Download Lecture 36.Thermodyn..

Thermodynamic Processes
Lecturer:
Professor Stephen T. Thornton
Reading Quiz
If you add some heat to a substance,
is it possible for the temperature of
the substance to remain unchanged?
A) yes
B) no
C) depends on Q
D) depends on W
Reading Quiz
A) yes
If you add some heat to a substance,
is it possible for the temperature of
the substance to remain unchanged?
B) no
C) depends on Q
D) depends on W
Yes, it is indeed possible for the temperature to stay the same. This is
precisely what occurs during a phase change – the added heat goes
into changing the state of the substance (from solid to liquid or from
liquid to gas) and does not go into changing the temperature! Once
the phase change has been accomplished, then the temperature of the
substance will rise with more added heat.
Follow-up: Does that depend on the substance?
Last Time
Heat
Internal energy
Specific heat
Latent heat
First Law of Thermodynamics
Today
Thermodynamic processes
Isothermal, adiabatic
Specific heat of gases
Equipartition
Conceptual Quiz:
A system absorbs heat Q and has an equal
amount of work done on it. What is the
change in the internal energy of the
system?
E  Q  W
A) Q
B) 2Q
C) -2Q
D) zero
E) Q/2
System gains heat
Q positive
System loses heat
Q negative
Work done by system W positive
Work done on system
W negative
Answer: B
System absorbs heat Q so it is
positive. System has work of equal
value done on it, so it is negative.
W = -Q
E = Q – W = Q – (-Q) = 2Q
A Constant-Pressure Process
Area of graph is W
System does work
to push piston in
cylinder at constant
pressure. Volume
expands.
F  PA
W  F x  PAx  PV
Area of graph = PV  W work done by system
In a general problem like this example,
the area under the curve is equal to the
work done by the system.
W   P dV
Area
here is
work
We add heat to a system at constant volume.
What is the work done?
W= PV = 0
Because volume doesn’t change, the work
done W must be zero.
W 0
E  Q  W  Q
Isotherms on a PV diagram
Isothermal process.
T is constant.
An Isothermal Expansion
PV  nRT  constant
PV  constant
hyperbola
W   P dV  nRT  dV
V
W = area under curve. We use calculus to show
 Vf 
 Vf 
W  nRT ln    NkT ln  
 Vi 
 Vi 
In an adiabatic process, the system is well
insulated thermally, and no heat flows (Q = 0).
When the piston compresses the volume, the
pressure and temperature must both go up.
Do demo
Adiabatic
Heating
If we push down
quickly, there is
no time for heat to
flow, and the
process is
adiabatic.
Temperature rises
quickly.
When the piston moves up, the
volume expands, and the pressure
and temperature decrease.
Adiabatic process occurs often when the process is
rapid, and there is no time for heat to flow.
Conceptual Quiz
Two equal-mass liquids, initially at the
same temperature, are heated for the same
A) the cooler one
time over the same stove. You measure
B) the hotter one
the temperatures and find that one liquid
has a higher temperature than the other.
Which liquid has a higher specific heat?
C) both the same
Conceptual Quiz
Two equal-mass liquids, initially at the
same temperature, are heated for the same
A) the cooler one
time over the same stove. You measure
B) the hotter one
the temperatures and find that one liquid
has a higher temperature than the other.
C) both the same
Which liquid has a higher specific heat?
Both liquids had the same increase in internal energy,
because the same heat was added.
But the cooler liquid
had a lower temperature change.
Because Q = mcT, if Q and m are both the same and T is
smaller, then c (specific heat) must be bigger.
Conceptual Quiz
The specific heat of concrete is
greater than that of soil. A baseball
field (with real soil) and the
surrounding parking lot are warmed
up during a sunny day. Which would
you expect to cool off faster in the
evening when the sun goes down?
A) the concrete parking lot
B) the baseball field
C) both cool off equally fast
Conceptual Quiz
The specific heat of concrete is
greater than that of soil. A baseball
field (with real soil) and the
surrounding parking lot are warmed
up during a sunny day. Which would
you expect to cool off faster in the
evening when the sun goes down?
A) the concrete parking lot
B) the baseball field
C) both cool off equally fast
The baseball field, with the lower specific heat, will change
temperature more readily, so it will cool off faster. The high specific
heat of concrete allows it to “retain heat” better and so it will not cool
off so quickly—it has a higher “thermal inertia.”
Monatomic Gas. One and one-half
moles of an ideal monatomic gas
expand adiabatically, performing
7500 J of work in the process. What is
the change in temperature of the gas
during this expansion?
Conceptual Quiz:
An ideal gas is heated so that it
expands at constant pressure. The gas
does work W. What heat is added to
the gas?
Hint: E = Q – W
A) Q = W
C) Q = 0
E) Q < W
B) Q = -W
D) Q > W
Answer: D
An ideal gas is heated so that it expands at
constant pressure. The gas does work W.
What heat is added to the gas?
Because the gas is heated, the
temperature will increase. Therefore the
internal energy E > 0. W > 0, so if
E = Q – W > 0, then Q = ΔE +W > W.
Thermodynamic Processes
and Their Characteristics
Constant pressure
W = PV Q = Eint + PV
Constant volume
W=0
Q = Eint
Isothermal (constant W = Q
temperature)
Adiabatic (no heat
W=–
flow)
Eint
Eint = Q – W
Eint = 0
Q=0
Conceptual Quiz:
A gas at point A compresses isothermally
along curve (ii). If the gas compresses
adiabatically, what curve does it follow?
A) Curve (i)
B) Curve (ii)
C) Curve (iii)
Answer: C
In the isothermal process, some heat flows out
of the system in order to keep the temperature
constant. In the adiabatic process, no heat can
flow out, so the temperature must rise. For the
same volume, therefore the pressure must rise,
compared to the isothermal case. Curve (iii)
must be correct.
Conceptual Quiz:
What happens if we compress a cylinder that is
thermally isolated?
E = Q – W
A) W > 0, Q > 0
B) W < 0, Q > 0, E = 0
C) W > 0, Q = 0, E > 0
D) W < 0, Q = 0, E > 0
Answer: D
The system is thermally isolated, so the
heat flow Q = 0. An external agent
pushes the piston down and does work
on the system. Therefore the system
does negative work, W < 0.
E = Q – W > 0.
Conceptual Quiz
Water has a higher specific
A) from the ocean to the beach
heat than sand. Therefore,
B) from the beach to the ocean
on the beach at night,
breezes would blow:
C) either way, makes no difference
Conceptual Quiz
Water has a higher specific
A) from the ocean to the beach
heat than sand. Therefore,
B) from the beach to the ocean
on the beach at night,
breezes would blow:

