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Chapter 18: Heat,Work,the First
Law of Thermodynamics
Thermodynamic systems
‰ Thermodynamic
system
Thermodynamic system is any collection of objects that is
convenient to regard as a unit, and that may have the potential
to exchange energy with its surroundings.
A process in which there are changes in the state of a thermodynamic system is called a thermodynamic process.
‰ Signs
for heat and work in thermodynamics
Q
Q is
Q is
W is
W is
+
–
+
–
W
if heat is added to system
if heat is lost by system
if work is done by the system.
if work is done on the system.
Work done during volume changes
‰ Work
done
Piston in a cylinder moves by dl
due to expansion of gas at pressure p
Force on piston: F = p·A
Incremental work done:
dW = F·dl = p·A dl = p·dV
Total work done : W = ∫ dW =
p
Vf
∫ pdV
Vi
Work done during volume changes (cont’d)
‰ Work
and pV diagram
p
Work done depends on the path taken.
ppAA
• Process A B :
If VB>VA , W > 0
• Process B A :
If VB>VA , W < 0
pB
• Process A B :
If VA=VB , W = 0
W=
VB
∫ pdV = area under p - V curve
VA
Work done during volume changes (cont’d)
‰ Work
and pV diagram (cont’d)
p
Work done depends on the path taken.
D
pA
pB
A C
:W=0
C B
: W = pB(VB-VA)
B C
C
A
: W = pB(VA-VB)
C
B : W = pB(VB-VA)
A D
: W = pA(VB-VA)
D B
:W=0
A
D
B : W = pA(VB-VA)
Internal energy and the first law of thermodynamics
‰ Internal
energy
Tentative definition:
The internal energy U of a system is the sum of the kinetic
energies of all of its constituent particles, plus the sum of
all the potential energies of interactions among these
particles.
‰ The
first law of thermodynamics
U 2 − U 1 = ∆U = Q − W
Now define the change in internal energy using the first law
of thermodynamics.
It is known from experiments that while Q and W depend on
the path, ∆U does not depend on the path.
Internal energy and the first law of thermodynamics
‰A
cyclic process
: is a process in which the initial
and final states are the same.
upper
lower
upper
WAlower
>
0
,
W
<
0
,
W
<
W
B→ A
A→ B
B→ A
→B
WA→ B→ A = W
lower
A→ B
+W
upper
B→ A
<0
A
pA
pB
B
VA
If the directions of the arrows are reversed, then
lower
WA→ B→ A = WAupper
+
W
→B
B→ A > 0
VB
Internal energy and the first law of thermodynamics
‰ Infinitesimal
changes in state
The first law of thermodynamics:
dU = dQ − dW
For the system we will discuss, the work dW is given by pdV
dU = dQ − pdV
Kinds of thermodynamics process
‰ Adiabatic
process
An adiabatic process is defined as one with no heat transfer into
or out of a system.
∆U = -W, Q=0
‰ Isobaric
process
An isobaric process is a constant-pressure process.
W = p(V2-V1) , p = constant
‰ Isothermal
process
An isothermal process is a constant-temperature process.
For a process to be isothermal, any heat flow into or out of the
system must occur slowly enough that thermal equilibrium is
maintained.
T = constant
Kinds of thermodynamics process (cont’d)
Work done by a gas depends on the path taken in pV space!
‰ Isochoric
process
An isochoric process is a
constant-volume process.
WA→ D→ B < WA→ B
Kinds of thermodynamics process (cont’d)
‰ Isothermal
vs. adiabatic process
Work done by expanding gases: path dependence
p
V2
W = ∫ pdV , p = nRT / V
V1
V2
⇒ W = nRT ∫
1
p1
V1
dV
V2
= nRT ln( )
V
V1
2
p2
Isotherm at T
V1
V2
V
Isothermal expansion (1 → 2 on pV diagram)
for ideal gas:
W = nRT ln (V2/V1)
Work done by expanding gases: path dependence
Expression for the work done in an isobaric
expansion of an ideal gas (3 → 2 on pV diagram).
p
p1
p2
W = p2(V2 – V1)
1
2
3
Isotherm at T
V1
V2
V
Internal energy of an ideal gas
‰ Free
expansion
the same final state
the same initial state
Controlled expansion
(The temperature is kept
constant and the expansion
process is done very slowly)
Work done
Heat exchanged
Uncontrolled expansion
= Free expansion
Different paths
No work done
No heat exchanged
Internal energy of an ideal gas
‰ Another
property of an ideal gas
From experiments it was found that the internal energy of
an ideal gas depends only on its temperature, not on its
pressure or volume.
