The Laws of Thermodinamics

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Transcript The Laws of Thermodinamics

The Laws of Thermodinamics
• Work in Thermodynamic
process
• Energy can be
transferred by heat and
by work done on the
system
• W=-FΔy =-PA Δy
• The work W done on a
gas at constant
pressure is given by:
W= -P ΔV
P-pressure through the
gas
ΔV- the change in volume
W= -P ΔV –can be used to
calculate the work done on
the system only when the
pressure of the gas remain ct.
during the expansion or
compression
ISOBARIC PROCESS- a
process in which the
pressure remain constant
• The area under the graph in
a PV diagram is equal in
magnitude to do work done
on a gas: A=P ΔV
• The first law of thermodynamics
(energy conservation law that relates
changes in internal energy): If a system
undergoes a change from an initial
state to a final state, where Q is the
energy transferred to the system by
heat and W is the work done on the
system, the change in the internal
energy of the system, ΔU is given by:
ΔU= Uf-Ui = Q+W
• Q>0- when energy is transferred into the system
by heat
• Q<0- when energy is transferred out of the
system by heat
• W>0 –when work is done on the system
• W<0 – when the system does work on its
environment
• The internal system of any isolated system
must remain constant, ΔU=0
• If the system isn’t isolated, ΔU=0if the system
goes through a cyclic process, which P, V, T, n
(nr of moles) return to original values
• U=3/2 n R T
• ΔU=3/2 n R ΔT
• Molar specific heat at V=ct. of a
monatomic ideal gas, Cv is: Cv =3/2 R
• ΔU=n Cv ΔT
• Q= ΔU-W= ΔU+ P ΔV
• P ΔV=n R ΔT
• Q= 3/2n R ΔT+ n R ΔT= 5/2 n R ΔT
• Molar heat capacity at P=ct., Cp is:
Cp=5/2R
• Q=n Cp ΔT
• Cp = Cv +R
• Adiabatic processno energy enters or
leaves the system
by heat (can still do
work)
• Q=0
ΔU=W
• For an ideal gas
undergoing an
adiabatic process:
PVγ =ct
γ= Cp / Cv
γ - adiabatic index of
the gas
• Isothermal process- the
temperature of the
system doesn’t change
• ΔT=0; ΔU=0; W=-Q
• For an ideal gas at T=ct
P=nRT/V
• The work done on the
environment during an
isothermal process is given
by: Wenv=nRTln(Vf/Vi)
• The PV diagram of a
typical isothermal process ,
contrasted with an
adiabatic process
• A heat engine takes energy
by heat and converts into
electrical or mechanical
energy
• A heat engine carries some
working substance through
cyclic process during which:
• 1. energy is transferred by
heat from a source at a high
temperature
• 2. work is done by the engine
• 3. energy is expelled by the
engine by heat to a source at
lower temperature
• The absorbs energy Qh from a hot reservoir,
does work Weng , that gives up energy Qc to the
cold reservoir
• W= -Weng (work is done on the engine)
• ΔU=0=Q+W
Q net=-W =Weng
• The work done by the engine= the net energy
absorbed by the engine
• Q net=|Qh|-|Qc|
• The thermal efficiency e of a heat engine is
the work done by the engine divided by the
energy absorbed during one cycle:
• e= Weng / |Qh|=|Qh|-|Qc|/ |Qh|=1-|Qc|/|Qh|
• e=1- engine has a 100% efficiency
The heat pumps and refrigerator
The coefficient of performance
for a refrigerator or an air
conditioner (the cooling
mode):
COP=|Qc|/W
• The coefficient of
performance of a heat
pump operating in a heat
mode :
COP=|Qh|/W
THE SECON LAW OF THERMODINAMICS:
No heat engine operating in a cycle can
absorb energy from a reservoir and use
it entirely for the performance of na
equal amount of work (e<1)
Ist law: we can’t get a greater amount of
energy out of a cyclic process that we put
in
2nd law: we can’t break it even
• Reversible process- every state along the
path in an equilibrium state, so the system
can return to its initial conditions by going
along the same path in the reverse
direction.
• Irreversible process- a process that
doesn’t satisfy this requirement
• Most natural processes are known to be
ireversible
• The Carnot Engine: a
heat engine operating
an ideal, reversible
cycle (the Carnot cycle)
between two reservoirs
is the most efficient
engine possible
• Carnot’s theorem: no
real engine operating
between 2 energy
reservoirs can be more
efficient than a Carnot
engine operating
between the same 2
reservoirs
Carnot
Cycles:
• An ideal gas is contained in a cylinder with a
movable piston at one end
• 1. The process A→B is an isothermal expansion
at temperature T (the gas is placed in thermal
contact with a hot reservoir, and the gas absorbs
energy Qh from the reservoir and does work WAB
in raising the piston)
• 2. The process B→C , the gas expends
adiabatically (energy enter or leave by heat).
The temperature falls from Th to Tc, and the gas
does work WBC to the raising piston
• 3. The process C→D, the gas is placed in
thermal contact with a cold reservoir at
temperature Tc and is compress
isothermally at temperatures Tc (the gas
expels energy Qc to the reservoir and the
work done on the gas is WCD)
• 4. The process D→A , the gas is
compressed adiabatically (the temperature
increases to Th, and the work done on the
gas is WDA)
• |Qc|/|Qh|=Tc/Th
• e=1-Tc/Th
• All carnot engines operating
reversibly between the same 2
temperatures have the same
efficiency
The third Law of Thermodynamics:
its impossible to lower the
temperature of a system to
absolute zero (such reservoirs
are not available)
All real engines operate irreversibly,
due to friction and the brevity of
their cycles, are less efficient than
the Carnot engine
• Entropy
• Let Qr be the energy absorbed or expelled
during a reversible, constant temperature
process between two equilibrium states.
Then the change in entropy during any
constant temperature process connecting
the two equilibrium states is defined as:
ΔS =Qr/T
SI unit: J/K (Joules/Kelvin)