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Thermodynamics
Chapter 12
Thermodynamics Vocabulary
• Thermo (heat) dynamics (transfer)
• Thermodynamic systems describe many many
particles (molecules) which obey Newton’s laws
for dynamics but which would be difficult to
analyze due to their numbers.
• We use macroscopic means for analysis of
these systems of many particles - involving
quantities such as pressure, volume and
temperature.
Thermodynamics Vocabulary
• System – a definite quantity of matter enclosed
by boundaries or surfaces. Boundaries need not
have definite shape or volume.
• Thermally Isolated System – system in which no
heat is transferred into or out of the system.
• Heat Reservoir – a large separate system with
unlimited heat capacity (any amount of heat can
be withdrawn or added without appreciably
changing the temperature).
State of a System
• Kinematics equations describe the motion of an
object using variables which change but are
related (displacement, velocity, acceleration).
Motion can be described at any moment using
these quantities.
• Thermodynamic systems can be described
similarly using relationships between pressure,
volume and temperature. Relationships
between these variables are called Equations of
State, and describe the state of the system at
any moment.
p-V Diagrams
• A pressure-volume diagram shows how
pressure changes as a function of volume.
• Each coordinate (V, p) specifies a state of
the system. Pressure and volume are
given and temperature may be found
indirectly.
• A process is any change in the state of a
system. Processes may be reversible or
irreversible.
p-V Diagrams
First Law of Thermodynamics
• First Law describes how heat transferred
to a system, Q, relates to internal energy
of the system, U, and work done by the
system, W.
• Recall, Q = heat added; W = work done;
U = internal energy (see page 360)
• First Law of Thermo: Q = ΔU + W
First Law of Thermodynamics
• Q = ΔU + W
• If a system absorbs 2000J of heat and
does 800J of work (expanding perhaps)
then the internal energy will increase by
1200J.
Example 1
• A 65 kg worker shovels coal for 3.0 hours.
During this process he does work at a rate
of 20 W and loses heat to the environment
at a rate of 480 W. Ignoring the loss of
water by perspiration, how much fat will
the worker lose? The energy value of fat
is 9.3 kcal/g/
Keep Track of Signs
• Positive work is done when force and
displacement are in the same direction
(expanding gas).
• Q is positive when heat is added, negative
when heat is lost.
Work from a p-V Diagram
• Work =
F·Δx
pA ·Δx
pΔV
• On a pressurevolume diagram, work
done by a system is
equal to the area
under the p-V curve.
Isothermal Process
• Isothermal means
constant temperature.
• Internal Energy
remains constant if
temperature remains
constant.
• Q = ΔU + W ; ΔU=0
• Q=W
Isothermal Process
First Law of Thermo:
Q = ΔU + W
But for isothermal,
Q=W
All added heat goes into
work done by expanding gas.
Work = nRT ln(Vf/Vi)
Isobaric Process
• A constant pressure
process is called
isobaric.
• Work in isobaric
process is
W = p(V2 – V1) = pΔV
• So Q = ΔU + pΔV
Example
• Two moles of a monatomic ideal gas
initially at 0° and 1.0 atm are expanded to
twice their original volume using two
different processes. They are expanded
isothermally and then, starting in the same
initial state, they are expanded
isobarically. During which process does
the gas do more work? Find the work
done in each case.
Isometric Process
• Isometric means
isovolumetric … it
describes a process at
constant volume.
• If volume doesn’t
change, no work is
done.
• Q = ΔU
Example
• Many ‘empty’ aerosol cans contain
remnant propellant gases under
approximately 1 atm of pressure and
20°C. They display the warning “Do not
dispose of this can in an incinerator”.
What is the change in internal energy of
such a gas if 500J of heat is added to it,
raising the temperature to 2000°F? What
is the final pressure of the gas?
Adiabatic Process
• Adiabatic means no heat
is transferred into or out
of the system.
