Chapter 18: Heat,Work,the First Law of...
Transcript of Chapter 18: Heat,Work,the First Law of...
Chapter 18: Heat,Work,the First Law of Thermodynamics
Thermodynamic systemsThermodynamic system
Thermodynamic system is any collection of objects that isconvenient to regard as a unit, and that may have the potentialto exchange energy with its surroundings.A process in which there are changes in the state of a thermo-dynamic system is called a thermodynamic process.
Signs for heat and work in thermodynamicsQ W
Q is + if heat is added to systemQ is – if heat is lost by systemW is + if work is done by the system.W is – if work is done on the system.
Work done during volume changesWork done
Piston in a cylinder moves by dldue to expansion of gas at pressure p
Force on piston: F = p·A
Incremental work done:dW = F·dl = p·A dl = p·dV
∫ ∫==f
i
V
V
pdVdWW :done work Total
p
Work done during volume changes (cont’d)
Work and pV diagram
p
pApA
pB
Work done depends on the path taken.
• Process A B :If VB>VA , W > 0
• Process B A :If VB>VA , W < 0
• Process A B :If VA=VB , W = 0
∫ ==B
A
V
V
pdVW curve V-punder area
Work done during volume changes (cont’d)
Work and pV diagram (cont’d)
Work done depends on the path taken.
C
A C : W = 0
C B : W = pB(VB-VA)
B C : W = pB(VA-VB)
D
A C B : W = pB(VB-VA)
A D : W = pA(VB-VA)
D B : W = 0
p
pA
pB
A D B : W = pA(VB-VA)
Internal energy and the first law of thermodynamics
Internal energyTentative definition:
The internal energy U of a system is the sum of the kineticenergies of all of its constituent particles, plus the sum ofall the potential energies of interactions among theseparticles.
The first law of thermodynamics
WQUUU −=∆=− 12
Now define the change in internal energy using the first lawof thermodynamics.
It is known from experiments that while Q and W depend onthe path, ∆U does not depend on the path.
Internal energy and the first law of thermodynamics
A cyclic processA
B
: is a process in which the initialand final states are the same.
pA
pBupperAB
lowerBA
upperAB
lowerBA WWWW →→→→ <<> ,0,0
VA VB0<+= →→→→upper
ABlower
BAABA WWW
If the directions of the arrows are reversed, then
0>+= →→→→lower
ABupper
BAABA WWW
Internal energy and the first law of thermodynamics
Infinitesimal changes in stateThe first law of thermodynamics:
dWdQdU −=
For the system we will discuss, the work dW is given by pdV
pdVdQdU −=
Kinds of thermodynamics processAdiabatic processAn adiabatic process is defined as one with no heat transfer intoor out of a system.
∆U = -W, Q=0
Isobaric processAn isobaric process is a constant-pressure process.
W = p(V2-V1) , p = constant
Isothermal processAn isothermal process is a constant-temperature process.For a process to be isothermal, any heat flow into or out of thesystem must occur slowly enough that thermal equilibrium ismaintained.
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 aconstant-volume process.
BABDA WW →→→ <
Kinds of thermodynamics process (cont’d)
Isothermal vs. adiabatic process
Work done by expanding gases: path dependence
V
p
Isotherm at T
1
2
V1 V2
p2
p1
Isothermal expansion (1 → 2 on pV diagram) for ideal gas:
W = nRT ln (V2/V1)
)ln(
/,
1
22
1
2
1
VVnRT
VdVnRTW
VnRTppdVW
V
V
V
V
==⇒
==
∫
∫
Work done by expanding gases: path dependence
V
p
Isotherm at T
1
2
V1 V2
p2
p1
W = p2(V2 – V1)
Expression for the work done in an isobaric expansion of an ideal gas (3 → 2 on pV diagram).
3
Internal energy of an ideal gasFree expansion
Uncontrolled expansion= Free expansion
Controlled expansion(The temperature is keptconstant and the expansionprocess is done very slowly)
the same final state
No heat exchangedHeat exchanged
the same initial state
No work doneWork done
Different paths
Internal energy of an ideal gasAnother property of an ideal gas
From experiments it was found that the internal energy ofan ideal gas depends only on its temperature, not on itspressure or volume.
However, for non-ideal gas some temperature change occur.The internal energy U is the sum of the kinetic and potentialenergies for all the particles that make up the system. Non-idealgas have attractive intermolecular forces. So if the internal energyis constant, the kinetic energies must decrease. Therefore as thetemperature is directly related to molecular kinetic energy, for anon-ideal gas a free expansion results in a drop in temperature.
