1

Chapter 4. Applications of the First Law

2

Introductory remarks:

work to heat:- no tax heat to work:- tax

work and heat are not forms of energy:- they are names for methods of transferring energy

Heat is not a noun! (nor an adjective)Heat converted to work means that energy was transferred

from a source by heating and then transferred by doing work.{To avoid circumlocutions we will often, in fact, use heat as a noun or an adjective!}We often wish to know how one quantity varies with a variation in another quantity under certain conditions. Tabulations are available for certain of these variations that are easily measurable. We can then often derive expressions for other variations that are not tabulated in terms of these tabulated quantities.

3

Heat CapacityWe consider a situation in which there is no change of phase. The heat capacity is defined as: C = (đQ /dT)

This is not a derivative of Q with respect to T as Q cannot be written as a function of T. (Misleading?)

The heat capacity is extensive. The intensive equivalent is called the specific heat capacity

c = đQ/(mdT) heat capacity per unit mass.c = đQ/(ndT) heat capacity per mole.

Since đQ depends upon the process, c is not specified until the process is specified. For a given dT, đQ could be +, - or 0.

4

Some minor comments:

Heat capacity is a misnomer because “heat’ is not “contained” in a body. It should be called an energy capacity.

When reference is made to a specific heat, keep in mind that it can be /mole or /kilomole or /mass or /volume.

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The most common specific heats are for constant pressure and for constant volume.

cv =( đQ/ndT)v c at constant V

cp = (đQ/ndT)p c at constant P

cp is usually measured as cv is difficult experimentally.

There are many relationships between thermodynamic quantifies which can be obtained by combining the definitions, laws and rules for partial derivatives. Here we will consider P,V,T systems (hydrostatic systems),

but there are similar relationships for other systems.

In a P,V,T system certain quantities are readily measured directly: P,V,T, volume expansivity β, and the isothermal compressibility κ.

6

Let us take T and V as independent variables.

VTU

VdTQ

V

TVU

VTU

TVU

VTU

C

)3(dVPdTQ

)2(VddTdU

)1(VPddUQ

VV T

UC

đ

đ

Cv can be determined

experimentally

This equation is true for any reversible process

If one has a model of a substance one can calculate from this model and compare with the experimentally

determined CV.

VT

U

đ

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Let us consider Cp:

Equation (3): dVPV

UdT

T

UQ

TV

đ

At constant pressure:

PTVP T

VP

V

U

T

U

dT

Q

đ

VPV

UCC

T

VP

P

V

CC

V

U VP

T

This is an example of an equation that relates the quantity on the left, which is usually not measured, to state functions which can be measured and are often tabulated.If we have a model of a substance, we can calculate the quantity on the left and compare the result with the experimentally determined quantities on the right hand side.

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EXAMPLE: Chapter 4The molar specific heat capacity of most substances (except at very low temperatures) can be satisfactorily expressed by the empirical formula 22 cTbTacP

(a) Find the heat required to raise the temperature of n moles of the substance at constant pressure from 21 TtoT

(b) Determine the specific heat capacity for Mg at 300K given

)(1027.313.3107.25 83

Kkmole

Jinccba P

(a) dTT

cbT2aqdTcq

dT

qc

2

1

T

T

2PP

12

21

2212

11)()(

TTcTTbTTaq

12

21

2212

11)()(

TTcTTbTTanQ

đ đ

9

(b)

28

23

)300(1

1027.3)300()13.3(2107.25Kkmole

KJK

Kkmole

J

Kkmole

JcP

Kkmole

JcP

41039.2 Mg at 300K

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EXAMPLE: Chapter 4From figure 4.1 (textbook) estimate the energy to heat one gram of Cu from 300K to 600K:(a) At constant volume(b) At constant pressure(c) Determine the change in internal energy of the Cu in each case.

(a) Constant V )constantV(dTncQdT

Q

n

1c V

V

V

For the specific heat take the midpoint (450K) Kkmole

JcV

31025

đ

kmolenkmolekg

kgnamuM 5

3

1057.1/5.63

105.63

)300(/

10251057.1

)(

35

0

KKkmole

JkmoleQ

TTncQdTncdQ ifV

T

T

V

Q f

i

Q=118J

constant V

đ

11

(b) Constant PKkmole

JcP

31026

)constantP(dTncQdT

Q

n

1c P

P

P

)300(10261057.1)( 35 KKkmole

JkmoleTTncQ ifP

Q=122J constant P

(c) At constant volume W=0 so JUQU 118

For constant P

)constantP(VdTdVT

V

V

1dVPW

P

V

V

f

i

f

i

f

i

f

i

V

V

T

T

T

T

TVVdTVVdTdV

Look up values for Cu 3315 1092.8102.5m

kgK

constant V

đđ

12

3733

3

1012.1/1092.8

10mV

mkg

kgmV

V

m

393715 1075.1)300(1012.1102.5 mVKmKTVV

JWmPaW 4395 1077.1)1075.1)(1001.1(

JUJJWQU 1221077.1122 4

constant P

(d) When heat enters a system at constant V, all the energy goes into increasing the internal energy. When heat enters at constant pressure, some of the energy is expended in doing work against the surroundings and there is a smaller temperature rise.