However, for non-ideal gas some temperature change occur.
The internal energy U is the sum of the kinetic and potential
energies for all the particles that make up the system. Non-ideal
gas have attractive intermolecular forces. So if the internal energy
is constant, the kinetic energies must decrease. Therefore as the
temperature is directly related to molecular kinetic energy, for a
non-ideal gas a free expansion results in a drop in temperature.
Heat capacities of an ideal gas
‰ Two
kinds of heat capacities
• CV : molar heat capacity at constant volume easier to measure
• Cp : molar heat capacity at constant pressure
But why there is a difference between these two capacities?
• For a given temperature increase, the internal energy change
∆U of an ideal gas has the same value no matter what the process.
constant-volume (no work)
dQ = nCV dT = dU + dW = dU ⇒ dU = nCV dT
constant-pressure
T1 and T2 are the same
dQ = nC p dT = dU + dW = dU + pdV
pV = nRT → pdV = nRdT ;
dU = nCV dT
⇒ nC p dT = nCV dT + nRdT
C p = CV + R
For an ideal gas
Heat capacities of an ideal gas
‰ Ratio
γ =
of heat capacities
Cp
CV
Example
C p = CV + R =
⇒γ =
5
= 1.67
3
C p = CV + R =
⇒γ =
3
5
R+R= R
2
2
for an ideal monatomic gas
5
7
R+R= R
2
2
7
= 1.40
5
for most diatomic gas
Adiabatic processes of an ideal gas
‰ p,
V, and γ
dU = nCV dT
( by definition)
dW = pdV
dU = − dW
⇒ nCV dT = − pdV
( adiabatic process)
nRT
nRT
( ideal gas)
dV ⇐ p =
V
V
dT R dV
⇒
+
=0
T CV V
⇒ nCV dT = −
R C p − CV
dT
dV
⇒
=
= γ −1 ⇒
+ (γ − 1)
=0
CV
CV
T
V
⇒ lnT + (γ − 1)lnV = const
⇒ ln(TV γ −1 ) = const ⇒ TV γ −1 = const
Adiabatic processes of an ideal gas
‰ p,
V, and γ (cont’d)
TV γ −1 = const
pV
( ideal gas)
nR
pV γ −1
⇒
V = const ⇒ pV γ = const
nR
T=
Q = 0,W = − ∆U
∆U = nCV (T2 − T1 )
( adiabatic process)
⇒
W = nCV (T1 − T2 )
pV = nRT
⇒
W=
For adiabatic process, ideal gas
( ideal gas)
CV
1
( p1V1 − p2V2 ) =
( p V − p2V2 )
γ −1 1 1
R
For adiabatic process, ideal gas
Exercises
Problem 1
p
Solution
(a)
Wab = 0, Wbc = pc (Vc − Va ),
b
c
pc
Wdc = 0, Wad = pa (Vc − Va )
pa
(b) Q = ∆U + W
Qab = U b − U a , Qbc = U c − U b + pc (Vc − Va )
Qdc = U c − U d , Qad = U d − U a + pa (Vc − Va )
a
d
Va
Vc
V
(c) Wabc = pc (Vc − Va ), Qabc = U b − U a + (U c − U b ) + pc (Vc − Va ) = (U c − U a ) + pc (Vc − Va )
Wadc = pa (Vc − Va ), Qadc = U d − U a + (U c − U d ) + pa (Vc − Va ) = (U c − U a ) + pa (Vc − Va )
So assuming
pc > pa ,
Qabc > Qadc and Wabc > Wadc .