• Q = 0 so first law of
thermodynamics
becomes
ΔU = -W
• If internal energy
increases, work has been
done
Adiabatic Process
• adiabatic expansion
• Pressure and volume are related along an adiabat by
p1V1γ = p2V2γ
• Work done in an adiabatic process can be found using
Wadiabatic = (p1V1 – p2V2) / (γ - 1)
γ = 1.67 for monotonic gas, 1.40 for diatomic
Summary
Process
Characteristic
Result
Isothermal T = constant ΔU = 0
First Law
Q=W
Isobaric
p = constant W = pΔV Q = ΔU + pΔV
Isometric
V = constant W = 0
Q = ΔU
Adiabatic
Q=0
ΔU = -W
Example
• Blowing Air
• A sample of helium expands to triple its initial
volume adiabatically in one case and
isothermally in another. It starts from the same
initial state in both cases. The sample contains
2.00mol of helium initially at 20 ºC and 1.0 atm.
a) during which process does the gas do
more work?
b) Calculate the work done in each process.
Second Law of
Thermodynamics
• Second Law provides a way to predict the
direction in which thermal processes
proceed.
• Heat flows spontaneously from a warmer
body to a cooler body
• Heat energy cannot be completely
transformed into mechanical work
Entropy
• Entropy is a measure of a system’s ability to do
work.
• Entropy is a measure of disorder.
• A system naturally moves towards greater
disorder. More order = lower entropy; more
disorder = higher entropy.
• ΔS>0 for natural processes
Entropy
• ΔS = Q / T [Joules / Kelvin] gives a
change in entropy at a constant
temperature, T.
• ΔS = increase in entropy
• Entropy increases if Q is positive;
temperature must be measured in Kelvin.
Examples
• While doing physical exercise at 34°C, an
athlete loses 0.400 kg of water per hour by
evaporation of perspiration from his skin.
Estimate the change in entropy of the
water as it vaporizes. Latent heat of
vaporization for perspiration is
24.2X105 J/kg
Heat Engines
• Heat engines convert
heat energy into useful
work
• Examples are electric
turbines (power plants)
and internal combustion
engines (automobiles).
• ‘Heat in’ generates work;
gas must be brought to
original state via a
compression which takes
work.
• Wnet is the enclosed area.
Heat Engines
• Net Work = Work done
by gas - Work done on
gas
• Wnet = Wexp – Wcom
• For a cycle, ΔU = 0
so…
• Wnet = Wexp – Wcom
Wnet = Qh – Qc
Efficiency
• Efficiency is the ratio
of net work out to
heat in.
• ε = Wnet / Qin
• ε = (Qh – Qc) / Qh
• ε = 1 – Qc/Qh
Efficiency
• No engine is perfectly efficient! More heat
(energy) is required (Qh) than useful work
created!!
• Why???
Example
• A leaf-blower engine absorbs 800J of heat
energy from a high temperature reservoir
(the ignited gas-air mixture) and exhausts
700J to a low temperature reservoir (the
outside air). What is the efficiency?
Example
• You have 0.100 mol of an
ideal monatomic gas that
follows the cycle shown.
The pressure and
temperature on the lower left
is 1.00 atm and 20 °C.
Assume that the pressure
doubles during the isometric
pressure increase and the
volume doubles during the
isobaric expansion. Find the
efficiency.
Otto Cycle – 4 Stroke Engine
Carnot Cycle
• A Carnot cycle is a cycle consisting of two
isotherms and two adiabats.
• Sadi Carnot discovered that efficiency can
be maximized by a cycle absorbing heat
from a constant high temperature
reservoir, Th (isotherm), and exhausting it
to a constant low temperature reservoir, Tc
(isotherm).
Carnot Cycle
• Efficiency of a Carnot
Cycle:
• ε
= (Th – Tc)/Th
= 1 – Tc/Th
= 1 – Qc/Qh