C) either way, makes no difference
Daytime
 sun heats both the beach and the water
» beach heats up faster
» warmer air above beach rises
» cooler air from ocean moves in underneath
» breeze blows ocean  land

csand < cwater
Nighttime
 sun has gone to sleep
» beach cools down faster
» warmer air is now above the ocean
» cooler air from beach moves out to the ocean
» breeze blows land  ocean
Why do we almost
always feel a
breeze at the
beach?
See if this works
this summer!
Water has high
thermal
conductivity!
Work Done by Thermal Systems
We have looked at specific heat for solids and
liquids. Let’s now look at specific heat for
gases. We have two important cases:
constant volume and constant pressure. Let
heat flow Q into the system at constant
volume and the temperature rises by T.
QV  mcV T , constant volume cV
CV is molar specific heat = McV  cV  CV / M
QV  nCV T , compare with QP  nCP T
CP  McP
First law, Q  Eint  W
For constant volume, W  0, and
3
QV  Eint , and because Eint  nRT ,
2
3
QV  Eint  nRT  nCV T
2
3
so CV  R
2
This is molar specific heat for a monatomic
ideal gas at constant volume.
Now let’s determine CP at constant pressure.
Consider an ideal gas.
W  PV  nRT
QP  Eint  W
3
5
 nRT  nRT  nRT
2
2
5
CP  R
2
And we have
CP  CV  R
CP – CV for Various Gases
Experimental Values
Helium (monatomic)
0.995 R
Nitrogen (diatomic)
1.00 R
Oxygen
1.00 R
Argon
1.01 R
Carbon Dioxide (triatomic)
1.01 R
Methane
1.01 R
Seems to be true for real gases, including
diatomic gases, not only monatomic gases.
A diatomic molecule can do more than just
move – it can rotate or vibrate, and both
contribute to its total kinetic energy.
The equipartition theorem states that each degree of
freedom contributes kT/2 to the average energy of a molecule.
We have degrees of freedom for
3
• velocity in x, y, and z - translation kT
2
• rotation about two different axes
2
kT
2
2
• vibration, which includes two contributions
For s degrees of freedom, we have
E  s kT
2
For N molecules, we have
Eint  N E  s NkT
2
2
kT
The average kinetic energy of translation (3 dimensions) is
E  3 kT
2
kT for each dimension
2
The addition of rotations about two axes adds two more
factors of ½ kT:
E  5 kT
2
Finally, the addition of vibration contributes two more factors
of ½ kT, one for the motion of the atoms and one for the
energy in the “spring”:
E  7 kT
2
E  s kT
2
For molecular hydrogen:
CV / R
In this table, we see that the molar specific heats for
gases with the same number of molecules are
almost the same, and that the difference CP – CV is
almost exactly equal to 2 (R) in all cases.
Copyright © 2009 Pearson Education, Inc.
Specific Heat of Gas. The specific
heat at constant volume of a
particular gas is
0.182 kcal/kg·K at room
temperature, and its molecular mass
is 34. (a) What is its specific heat at
constant pressure? (b) What do you
think is the molecular structure of
this gas?
Piston & Cylinder. When 6.3 x 105 J
of heat is added to a gas enclosed in a
cylinder fitted with a light frictionless
piston maintained at atmospheric
pressure, the volume is observed to
increase from 2.2 m3 to 4.1 m3.
Calculate (a) the work done by the gas,
and (b) the change in internal energy of
the gas. (c) Graph this process on a PV
diagram.