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.


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


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 −=

chapter 18: heat,work,the first law of...

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