Transcript THERMODYNAMICS

Work in Thermodynamic Processes
 Energy can be transferred to a
system by heat and/or work
 The system will be a volume
of gas always in equilibrium
 Consider a cylinder with a
movable piston
 As piston is pressed a
distance Δy, work is done on
the system reducing the
volume
 W = -F Δy = - P A Δy
Work in Thermodynamic Processes
– Cont.
 Work done compressing
a system is defined to be
positive
 Since ΔV is negative
(smaller final volume)
& A Δy = V
 W = - P ΔV
 Gas compressed  Won
gas = pos.
 Gas expands  Won gas =
neg.
Work in Thermodynamic Processes
– Cont.
 Can only be used if gas is
under constant pressure
 An isobaric process (iso =
the same) P1 = P2
 Represented on a pressure
vs. volume graph – a PV
diagram
 Area under any curve =
work done on the gas
 If volume decreases – work
is positive (work is done
on the system)
THERMODYNAMICS
First Law of Thermodynamics
 Energy is conserved
 Heat added to a system goes
into internal energy, work or
both
ΔU = Q + W
 Heat added to system 
internal energy  Q is
positive
 Work done to the system 
internal energy  W is
positive (again)
First Law – Cont.
 A system will have a
certain amount of
internal energy (U)
 It will not have certain
amounts of heat or work
 These change the system
 U depends only on state
of system, not what
brought it there
 ΔU is independent of
process path (like Ug)
Isothermal Process
 Temperature remains
constant
 Since P = N kB T / V =
constant / V
 An isotherm (line on
graph) is a hyperbola
Isothermal Process – Cont.
 Moving from 1 to 2,
temperature is constant,
so P & V change
 Work is done = area
under curve
 Internal energy is
constant because
temperature is constant
 Q = -W
 Heat is converted into
mechanical work
Isometric (isovolumetric) Process
 The volume is not
allowed to change
 V1 = V2
 Since no change in
volume, no work is done
 ΔU = Q
 Heat added must go into
internal energy  it
 Heat extracted is at the
expense of internal
energy  it
Isometric Process – Cont.
 The PV diagram
representation
 No change in volume
 Area under curve = 0
 That is, no work done
 The process moves from
one isotherm to another
Isobaric Process
 As heat is added to system,
pressure is required to be
constant
 The ratio of V / T = constant
 Some of the heat does work
and the rest causes a change
in temperature
 Thus moving to another
isotherm
 Recall: changes in
temperature = changes in
internal energy
 ΔU = Q + W
Adiabatic Process
 No heat is transferred
into or out of system
Q=0
 ΔU = W
 All work done to a system
goes into internal energy
increasing temperature
 All work done by the
system comes from
internal energy & system
gets cooler
Adiabatic Process – Cont.
 Either system is
insulated to not
allow heat
exchange or the
process happens
so fast there is
no time to
exchange heat
Adiabatic Process – Cont.
 Since temperature
changes, we move
isotherms
The Second Law of
Thermodynamics & Heat Engines
 Heat will not flow
spontaneously from a
colder body to a warmer
 OR: Heat energy cannot
be transferred
completely into
mechanical work
 OR: It is impossible to
construct an operational
perpetual motion
machine
Heat Engines
 Any device that
converts heat energy
into work
 Takes heat from a
high temperature
source (reservoir),
converts some into
work, then transfers
the rest to
surroundings (cold
reservoir) as waste
heat
Heat Engines – Cont.
 Consider a cylinder and piston
 Surround by water bath & allow
to expand along an isothermal
 The heat flowing in (Q) along
AC equals the work done by the
gas as it expands (W) since ΔU =
0
 To return to A along same
isothermal, work is done on the
gas and heat flows out
 Work expanding = work
compressing
Heat Engines – Cont.
 A cycle naturally can have
positive work done
 In going from A to B work is
done by gas, temperature 
(ΔU ) and heat enters
system
 B to C
 No work done, T , ΔU , &
heat leaves system
 C to A
 ΔU = 0, heat leaving = work
done
 The work out = the net heat
in (ΔU = 0)
Heat Engines – Cont.
 Thermal Efficiency
 Used to rate heat engines
 efficiency = work out / heat in
 e = Wout / Qin
 Qin = heat into heat engine
 Qout = heat leaving heat engine
 For one cycle, energy is
conserved
 Qin = W + Qout
 Since system returns to its
original state ΔU = 0
The Carnot Engine
 Any cyclic heat engine
will always lose some
heat energy
 What is the maximum
efficiency?
 Solved by Sadi Carnot
(France) (Died at 36)
 Must be reversible
adiabatic process
The Carnot Engine – Cont.
 Carnot Cycle
 A four stage reversible
process
 2 isotherms & 2 adiabats
 Consider a hypothetical
device – a cylinder & piston
 Can alternately be brought
into contact with high or
low temperature reservior
 High temp – heat source
 Low temp – heat sink –
heat is exhausted
The Carnot Engine – Cont.
 Step 1: an
isothermal
expansion, from A
to B
 Cylinder receives
heat from source
 Step 2: an adiabatic
expansion, from B
to C
The Carnot Engine – Cont.
 Step 3: an isothermal
compression, C to D
 Ejection of heat to sink at
low temp
 Step 4: an adiabatic
compression, D to A
 Represents the most
efficient (ideal) device
 Sets the upper limit
Entropy
 A measure of disorder
 A messy room > neat
room
 Pile of bricks > building
made from them
 A puddle of water > ice
came from
 All real processes
increase disorder 
increase entropy
  of entropy of one
system can be reduced at
the expense of another
Entropy – Cont.
 Entropy of the universe
always increases
 The universe only moves
in one direction –
towards  entropy
 This creates a “direction
of time flow”
 Nature does not move
systems towards more
order
Entropy – Cont.
 As entropy , energy is
less able to do work
 The “quality” of energy
has been reduced
 Energy has “degraded”
 Nature proceeds towards
what is most likely to
happen