VP

P

P

V

V CCT

QC

T

QC

smaller

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Free (Joule) expansion of a real gas. (System isolated)

In this process Q=0, W=0 so U=constant. Does T change during this process? The Joule coefficient is introduced. Is it zero?

UV

T

This coefficient is extremely difficult to measure directly, so indirect methods are used to answer the question.Considering U=U(T,P):

dPP

UdT

T

UdU

TP

If, in a free expansion, dT=0 and since dU=0 then 0

TP

U

Experiments by Rossini and Framdsen found that 0)(

TfP

U

T

Hence, for real gases, in a Joule expansion. 0dT

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Ideal Gas:We now refine our definition of an ideal gas as one that satisfies the relationship PV=nRT and U=U(T) only. For a quasi-static process of a hydrostatic system: đQ=dU+PdVWe have and since, for an ideal gas, U=U(T) only,

this gives so (ideal gas)

V

V T

UC

dT

dUCV PdVdTCQ V

Using the equation of state: PdV+VdP=nRdT and so

)1(VdPnRdTdTCQ V Dividing by dT and considering processes at constant P,

nRCCornRCdT

QVPV

P

(Mayer’s Equation)

Substituting into (1) gives (ideal gas)VdPdTCQ P đ

đ

đ

đ

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Continuing with an ideal gas:

Starting with U(T,V) dVV

UdT

T

UdU

TV

dT

dV

V

U

T

U

dT

dU

TV

PTVP T

V

V

U

T

U

T

U

But, for an ideal gas (only!), 0V

U

T

VVP

CT

U

T

U

Hence, for an ideal gas (only!) we can write irrespective of the process under consideration dT

dUCV

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We introduce a very useful ratio:

V

P

c

c

Although the molar specific heats vary somewhat with temperature, at temperatures not too far from ordinary temperatures, experimental results give:

Monatomic gases:

Diatomic gases :

67.125

soRcP

40.127

soRcP

For a phase change:VaporLiquidSolid

System gives off heat

System absorbs heat

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We are now ready to introduce the concept of enthalpy and I would like to provide a reason for doing so.

It is not always convenient, and may even be dangerous, to carry out a process at constant volume. (This is particularly true in chemistry!) It is often preferable to carry out experiments at constant pressure (perhaps atmospheric). For these latter processes, the concept of enthalpy is useful.

As an example consider a liquid at its BP in a cylinder and the piston is held in place by atmospheric pressure. Now consider energy supplied to vaporize the liquid, the pressure remaining atmospheric. The result will be an increase in the volume and an increase in internal energy. But to achieve this transformation we must supply not only the internal energy increase, but also the energy required for the gas to push the piston and since the pressure is constant ΔW=PΔV and so the energy that must be supplied is

ΔU+PΔV and we then write ΔU+PΔV =ΔH

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The quantity H is called the enthalpy.

Even absent the piston this energy must be supplied as the atmosphere must be pushed aside.

Such a situation is very common in chemical processes.Hence there exist extensive tabulations of ΔH.

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Enthalpy and heats of transformation:

Isobaric processes often occur, such as in a change of phase (isothermal and isobaric). Keeping track of the work can be annoying.We can sidestep this problem by considering the enthalpy, rather than the energy in these processes. In a phase change )vv(PW 12 From the first law: vPdqddu Consider a finite change and introduce , the latent heat of transformation per kilomole.

q

)Pvu()Pvu(or)vv(P)uu( 11221212

We introduce the specific enthalpy h=u+Pv

Hence 12 hh

We now consider the possible phase changes. Notation:1=solid 2=liquid 3=vapor

1 prime(′) =solid 2 primes(″)=liquid 3 primes(′′′)=vapor

is the heat involved in a phase change

(for a kilomole)

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We can now write, in terms of this notation:hh 12hh 23hh 13

solid liquid (fusion)liquid vapor (vaporization)

solid vapor (sublimation)

h is a state function! Hence 0dh

We consider a small cycle that encloses the triple point. (see next slide)We assume that the circle is so small that the enthalpy changes appreciably only during a change of phase. The cycle is

solid vapor liquid solid

solid vapor (heat flows in) vapor liquid (heat flows out) liquid solid (heat flows out)

131 h23322 h12213 h

1223133210 hhhdh

This gives 231213 (all positive quantities)

21

P

CP

TP

solid liquid

vapor

S-L

S-V

L-V

T

22

The enthalpy of a system is then H=U+PVIf we could annihilate the system we could, in principle, not only extract the energy, but also the work done (PV) by the atmosphere as it fills in the vacuum left when the system disappears.

