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Physics 231 Lecture 33
•
•
Main
M
i points
i off today’s
d ’ lecture:
l
Work in thermodynamic
processes:
Wsystem = PΔV
•
First Law of Thermodynamics:
ΔU = ΔQ − PΔV
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•
Specific heat at constant pressure
Processes
– cyclic:
ΔP = ΔV = ΔT = 0
– isobaric:
ΔP = 0
– isovolumetric: ΔV = 0
– isothermal:
ΔT = 0
– adiabatic:
ΔQ = 0
Conceptual question
•
Which
Whi
h gives
i
th
the llargestt average radiation
di ti absorbed
b b d on a 1 m2
area at that distance?
– 1. a 50-W source at a distance R.
– 2. a 100-W source at a distance 2R.
– 3. a 200-W source at a distance 4R.
Pabsorbed
b b d = fraction of area covered eemitted power
Pabsorbed
1m 2
=
eP
2
4π r
1m 2
P1 =
e50
2
4π R
1m 2
e100
P2 =
2
4π (2R)
1m 2
P2 =
e200
2
4π (4R)
Work and the 1st Law of thermodynamics
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Consider the ideal gas contained in the volume under the
cylinder at right. The piston compresses the gas very
slowly, moving downwards with a constant velocity.
The gas exerts a force Fggas=PA upward on the piston and
conversely, the piston exerts a force Fpiston=-PA downward
on the gas.
The work done by the piston on the gas is:
W piston = Fpiston Δy = PA Δy = − PΔV
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The work done by the gas on the piston is:
Wgas = − Fgas Δy = − PA Δy = PΔV
•
If U denotes the internal energy of the system, the
conservation of energy
gy dictates:
ΔU = ΔE th = ΔQ + Wpiston = ΔQ − Wgas = ΔQ − PΔV 1st Law of Thermodynamics
Here I have changed notation to U, because other systems,
besides the ideal gas, can have potential energy.
Why does work increase thermal energy
•
•
Recall
R
ll that
th t the
th thermal
th
l energy U is
i given
i
by
b U=3/2Nk
U 3/2NkBT
If work done on the gas increases U it must increase T.
•
Positive work done on the gas increases U because the collisions with
the piston moving downward increases the velocity of the molecules.
Negative work done on the gas decreases U because the collisions with
the piston moving upward decreases the velocity of the molecules.
•
Example
•
The workk done
Th
d
to
t compress one mole
l off a monatomic
t i ideal
id l gas is
i 6200
J. The temperature of the gas changes from 350 K to 550 K. How
much heat flows between the gas and its surroundings? Determine
whether the heat flows into or out of the gas.
gas
Work done by the gas is : Wgas = −6200J
3
Internal energy change in the gas is: ΔU = nR ( ΔT )
2
3
3
Q = ΔU + Wgas = nR ( ΔT ) − 6200J = (8.31J / K )( 550 − 350K ) − 6200J
2
2
Q = −3700J
Computation of work:
•
•
•
Whatt is
Wh
i the
th workk done
d
b the
by
th gas going
i ffrom (Pi ,V
Vi) tto (Pf ,V
Vf) ?
– a) Pi(Vi-Vf)
– b) Pi(Vf-Vi)
– c) 0
– d) Vi(Pi-Pf)
– e) Vf(Pi-P
Pf)
The work done by the gas is the mathematical area under the curve Pi(Vf-Vi). It
is negative if the volume is decreasing. It is pushing to increase the volume, but
the volume is being
g decreased.
No work is done on the isovolumetric (constant volume) part of the path!
Computation of work:
•
•
•
Whatt is
Wh
i the
th workk done
d
on the
th gas going
i ffrom (Pi ,V
Vi) tto (Pf ,V
Vf) ?
– a) Pi(Vi-Vf)
– b) Pi(Vf-Vi)
– c) 0
– d) Vi(Pi-Pf)
– e) Vf(Pi-P
Pf)
Note that the work done on the gas is (-1) times the mathematical area under the
curve, -Pi(Vf -Vi). The mathematical area is negative if the volume is decreasing.
