Download Thermodynamics – The Laws of Thermodynamic

The Laws of Thermodynamic
Zeroth Law
If A is in thermal equilibrium with C and B in thermal
equilibrium with C then A and B have to be in thermal
equilibrium. No heat flows!
First Law of Thermodynamics (or energy conservation)
Principle of Conservation of Energy:
Energy can neither be created nor destroyed, only transformed
Energy may be transformed from one form to another, but the total
energy of any body or system of bodies is a quantity that can be
neither increased nor diminished
How can we change the internal energy of a system?
1. Add or remove heat
2. Work done on the system or from system on the external world
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First Law of Thermodynamics
The change in a systems internal energy is related to the heat and the
work.
∆U= U2 - U1 = Q - W
Where:
U1 = internal energy of system @ start
U2 = internal energy of system @ end
Q = net thermal energy flowing into system
during process
Positive when system gains heat
Negative when system loses heat
W = net work done by the system
Positive when work done by the system
Negative when work done on the system
The state of a system is described by its pressure, temperature and
volume.Thermal processes can change the state of a system.
We assume that thermal processes have no friction or other
dissipative forces.
In other words: All processes are reversible
(Reversible means that it is possible to return system and surroundings to the
initial states)
REALITY: irreversible
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Processes with constant pressure (Isobar)
F = P0 A
W = Fs = P0A(xf-xi) = P0(Vf-Vi)= P0∆V (=area under PV-plot)
∆U= Q - P∆V
Processes with constant volume (Isochore)
W=0 (no change of volume)
∆U= Q
Processes with constant temperature (Isotherm)
T const. Æ no change in the internal energy
PV = NkT = constant
Area under PV-curve (calculus)
W = NkT ln(Vf/Vi) = nRT ln(Vf/Vi)
Q=W
∆U= 0
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Adiabatic (“not passable”) processes
(no heat is gained or lost by the system Q=0, i.e. system perfectly isolated )
Q=0
∆U= -W
PV = constant (isothermal)
PVγ = constant (adiabatic)
Specific Heats under constant pressure and constant volume
(amount of heat depends on type of process)
Specific heat
Q = m c ∆T
Molar specific heat
Q = n C ∆T
Constant Volume: CV
Constant Pressure : CP
Which one is larger?
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With ∆U= = Q – W and Q = n C ∆T
e.g for a monoatomic gas
∆U = 3/2 n R ∆T
Constant volume: ∆U= Q
3/2 n R ∆T = n CV ∆T
CV= 3/2 R
Constant pressure: Q = ∆U + W
n CV ∆T = 3/2 n R ∆T + n R ∆T
CP= 5/2 R
CV – CP = R (always valid for any ideal gas)
Calculus shows that γ = CP/CV
For a monoatomic gas therefore γ = 5/3
PVγ = constant
Or
P = const. 1/ Vγ
Summary – Thermal processes
Constant pressure
W = P∆V
Q = ∆U + P∆V
Constant volume
W=0
Q = ∆U
Isothermal (constant temperature) W = Q
∆U = 0
Adiabatic (no heat flow)
Q=0
W = –∆U
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The second law of thermodynamics
(the entropy law or law of entropy)
Heat can never pass spontaneously from a colder to a hotter body. As a result of
this fact, natural processes that involve energy transfer must have one direction, and
all natural processes are irreversible. This law also predicts that the entropy of an
isolated system always increases with time. Entropy is the measure of the disorder
or randomness of energy and matter in a system. Because of the second law of
thermodynamics both energy and matter in the universe are becoming less useful as
time goes on. Perfect order in the Universe occurred the instance after the Big Bang
when energy and matter and all of the forces of the Universe were unified.
As we know from other processes
Physicist Lord Kelvin stated it technically as follows: "There is no
natural process the only result of which is to cool a heat reservoir and do
external work."
Convert heat into mechanical work:
Heat engines
W = Qh – QC
Efficiency: e = W/Qh = (Qh – QC)/ Qh = 1 - QC/ Qh
(e.g. a efficiency of e=0.8 means 80% of the heat is converted to mechanical
work)
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Carnot’s Theorem/Principle
To get maximal efficiency is must be a heat engine where all processes are
reversible. An irreversible heat engine operating between two heat reservoirs at
constant temperatures cannot have efficiency greater than that of a reversible
heat engine operating between the same two temperatures. Reversible engines
operating between the same temperatures have the same efficiency.
Maximum efficiency of a heat engine: e = = 1 - TC/ Th
Maximum work done by a heat engine: Wmax=emaxQh = (1 - TC/ Th) Qh
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