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Chapter 15. Work, Heat, and the First Law of
Thermodynamics
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Heat and Work in Ideal-Gas Processes
• Consider a gas cylinder sealed at
one end by a moveable piston.
• Assume KE =0 (cylinder not
moving) and PE = 0 (cylinder
position at y = 0).
• But since T > 0 kelvin, Uint ≠ 0.
•For a monatomic gas:
U = 3/2 nRT
• For non-monatomic gases, U is
still proportional to T:
UαT
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Heat and Work in Ideal-Gas Processes
•How can we change the amount
of internal energy in our system of
an ideal gas?
• We have studied 2 energy
transfer mechanisms, heat (Q) and
work (W). Let’s look at both of
these mechanisms.
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Review: What is the best definition of heat?
A. the amount of thermal energy in an object.
B. the energy that moves from a hotter
object to a colder object.
C. how high the temperature of an object is.
D. all of the above
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
What is the best definition of heat?
A. the amount of thermal energy in an object.
B. the energy that moves from a hotter
object to a colder object.
C. how high the temperature of an object is.
D. all of the above
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Heat, Temperature, and Thermal Energy
• Thermal energy U is an energy of the system due to
the motion of its atoms and molecules. Any system has
a thermal energy even if it is isolated and not
interacting with its environment. The units of U are
Joules.
• Heat Q is energy transferred between the system and
the environment as they interact due to a difference in
temperature. The units of Q are Joules.
• Temperature T is a state variable that quantifies
the “hotness” or “coldness” of a system. A
temperature difference is required in order for heat to be
transferred between the system and the environment. The
units of T are degrees Celsius or Kelvin.
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If heat is the only energy transfer mechanism
ΔU = Uf -Ui = Q
Q is positive when the
system gains energy.
This means that the
environment has a
higher temperature than
the system.
Q is negative when the
system loses energy.
This means that the
environment has a
lower temperature than
the©system.
Copyright
2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Work done by the system and on the system
W = |F| |Δx| cos θ
• If the piston moves to the right:
• Work done by the gas molecules
on the piston is positive (force is to
the right, piston moves to the right).
• energy is added to the piston,
energy is taken away from the gas
and the gas expands.
•If the piston moves to the left:
•Work done by the gas molecules on the piston is
negative (force is to the right, gas molecules move left).
•Energy is added to the gas and the gas compresses.
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15.3 The First Law of Thermodynamics
THE FIRST LAW OF THERMODYNAMICS
The internal energy of a system changes due to heat and work:
U  U f  U i  Q  Wby
Heat is positive when the system gains heat and negative when the
system loses heat.
Work is positive when it is done by the system and negative when it
is done on the system.
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A gas cylinder and piston are covered with heavy
insulation so there can be no heat exchange with the
environment. The piston is pushed into the cylinder,
compressing the gas. According to the 1st Law of
Thermodynamics, the gas temperature:
A. decreases.
B. increases.
C. doesn’t change.
D. There’s not sufficient
information to tell.
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ΔU = Q – Wby
• insulation implies no heat exchange with
environment
•Work done by gas is negative. Gas pushes
left, piston compresses gas to the right:
ΔU = Q – Wby
ΔU = – – Wby
ΔU is proportional to temperature increase.
Therefore, temperature increases.
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EOC # 1
The internal energy of a system changes because the
system gains 165 J due to heat transfer and does 312 J
of work. In returning to its initial state *, the system
loses 114 J of heat.
a. How much work is involved during the return
process?
b. Is the work done by the system, or on the system?
* Initial state (for a gas) means same voloume,
pressure, temperature, and internal energy.
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EOC # 1-Answer
The internal energy of a system changes because the
system gains 165 J due to heat transfer and does 312 J
of work. In returning to its initial state *, the system
loses 114 J of heat.
a. How much work is involved during the return
process? 261 J involved
b. Is the work done by the system, or on the system?
Work is done on the system.
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15.3 The First Law of Thermodynamics
Example 2 An Ideal Gas
The temperature of three moles of a monatomic ideal gas is reduced
from 540K to 350K as 5500J of heat flows into the gas.
Find (a) the change in internal energy and (b) the work done by the
gas.
U  U f  U i  Q  W
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U  32 nRT
15.3 The First Law of Thermodynamics
(a)
U  32 nRT f  32 nRTi