Heat capacities of an ideal gasTwo 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)
constant-pressure
dTnCdUdUdWdUdTnCdQ VV =⇒=+==
nRdTdTnCdTnCdTnCdUnRdTpdVnRTpV
pdVdUdWdUdTnCdQ
Vp
V
p
+=⇒==→=
+=+==
;
RCC Vp +=
T1 and T2 are the same
For an ideal gas
Heat capacities of an ideal gasRatio of heat capacities
V
p
CC
=γ
Example
67.135
25
23
==⇒
=+=+=
γ
RRRRCC Vp
for an ideal monatomic gas
40.157
27
25
==⇒
=+=+=
γ
RRRRCC Vp
for most diatomic gas
Adiabatic processes of an ideal gasp, V, and γ
constTVconstTVnconstnVnT
VdV
TdT
CCC
CR
VdV
CR
TdT
VnRTpdV
VnRTdTnC
pdVdTnCdWdU
pdVdWdTnCdU
V
Vp
V
V
V
V
V
=⇒=⇒
=−+⇒
=−+⇒−=−
=⇒
=+⇒
=⇐−=⇒
−=⇒−=
==
−− 11)()1(
0)1(1
0
γγ
γ
γγ
l
ll
( adiabatic process)
( by definition)
( ideal gas)
Adiabatic processes of an ideal gasp, V, and γ (cont’d)
constpVconstVnRpVnRpVT
constTV
=⇒=⇒
=
=
−
−
γγ
γ
1
1
( ideal gas)
)(1
1)(
)(
)(,0
22112211
21
12
VpVpVpVpR
CW
nRTpVTTnCW
TTnCUUWQ
V
V
V
−−
=−=
⇒=
−=⇒
−=∆∆−==
γ
( adiabatic process)
( ideal gas)For adiabatic process, ideal gas
For adiabatic process, ideal gas
ExercisesProblem 1
Solutionp
b cpc
a d
(a)
pa)(,0),(,0
acaaddc
accbcab
VVpWWVVpWW
−==−==
)(,)(,
acaadaddcdc
accbcbcabab
VVpUUQUUQVVpUUQUUQ
WUQ
−+−=−=−+−=−=
+∆=(b) VVcVa
(c) )()()()(),( accacaccbcababcaccabc VVpUUVVpUUUUQVVpW −+−=−+−+−=−=
)()()()(),( acaacacadcadadcacaadc VVpUUVVpUUUUQVVpW −+−=−+−+−=−=
,ac pp > adcabc QQ > .adcabc WW >So assuming and
ExercisesProblem 2
Solution
pb
a cVVa Vc
pc
pa
Three moles of an ideal gas aretaken around the cycle acb shownin the figure. For this gas, Cp=29.1J/(mol K). Process ac is at constantpressure, process ba is at constantvolume, and process cb is adiabatic.Ta=300 K,Tc=492 K, and Tb=600 K.Calculate the total work W for the cycle.
.1079.4)300492)](/(315.8)[3()( 3 JKKmolKJmolTTnRTnRVpW acac ×=−=−=∆=∆=Path ac is at constant pressure:
Path cb is adiabatic (Q=0):
.1074.6)492600)])/(315.8)/(1.29)[3(
))((,
3 JKKmolKJmolKJmol
TTRCnWRCCTnCUUQW
cbpcb
pVVcb
×−=−−−=
−−−=→
−=∆−=∆−=∆−=
Path ba is at constant volume: 0=baW
.1095.1 3 JWWWW bacbac ×−=++=
ExercisesProblem 3
You are designing an engine that runs on compressed air. Air entersthe engine at a pressure of 1.60 x 106 Pa and leaves at a pressure of2.80 x 105 Pa. What must the temperature of the compressed air befor there to be no possibility of frost forming in the exhaust ports of theengine? Frost will form if the moist air is cooled below 0oC.
Solution1
221
111 −−− =→= γγγ VTVTconstTV
γγγ2211 VpVpconstpV =→=
γγγγγ11
2
1212
121
11 )(
−−− =→=
ppTTTpTp
.176449)1080.21060.1)(15.273( 7
2
5
6
1 CKKT °==××
=Using γ=7/5 for air,
ExercisesProblem 4
An air pump has a cylinder 0.250 m long with a movable piston. The pump isused to compress air from the atmosphere at absolute pressure 1.01 x 105 Painto a very large tank at 4.20 x 105 Pa gauge pressure. For air, CV=20.8J/(mol K). (a) The piston begins the compression stroke at the open end of thecylinder. How far down the length of the cylinder has the piston moved whenair 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 pumpdo 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 and the distance the piston hasmoved is:
γ/12112 )/( ppLL =
.173.0])1021.51001.1(1)[250.0(])(1[ 400.1/1
5
5/1
2
1121 m
PaPam
ppLLL =
××
−=−=− γ
Eq.19.24 γγγ2211 VpVpconstpV =→=
ExercisesProblem 5
Solution
(b) Raising both side of Eq.1 T1V1γ-1=T2V2
γ−1 to the power γ and both sides ofEq.2 p1V1
γ = p2V2γ to power γ-1, dividing to eliminate the term and
solving for the ratio of the temperatures:
Using the result for part (a) to find L2 and the using Eq.1 gives thesame result.
(c) Of the many possible ways to find the work done, the most straightforwarded is to use the result of part (b) in Eq.3 W=nCV(T1-T2) :
where an extra figure was kept for the temperature difference.
)1(1
−γγV)1(
2−γγV
.206480)1001.11021.5)(15.300()( )400.1/1(1
5
5)/1(1
2
112 CK
PaPaK
ppTT °==
××
== −− γ
.1045.7)0.179)]((8.20)[0.20( 4 JKKmolJmolTnCW V ×=⋅=∆=
Eq.1 122
111
1 −−− =→= γγγ VTVTconstTVγγγ
2211 VpVpconstpV =→=)( 21 TTnCW V
Eq.2Eq.3 −=