Announcements 9/20/10
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Demo: constant volume change (hopefully
working today)
Do this “Quick Writing” assignment while
you’re waiting:
Ralph is confused because he knows that
when you compress gases, they tend to heat
up (think of bicycle pumps). So, how are
“isothermal” processes possible? How can
you compress a gas without its temperature
increasing?
Thought question (ungraded)
How will the temperature
of the gas change during
this process from A to B?
a. Increase
b. Decrease
c. First increase, then
decrease
d. First decrease, then
increase
e. Stay the same
A
2.0
1.8
1.6
Pressure (atm)

1.4
1.2
1.0
B
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
3
Volume (m )
1.4
1.6
1.8
2.0
Reading quiz (graded)

What is “CV”?
a. heat capacity
b. mass-pacity
c. molar heat capacity
d. molar heat capacity, but only for constant
volume changes
e. a detailed resumé
Thought question (ungraded)

Which should be larger, the molar heat
capacity for constant volume changes or the
molar heat capacity for constant pressure
changes? (Hint: Think of the First Law.)
a. constant volume
b. constant pressure
c. they are the same
d. it depends on the temperature
CV and CP
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
Constant volume change (monatomic):
W=0
Eint = Qadded
(3/2)nRT = Qadded
Compare to definition of C: Qadded = nCVT
(like Qadded = mcT)
CV = (3/2)R
Constant pressure change
a. What’s different?
b. result: (monatomic) CP = (5/2)R
What would be different for gases with more
degrees of freedom?
Reading quiz (graded)

What does gamma equal in the equation for an

adiabatic process: PV  constant
a. CP + CV
b. CP - CV
c. CV - CP
d. CV / CP
e. CP / CV
Isothermal vs Adiabatic
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
Isothermal: PV  constant

Adiabatic: PV  constant
 steeper curves for adiabatic
Thought question

How much do you think the temperature of
the air in this room would change by if I
compressed it adiabatically by a factor of 10?
(Vf = V0/10)
a. less than 0.1 degree C
b. about 0.1 degrees C
c. about 1 degree C
d. about 10 degrees C
e. more than 10 degrees C
Demo/Video
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Video: adiabatic cotton burner
Demo: freeze spray
Video: adiabatic expansion
Derivation of PV (for Monatomic)
Eint = Qadded + Won
(3/2) nRT = -integral(PdV)
(3/2) nRdT = -PdV
(3/2) nR d(PV/nR) = -PdV
(3/2) (PdV + VdP) = -PdV
What’s different
if diatomic?
(3/2) VdP = -(5/2) PdV
dP/P = -(5/3) dV/V
lnP = (-5/3)lnV + constant
lnP = ln(V-5/3) + constant
P = constant  V-5/3
(it’s a different constant)
P V5/3 = constant
Thought question

Which of the curves on the PV diagram below is
most likely to represent an isothermal compression,
followed by an adiabatic expansion back to the
initial volume?
a.
b.
c.
d.
e.
Thought questions
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
What would be the molar specific heat for an
adiabatic process? (Hint: think of Q = nCT.)
a. CV
b. CV + R
c. CV + 2R
d. CV - R
e. none of the above
What would be the molar specific heat for an
isothermal process? (Same hint.)
a. CV
b. CV + R
c. CV + 2R
d. CV - R
e. none of the above
Water/steam “saturation curve”: ideal gas?
30000
bk = actual data for water/steam
Pressure (kPa)
25000
20000
All Gas
15000
10000
5000
All Liquid
0
0.00
adding heat
energy, T=325C
Part Liquid,
Part Gas
0.01
0.02
0.03
0.04
0.05
3
Volume, divided by mass (m /kg)
0.06
1 atm
= 101 kPa
Water/steam “saturation curve”: ideal gas?
30000
bk = actual data for water/steam
rd = ideal gas, T=400
bl = ideal gas, T=500
gn = ideal gas, T=600
Pressure (kPa)
25000
20000
All Gas
15000
10000
5000
All Liquid
0
0.00
adding heat
energy, T=325C
Part Liquid,
Part Gas
0.01
0.02
0.03
0.04
0.05
3
Volume, divided by mass (m /kg)
0.06
1 atm
= 101 kPa
Water/steam “saturation curve”: ideal gas?
30000
25000
Pressure (kPa)
"critical point"
bk = actual data for water/steam
rd = ideal gas, T=400
bl = ideal gas, T=500
gn = ideal gas, T=600
20000
All Gas
15000
10000
5000
All Liquid
0
0.00
adding heat
energy, T=325C
Part Liquid,
Part Gas
0.01
0.02
0.03
0.04
0.05
3
Volume, divided by mass (m /kg)
0.06
1 atm
= 101 kPa
Water/steam “saturation curve”: ideal gas?
30000
????
25000
Pressure (kPa)
"critical point"
bk = actual data for water/steam
rd = ideal gas, T=400
bl = ideal gas, T=500
gn = ideal gas, T=600
20000
All Gas
15000
10000
5000
All Liquid
0
0.00
adding heat
energy, T=325C
Part Liquid,
Part Gas
0.01
0.02
0.03
0.04
0.05
3
Volume, divided by mass (m /kg)
0.06
1 atm
= 101 kPa