(b)
3
2
3.0 mol 8.31 J mol  K 350 K  540 K   7100 J
W  Q  U  5500 J   7100 J   12600 J
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15.4 Thermal Processes
Ideal Gas Processes
isobaric: constant pressure
isochoric: constant volume
isothermal: constant temperature
adiabatic: no transfer of heat
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15.4 Thermal Processes
An isobaric process is one that occurs at
constant pressure.
According to the 1st Law of
Thermodynamics:
ΔU = Q – Wby
Wby  Fs  P As   PV

Wby  PV  P V f  Vi

During an isobaric process:
ΔU = Q – P ΔV: energy is transferred by
both work and heat.
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The PV diagram for an work done during an
isobaric process is a horizontal line

W  PV  P V f  Vi
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
An isobaric process
• During an isobaric process, a system gains
1500 J of heat. The internal energy of the
system increases b y 4500 J and the volume
decreases by 0.01 m3 . Find the pressure of
the system.
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An isobaric process
• During an isobaric process, a system gains
1500 J of heat. The internal energy of the
system increases b y 4500 J and the volume
decreases by 0.01 m3 . Find the pressure of
the system.
• Answer: 3.0 x 105 Pa
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The work done by the gas equals the area under
the PV curve

W  PV  P V f  Vi
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
15.4 Thermal Processes
Example 4 Work and the Area Under a
Pressure-Volume Graph
Determine the work for the process in
which the pressure, volume, and temperature of a gas are changed along the
straight line in the figure.
The area under a pressure-volume graph is
the work for any kind of process.
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15.4 Thermal Processes
Since the volume increases, the work
is positive.
Estimate that there are 9colored
squares in the drawing.


W  9 2.0 105 Pa 1.0 10 4 m 3
 180 J

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Two process are shown that take an ideal
gas from state 1 to state 3. Compare the
work done by process A to the work done
by process B.
A. WA > WB
B. WA < WB
C. WA = WB = 0
D. WA = WB but neither is zero
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Two process are shown that take an ideal
gas from state 1 to state 3. Compare the
work done by process A to the work done
by process B.
A. WA > WB
B. WA < WB
C. WA = WB = 0
D. WA = WB but neither is zero
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15.4 Thermal Processes
isochoric: constant volume
U  Q  W  Q
But W = PΔV = 0:
during an isochoric
process, energy is
transferred by heat only
For the isochoric process the area under the
curve is equal to zero.
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15.5 Thermal Processes Using and Ideal Gas
ISOTHERMAL EXPANSION OR
COMPRESSION
Isothermal
expansion or
compression of
an ideal gas
 Vf
W  nRT ln 
 Vi



