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Transcript
Chapter 10: Thermodynamics
10-1 Relationship Between Heat and
Work
• In a closed system there’s a direct relationship
between heat and work. Heat and work both
transfer energy to or from a system.
• Key idea: A system never has “heat” or
“work”, it has internal energy which is
affected by heat in/out or work done on/by the
system
Transfer of heat and work
• System: a set of particles or interacting
particles considered to be a distinct physical
entity
• Environment: the combination of conditions
and influences outside a system that affects the
behavior of the system
• Ex of closed systems: a gas confined in a
cylinder by a piston, a calorimeter, a
thermos…
Work Done on or By a Gas
• Is represented in the equation:
W = PΔV
W - in Joules (J)
P = pressure in Pascal (Pa)
1 Pa = 1 N/m2
ΔV= volume change in (m3)
Sample Problem
• Gas in a container is at a pressure of 1.6x105
Pa and a volume of 4.0 m3. What is the work
by the gas if it expands at a constant pressure
to twice its initial volume?
Solution
P= 1.6x105 Pa =1.6x105 N/m2
∆V=Vf – Vi = 8.0 m3 - 4.0 m3 =4.0 m3
W= P ∆ V
W = (1.6x105 N/m2)(4.0 m3)
W= 6.4x105 J
Thermodynamic Processes
• Isovolumetric process: a
thermodynamic process that takes
place at a constant volume so that
no work is done on or by the
system, ex: a car with closed
windows parked in a hot garage.
• Isothermal process: a
thermodynamic process that takes
place at a constant temperature, ex
usually a slow process like a
balloon expanding slowly during
the day.
• Adiabatic process: a
thermodynamic process during
which heat energy is transferred to
or from the system. ex: usually a
fast process like filling a tank
• Isobaric process: a process that
takes place at a constant pressure.
ex: heating an open pot of water
10.2 The First Law of Thermodynamics
• The first law is a statement of conservation of
energy that takes into account a system’s
internal energy (U) as well as the energy
transfer to/from the system by work and heat.
• It is expressed as:
Signs of Q and W For a System
ΔQ = positive if heat is added to a system
ΔQ = negative if heat is released from a system
ΔW = positive if work is done by the system
ΔW = negative if work is done on the system
First Law – Isovolumetric Process
ΔU=Q–W
ΔV = 0
Since W = P ΔV, W = 0
therefore, Δ U = Q
First Law – Isothermal Process
ΔU = Q – W
since ΔT = 0 , ΔU = 0
therefore Q = W
First Law – Adiabatic Process
ΔU=Q–W
Q = 0, therefore
ΔU=–W
First Law – Isobaric Process
ΔU=Q–W
since W = PΔV
Δ U = Q – PΔV
First Law – Isolated System
ΔU=Q–W
since Q = W = 0
ΔU=0
Sample Problem
• A total of 135 J of work is done on a gaseous
refrigerant as it undergoes compression. If the
internal energy of the gas increases by 114 J
during the process, what is the total amount of
energy transferred as heat?
Solution
W= -135 J (work done on the system is -)
∆ U= 114 J
∆U=Q-W
Q= ∆U +W
Q= 114 J + (-135 J)= -21 J
Q= -21 J
In this problem, energy is removed from the gas as
heat, which is indicated by the negative sign on the Q
value ( Q < 0 ).
Cyclic Processes
• A thermodynamic process in which a system
returns to the same conditions under which it
started (no change in system’s energy)
ΔUnet = 0 and Qnet = Wnet
• Resembles an isothermal process in that all
energy is transferred as work and heat.
The Heat Engine
• Any device that exploits a
temperature difference to do
mechanical work
• The net work done is equal to
the difference in energy taken
in as heat from a high-temp.
reservoir (Qh) and the energy
expelled as heat to the low
temp. reservoir(Qc).
• Wnet = Qnet = Qh − Qc
10.3 The Second Law of Thermodynamics
• States that no cyclic process that converts
heat entirely into work is possible.
• Includes the requirement that a heat engine
give up some energy at a lower temperature
in order to do work.
• So a heat engine cannot transfer all energy as
heat to do work.
Second Law of Thermodynamics
• No cyclic process that converts heat entirely
into work is possible
Wnet  Qnet  Qh  Qc
Efficiency of a Heat Engine
net work done by engine
efficiency 
energy added as heat
Wnet Qh  Qc
Qc
eff 

 1
Qh
Qh
Qh
• The smaller the fraction of usable energy that an
engine can provide, the lower its efficiency is.
Sample Problem
• Find the efficiency of a gasoline engine that,
during one cycle, receives 204 J of energy
from combustion and loses 153 J as heat into
the exhaust.
Solution
Qh= 204 J
Qc= 153 J
eff= 1- Qc
Qh
eff= 1- 153 J
204 J
eff= 0.250 J
Entropy
• Entropy: a measure of the randomness or
disorder of a system
• A greater disorder means there is less energy
to do work
• The motion of the particles of a system is not
well ordered and therefore is less useful for
doing work
• Once a system has reached a state of the
greatest disorder, it will tend to remain in that
state and have maximum entropy.
• The second law of thermodynamics states that
the entropy of the universe increases in all
natural processes.