const. Hence the work integral is:
Ti
Tf
Vi
Vf
This result is consistent with W = -ΔU as it
had to be!
Example: Isothermal and Adiabatic Expansion
Consider expanding .05 mol of He from
2.5atm to 1atm. Starting at T=295K. Find
the final temperature and volume for
both an adiabatic and isothermal process.
From the ideal gas law the initial volume is:
For He γ = 5/3.
For isothermal expansion the volume increases by 2.5 or Vf = 1.21L.
For adiabatic expansion:
Specific Heats of an Ideal Gas
From kinetic theory we showed the average (translational) kinetic energy
per molecule is:
For n moles the
internal energy due
to KE is:
This means we
can find Cv :
For inert gases we have:
Specific Heats of an Ideal Gas
However for diatomic gases,
nitrogen, oxygen, etc
What’s up??
Internal energy may be more than just translational kinetic energy!
Equipartition Theorem:
In thermodynamic equilibrium, the
average energy per molecule is 1/2kT
for each degree of freedom
This means Cv=R/2 for each degree of freedom
For a diatomic molecule there are 5 degrees of freedom
Summary of the First Law of Thermodynamics
First Law of Thermodynamics
Thermodynamic processes,
Quasi-static - work is given by
Isothermal – constant temperature
Isochoric – constant volume
Isobaric – constant pressure
Adiabatic – no heat transfer
Equipartition theorem –
1/2kT average molecular energy
for each degree of freedom
Cyclic Processes
What is so special about them?
The system returns to the same point (state)
P
What can we say about change ΔU, Q and W?
Since the system returns to the same state and
V
U is a function of state only, ΔU = 0!
By the 1st law ΔU = Q – W and we have
Q = W.
A
B
That is the system gets heat Q from the
outside and does an equal amount of work W.
Or the other way around! Both Q and W may
be negative!
What simple thermodynamic process are cyclic processes similar to?
In what are they different?
Cyclic Processes
Since ΔU = 0, all we have to do is calculate W. How do we do that?
In the simplest case we can divide the process into two parts:
expansion A → B, with positive work done;
contraction B → A with negative work done.
WAB > 0
WBA < 0
Work Done in a Cycle
Determine the work done by an
Ideal gas in the cycle ACB.
Examine this cycle one process at
a time and them sum them up.
Example: Work by a Gas with Isothermal Expansion
At point B an ideal gas under a pressure of
250kPa has a volume of 1L. The gas expands
along an Isotherm to a volume of 5L.
Assume γ=1.4.
Find the work done by the gas in the cycle BACB. First the work done along BA,
The work done by the gas along AC,
Since the work along CB is zero the total work during the cycle is
Example: Work by a Gas with Adiabatic Expansion
At point B an ideal gas under a pressure of
250kPa has a volume of 1L. The gas expands
along an adiabat to a volume of 5L.
Assume γ=1.4.
First we need the pressure at A.
The work along BA is
The net work is
An Adiabatic, Isothermal Cyclic Process
A gas occupies 4L at 300K and
100kPa. It is compressed adiabatically
to 1L (red) and then cooled at
constant volume to 300K (black).
Then it is allowed to expand
isothermally to 4L (green). How much
work is done by the gas in the cyclic
process ABCA?
Adiabatic compression:
An Adiabatic, Isothermal Cyclic Process
A gas occupies 4L at 300K and
100kPa. It is compressed adiabatically
to 1L (red) and then cooled at
constant volume to 300K (black).
Then it is allowed to expand
isothermally to 4L (green). How much
work is done by the gas in the cyclic
process ABCA?
Isothermal expansion:
The total work done by the gas in the cycle is:
Why is it negative??
A Circular Cyclic Process
An ideal gas is taken through a
circular cyclic process, shown.
(a) How much work does the gas do?
(b) If there are 1.3 moles of gas, what are
the max and min temperatures?
The expressions for the pressure and volume are:
The maximum and minimum temperatures occur the largest and smallest
distance from the origin. This occurs when θ = π/4 and 5π/4 respectively or:
Question:
The gas goes from the state i to the state f.
What is the minimum of work that the gas
can do?
What if the pressure is not allowed to drop
below Pf ?
Any other suggestions?
1
2
We can start at i , go to 1, make, say 10 circles,
go to 2 and only then continue to f.
10 loops
1
W1→2→1 < 0
2
Wi→1→2→ f > 0
Wnet = 10 ⋅ W1→2→1 + Wi→1→2→ f < 0
With 10 loops, where the work is negative, the net work is
certain to be negative!
To convert internal energy or heat into work?
We build a heat engine.
ΔU = Q − W
PV = nRT W = ∫ PdV
Isothermal engine
ΔU = 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
reservoir.
Adiabatic Engine
W = − ΔU Q = 0
W = ∫ PdV
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
Ti
Tf
W>0
Vi
Vf
We need an engine working in cycles and converting heat supplied
from the outside into mechanical work with a possibly 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://perso.orange.fr/olivier.granier/thermo/simul/carnot/simul.htm
Carnot Cycle
What is the net work and
efficiency of a Carnot cycle?
Wtotal = WAB + WCD
A
B
D
C
Since ThVBγ−1=TcVCγ-1
and ThVAγ-1=TcVDγ-1
Qc Tc
=
Qh Th
Efficiency
We have VB / VA = VC / VD and
W Qh − Qc Th − Tc
=
=
e=
Qh
Qh
Th
Carnot Cycle
Efficiency
W Qh − Qc Th − Tc
e=
=
=
Qh
Qh
Th
Irreversible engines are necessarily less efficient,
but so are many reversible engines!
In a Carnot cycle all of the heat exchange takes place
only between the highest and lowest temperatures,
hence the Carnot cycle is the most efficient cycle.
Carnot Cycle
Efficiency
W Qh − Qc Th − Tc
=
=
e=
Qh
Qh
Th
• For highest efficiency we would want to run our engine
between a very hot and a very cold reservoirs.
• Large temperature difference, Th-Tc, and low temperature of
the cold reservoir, Tc, are very helpful.
• Efficiency can in principle reach 100% for Tc = 0, but we
normally do not have such reservoirs available…
Example: Carnot Cycle
During one cycle a Carnot engine
extracts 890J from a 550K reservoir
and rejects 470J to a cooler reservoir.
(a) How much work does the engine do during each cycle?
(b) What is the efficiency of the engine?
(c) What temperature is the cool reservoir?
Example: Carnot Cycle
A Carnot engine operates between
Helium’s melting point and its
boiling point at 4.25K. It has an
efficiency of 77.7%.
At what temperature does Helium melt?