ΔU = Q – Wby
But ΔU α ΔT
If T does not change, ΔU = ΔT = 0!
And Q = W
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EXAMPLE The work of an isothermal
compression
QUESTION:
A cylinder contains 7.0 g of N2 gas. The gas compresses to half its
volume at a constant temperature of 80˚C.
a. How much work must be done by the gas?
b. By how much does the internal energy of the gas change?
c. How much heat was added or taken away from the gas?
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EXAMPLE The work of an isothermal
compression
Answer:
A cylinder contains 7.0 g of N2 gas. The gas compresses to half its
volume at a constant temperature of 80˚C.
a. How much work must be done by the gas? -508 J (work was
done on the gas to compress it)
b. By how much does the internal energy of the gas change? 0J
c. How much heat was added or taken away from the gas? 508 J of
heat was taken away from the gas.
Q – Wby = 0
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15.5 Thermal Processes Using and Ideal Gas
ADIABATIC EXPANSION/COMPRESSION
According to the 1st Law of
Thermodynamics: ΔU = Q – Wby
but Q = 0, since walls are insulated
ΔU = – Wby
For a monatomic ideal gas:
Wby  U  32 nRTi  T f 
The red curve shows an adiabatic
expansion of an ideal gas.
The blue curves are isotherms at Ti and
Tf.
Adiabatic curves can be approximated
as linear.
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Adiabatic Processes without adiabatic
(insulating) walls
• Rapid expansion or
compression does not
allow the gas and
surroundings to come to
equilibrium
– bicycle pump
– air being forced to rise
or sink due to
atmospheric conditions
– loading/ unloading a
pneumatic lift.
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Mass Lifter Lab – thermodynamic process with 4 steps
1.
2.
3.
4.
mass is loaded onto movable piston of a cylinder of gas kept at a low
temperature. This is a “very rapid” event. Energy transfer due to
temperature difference between gas and surroundings assumed to be
negligible.
gas of cylinder is put in contact with room temperature water and
energy transfer allowed. The mass rises.
mass is unloaded at a “rapid” pace.
gas of cylinder put in contact with ice water. The movable piston falls.
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Molar specific heat capacity
• For a solid or liquid: Q = cm ∆T, where c is
the specific heat capacity in Joules/kg kelvin.
• For a gas, we use number of moles (n) instead
of mass: Q = Cn ∆T where C is the molar
specific heat capacity in Joules/mol kelvin.
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Molar specific heat capacity for a monatomic
ideal gas
• At constant pressure Cp = 5/2R
• At constant volume Cv = 3/2R.
• Q = Cn ∆T ; so for the same amount of heat
(Q) added to a gas, the gas at constant volume
will show a greater temperature change
because it is not losing energy by expanding.
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Molar specific heat capacity
Two containers hold equal masses of helium gas
(monatomic) at equal temperature. You supply 10 J
of heat to container A while not allowing the volume
to change. You supply 10 J of heat to container B
while not allowing the pressure to change. Is TfA
greater than, less than or equal to TfB? Explain,
using the 1st Law of Thermodynamics.
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Molar specific heat capacity
For an isochoric process:
QA = Cvn ΔT = (3/2R)n ΔTA = 10J
For an isobaric process:
Q = Cpn ΔT = (5/2R)n ΔTB = 10J
ΔTA is greater than ΔTB
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EOC # 33
Heat, Q, is added to a monatomic ideal gas at
constant pressure. As a result, work, W is
done by the gas. What is Q/W, the ratio of
heat added to work done by the gas? This is a
numerical value, with no variables.
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EOC # 33
Heat, Q, is added to a monatomic ideal work at
constant pressure. As a result, work, W is
done by the gas. What is Q/W, the ratio of
heat added to work done by the gas?
Q = Cp nR ∆T = 5/2 nR ∆T
W = P ∆V = PVf – PVi but PV = nR ∆T
so W = nR ∆T
Q/W = 5/2/1 = 2.5.
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Which of the following processes involve
heat (energy transfer due to ∆T)?
A. The brakes in your car get hot when you
stop.
B. You push a rigid cylinder of gas across a
frictionless surface.
C. A steel block is placed on top of a candle.
D. You push a piston into a cylinder of gas,
increasing the temperature of the gas.
E. All of the above
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Which of the following processes involve
heat?
A. The brakes in your car get hot when you
stop.
B. You push a rigid cylinder of gas across a
frictionless surface.
C. A steel block is placed on top of a candle.
D. You push a piston into a cylinder of gas,
increasing the temperature of the gas.
E. All of the above
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EOC #26
The drawing refers to 1 mole of a monatomic gas and a 4-step process.
Complete the following table:
∆U
Wby
Q
A to B
B to C
C to D
D to A
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EOC #26
The drawing refers to 1 mole of a monatomic gas and a 4-step process.
Complete the following table:
∆U
Wby
Q
A to B
4990J
3320J
8310J
B to C
-4990J
0J
-4990J
C to D
-2490J
-1660J
-4150J
D to A
2490J
0J
2490J
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