Problem 2
Exercises
p
Three moles of an ideal gas are
taken around the cycle acb shown
pc
in the figure. For this gas, Cp=29.1
J/(mol K). Process ac is at constant
pa
pressure, process ba is at constant
volume, and process cb is adiabatic.
Ta=300 K,Tc=492 K, and Tb=600 K.
Calculate the total work W for the cycle.
Solution
Path ac is at constant pressure:
b
a
c
Va
Vc
V
Wac = p∆V = nR∆T = nR(Tc − Ta ) = (3mol )[8.315 J /( molK )]( 492 K − 300 K ) = 4.79 × 103 J .
Path cb is adiabatic (Q=0):
Wcb = Q − ∆U = − ∆U = −nCV ∆T , CV = C p − R
→ Wcb = −n (C p − R )(Tb − Tc )
= −(3mol )[29.1J /( molK ) − 8.315 J /( molK )])600 K − 492 K ) = −6.74 × 103 J .
Path ba is at constant volume: Wba = 0
W = Wac + Wcb + Wba = −1.95 × 103 J .
Problem 3
Exercises
You are designing an engine that runs on compressed air. Air enters
the engine at a pressure of 1.60 x 106 Pa and leaves at a pressure of
2.80 x 105 Pa. What must the temperature of the compressed air be
for there to be no possibility of frost forming in the exhaust ports of the
engine? Frost will form if the moist air is cooled below 0oC.
Solution
TV γ −1 = const → T1V1γ −1 = T2V2γ −1
pV γ = const → p1V1γ = p2V2γ
1
p1 1− γ
γ −1 γ
γ −1 γ
p1 T1 = p2 T2 → T1 = T2 ( )
p2
Using γ=7/5 for air,
1.60 × 106 72
T1 = ( 273.15K )(
) = 449 K = 176°C.
5
2.80 × 10
Problem 4
Exercises
An air pump has a cylinder 0.250 m long with a movable piston. The pump is
used to compress air from the atmosphere at absolute pressure 1.01 x 105 Pa
into a very large tank at 4.20 x 105 Pa gauge pressure. For air, CV=20.8
J/(mol K). (a) The piston begins the compression stroke at the open end of the
cylinder. How far down the length of the cylinder has the piston moved when
air first begins to flow from the cylinder into tank? Assume that the
compression is adiabatic. (b) If the air is taken into the pump at 27.0oC, what
is the temperature of the compressed air? (c) How much work does the pump
do in putting 20.0 mol of air into the tank?
Solution
(a) For constant cross-section area, the volume is proportional to the length,
and Eq.19.24 becomes L2 = L1 ( p1 / p2 )1/ γ and the distance the piston has
moved is:
p1 1 / γ
1.01 × 105 Pa 1 / 1.400
L1 − L2 = L1[1 − ( ) ] = (0.250m )[1 − (
)
] = 0.173 m.
5
p2
5.21 × 10 Pa
Eq.19.24
pV γ = const → p1V1γ = p2V2γ
Problem 5
Exercises
Solution
(b) Raising both side of Eq.1 T1V1γ-1=T2V2γ−1 to the power γ and both sides of
γ ( γ −1)
and
Eq.2 p1V1γ = p2V2γ to power γ-1, dividing to eliminate the term V1
γ ( γ −1)
V2
solving for the ratio of the temperatures:
5.21 × 105 Pa 1−(1 / 1.400 )
p1 1−(1 / γ )
)
T2 = T1 ( )
= 480 K = 206°C.
= (300.15K )(
1.01 × 105 Pa
p2
Using the result for part (a) to find L2 and the using Eq.1 gives the
same result.
(c) Of the many possible ways to find the work done, the most straight
forwarded is to use the result of part (b) in Eq.3 W=nCV(T1-T2) :
W = nCV ∆T = ( 20.0mol )[20.8 J ( mol ⋅ K )](179.0 K ) = 7.45 × 104 J .
where an extra figure was kept for the temperature difference.
Eq.1
TV γ −1 = const → T1V1γ −1 = T2V2γ −1
Eq.2
pV γ = const → p1V1γ = p2V2γ
Eq.3
W = nCV (T1 − T2 )