The work done on the g
gas is positive,
p
, you
y are compressing
p
g it and thereby
y
exerting a force in the direction of that it is being compressed.
No work is done on the isovolumetric (constant volume) part of the path!
Conceptual quiz
•
A gas is taken from an initial state of pressure and volume, PA
VA,to a final, different, state, PB VB. As it changes work is done
on the gas.
g
– a) The amount of work depends on the path taken from A to
B
– b) The amount of work does not depend on the path taken
from A to B
quiz:
•
•
•
Whatt is
Wh
i the
th workk done
d
on the
th gas going
i ffrom (Pi ,V
Vi) tto (Pf ,V
Vf) ?
– a) Pf(Vi-Vf)
– b) Pf(Vf-Vi)
– c) 0
– d) Vi(Pi-Pf)
– e) Vf(Pi-P
Pf)
Note that the work done on the gas is the negative of the mathematical area
under the curve. The mathematical area is negative if the volume is
decreasing. We are compressing the gas and the volume is decreasing and we
are pushing in the direction to decrease it.
No work is done on the isovolumetric part of the path!
Computation of work:
•
Whatt is
Wh
i the
th workk done
d
on the
th gas going
i ffrom (Pi ,V
Vi) tto (Pf ,V
Vf) ?
– a) Pi(Vi-Vf)
– b) Pf (Vi-Vf)
– c) 0
– d) (Pi+ Pf)(Vi-Vf)/2
– e)) -(P
( i+ Pf)(
)(Vi-Vf))/2
Heat capacity at constant pressure
•
Consider a cylinder
y
filled with a n=2 moles of
monatomic gas and plugged at the top by a frictionless
piston. Above the piston is atmospheric pressure. This
ensures the piston will be at a constant pressure
P=299kPa. The bottom of the cylinder piston is a heat
source that can heat the gas. The sides of the cylinder
and the piston have zero thermal conductivity. What is
th heat
the
h t Q required
i d to
t raise
i the
th temperature
t
t
off the
th gas
from initial temperature Ti =300 K to final temperature
Tf =400K?
Heat source
Q = ΔU + W
3
ΔU = n RΔT
2
W = pΔV
3
= n R ( Tf − Ti )
2
= p ( Vf − Vi ) = pVf − pVi
= nRTf − nRTi = nR ( Tf − Ti )
3
5
Q = n R ( Tf − Ti ) + nR ( Tf − Ti ) = n R ( Tf − Ti ) = 28.31*(400 − 300)J = 1.66kJ
2
2
3
5
c p = c v + R = R + R = R is the molar heat capacity at constant pressure
2
2
Example
•
A person takes in a breath of 0°C air and holds it until it warms to
37 0°C
37.0
C. The air has an initial volume of 0
0.600
600 L and a mass of 7
7.70
70 x
10–4 kg. Determine (a) the work done by the air on the lungs if the
pressure remains constant at atmospheric pressure, (b) the change in
internal energy
gy of the air,, and (c)
( ) the energy
gy added to the air by
y heat.
Model the air as if it were a diatomic gas: cV =5/2R., cp=7/2R.
Pf = Pi = Patm = 1.01x105 Pa
Wgas = Patm ( Vf − Vi ) = nR ( Tf − Ti )
 nRTf 
 Tf 
 Patm Vf 
− 1 = Patm Vi  − 1
= Patm Vi 
− 1 = Patm Vi 
 nRTi 
 Ti 
 Patm Vi 
 310 
Wgas = 1.01x105 Pa .0006m3 
− 1 = 8.2J
 273 
Assume diatomic gas:
 Tf 
5
5
5
Wgas = 20.5J
=
ΔU = nR ( Tf − Ti ) = nRTi  − 1
2
2
2
 Ti 7 
ΔQ = ΔU + Wgas = nR ( Tf − Ti ) = 28.7J
2
(
)(
)