lecture 8 - first law

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Announcements 9/16/11
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Prayer
Still at least three unregistered clickers: 14710762, 16488CD2,
1DAE9D2E
“Real” thermodynamics (more unified, fewer disjointed topics):
a. Today
– PV diagrams
– work
– isothermal contours
– Internal energy
– First Law of Thermodynamics
b. Continues for the next 4 lectures after today. Then one more
lecture. Then exam!
Pearls
Before
Swine
Reading quiz (graded)
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Which of the following is NOT true of the work
done on a gas as it goes from one point on a
PV diagram to another?
a. It cannot be calculated without knowing n
and T.
b. It depends on the path taken.
c. It equals minus the integral under the curve.
d. It has units of Joules.
e. It is one of the terms in the First Law of
Thermodynamics.
Work done by an expanding gas
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1 m3 of an ideal gas at 300 K
supports a weight in a piston such
that the pressure in the gas is
200,000 Pa (about 2 atm). The gas
is heated up. It expands to 3 m3.
Plot the change on a graph of
pressure vs. volume (a P-V diagram)
How much work did the gas do as it
expanded?
a. How do you know it did work?
W  F  distance
  P  Area   distance
 PV
= 400,000 J
More on Work…
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PV diagrams
What if pressure doesn’t
stay constant?
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Won gas   PdV
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Work done on gas vs work
done by gas
Thought question (ungraded)
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A gas in a piston
expands from point A
to point B on the P-V
plot, via either path 1
or path 2. Path 2 is a
“combo path,” going
down first, then over.
The gas does the most
work in:
a. path 1
b. path 2
c. same work
Quick Writing
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Describe with words how
you could actually make
a gas (in some sort of
container) change as in
path 2.
Internal Energy, Eint (aka U)
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Eint = Sum of all of the microscopic kinetic energies.
(Also frequently called “U”.)
Return to Equipartition Theorem:
a. “The total kinetic energy of a system is shared equally
among all of its independent parts, on the average,
once the system has reached thermal equilibrium.”
b. Each “degree of freedom” of a molecule has kinetic
energy of kBT/2
c. Monatomic molecules  3 d.o.f.
d. At room temperatures, diatomic  5 d.o.f.
(3 translational, 2 rotational)
Internal Energy
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Monatomic: Eint = N  3 kBT/2
= (3/2)nRT
Eint  32 nRT
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Diatomic: (around room temperature)
Eint = N  5 kBT/2
= (5/2)nRT
Eint  52 nRT
Thought question (ungraded)
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The process in which
Eint is the greatest
(magnitude) is:
a. path 1
b. path 2
c. neither; it’s the same
Isothermal Contours
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A gas changes its volume and pressure
simultaneously to keep the temperature
constant the whole time as it expands to twice
the initial volume. What does this look like on a
PV diagram?
PV  nRT  xy  constant
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What if the temperature is higher? Lower?
“First Law”
Eint = Qadded + Won system
 What does that mean? You can add internal
energy, by…
a. …adding heat
b. …compressing the gas
 Possibly more intuitive version:
Qadded = Eint + Wby system
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When you add heat, it can either
…increase internal energy (temperature)
…be used to do work (expand the gas)
Three Specific Cases
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Constant pressure, “isobaric”
a. Work on = ? –PV
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Constant volume, “isovolumetric”
a. Work on = ? 0
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Constant temperature, “isothermal”
a. Work on = ?  PdV   nRT dV  nRT
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V
 nRT ln V2 V1
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dV
V
Worked Problems
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For each problem, draw the process on a P-V diagram,
state what happens to the temperature (by visualizing
contours), and calculate how much heat is added/removed
from gas via the First Law.
a. A monatomic gas (1.3 moles, 300K) expands from 0.1
m3 to 0.2 m3 in a constant pressure process.
T increases, Q = Eint + PV = 8102 J added
b. A diatomic gas (0.5 moles, 300K) has its pressure
increased from 100,000 Pa to 200,000 Pa in a constant
volume process.
T increases, Q = Eint = 3116 J added
c. A diatomic gas (0.7 moles, 300K) gets compressed from
0.4 m3 to 0.2 m3 in a constant temperature process.
T stays constant, Q = –Won gas = –1210 J (i.e., 1210 J of heat removed from gas)
Quick Answers From Students
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Eint will be positive if ______________
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Qadded will be positive if ______________
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Won system will be positive if ______________
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