For an isobaric process dH=dU+PdVand the enthalpy can increase for two reasons:

• the energy of the system increases• the system expands

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Relationships involving enthalpy:Pdvduq

vdPPdvdudhPvuh Hence vdPdhq

But we can write dPP

hdT

T

hdh)P,T(hh

TP

dPvP

hdT

T

hq

TP

Now for P constant dTT

hq

P

P

P

PP T

hcso

T

h

dT

q

Vc Pc is an “energy capacity”, while is an “enthalpy capacity”

đ

đ

đ

đ

đ

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It should be mentioned that there does not have to be any heat involved as energy or enthalpy can be changed by other means, such as in a microwave oven.

In first order phase transitions (melting, boiling, sublimation)the change in enthalpy is called “latent heat”. This term is used because there is no change in temperature.

Extensive tables of enthalpy changes exist. For example we can find that the change in enthalpy when 1 mole of water at 1 atmosphere and 373K is converted to steam is 40,660J. Since a mole is about 18 grams the change in enthalpy per kilogram is 2.26MJ/kg, which is the latent heat of vaporization of water. Not all of the energy ends up in the vapor. Some work must be performed to push the atmosphere away.

Neglecting the volume of the liquid and treating the vapor as ideal and for 1 mole of water, PVvapor= RT

PVvapor= (8.31J/K)(373K)=3100J

This is about 8% of the 40,660J

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Note on enthalpy and review:As a result of the Joule-Thomson experiment (discussed later), it can be shown that, for an ideal gas: 0

TP

H

PdVdTCQsodT

dUCgasidealPdVdUQ VV

Consider H=H(T,P) dPP

HdT

T

HdH

TP

=0 for an ideal gas

Hence dTCdHCT

H

dT

dHdT

T

HdH PP

PP

VdPPdVPdVQVdPPdVdUdHPVUH

VdPdTCQVdPQdTC PP (ideal gas)

We will soon use these two expressions for an ideal gas.

đ đ

đ

đ đ

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For an ideal gas f

i

P

f

i

P dTCdHdTCdH

f

i

Pif dTCHH for all processes involving an ideal gas

u h

reversible process vdPqdhPdvqdu

PP

VV T

hc

T

uc

ideal gas vdPdTcqPdvdTcq PV

0P

h0

V

u

TT

f

i

Pif

f

i

Vif dTchhdTcuu

đđ

đđ

27

Quasi-static adiabatic processes (ideal gas):

We had vdPdTcqandPdvdTcq PV

In an adiabatic process q=0 and so, upon division of the two equations, we obtain

v

dv

P

dP

For modest temperature changes, is constant, and upon integrating we obtain:

constantPv (adiabatic)

This is for an ideal gas!

đ đ

đ

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Example: Adiabatic expansion of an ideal gas. (The hard way.)?,, UVPVP ffii

f

i

f

i

V

V

ii

V

V

iiV

dVVPWPVVPPdVW

111

111

11

1

if

ii

V

V

iiVV

VP

VVPW

f

i

)1(11

111

1

11

1

f

iii

if

iii

V

VVP

VV

VVPW

To find ratio of volumes )2(fi

if

f

i

TP

TP

V

VnRTPV

i

f

f

i

f

i

f

i

f

i

f

i

i

fffii T

T

V

Vor

T

T

V

V

V

Vinso

V

V

P

PVPVP

1

)2(

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Placing this in (1)

][1

11

11 if

i

fi

i

fii TTnR

WT

TnRT

T

TVPW

Since the gas is ideal

VVVVVP CnR

CCCnRnRCC

1

)1(

][ ifV TTCW ][ ifV TTCUWWQU

or, since for an ideal gas U=U(T) only dT

dUCV

][ ifVV TTCUdTCdU

(The first approach was just to show how much fun algebra can be and to practice some techniques.)

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EXAMPLE:

Q

000 ,,, TVPn 000 ,,, TVPn

0827, Pn

LTn, RT

Ideal gas5.1

adiabatic piston adiabatic walls

A cylinder with thermally insulated walls contains a moveable frictionless thermally insulated piston. On each side of the piston are n moles of an ideal gas. The initial state variables are the same on both sides. By means of a heating coil in the gas on the left side, heat is slowly supplied. The gas on the left expands and compresses the gas on the right hand side until its pressure has increased toIn terms of determine: (a) How much work is done on the gas on the right side.(b) The final temperature of the gas on the right side.(c) The final temperature of the gas on the left side.(d) How much heat flows into the gas on the left side.

0)8/27( P0,, Tcn V

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(a) On the right side no heat enters so we have an adiabatic compression.

0

/1

R0RR000 V27

8VV

27

8VVP

8

27VPconstantPV

000

3/2

914

94

278

VVsoVVVV LRR

3/23/2

8

27

8

2700027

808

271

nR

VPVP

nRRT

RnRT

RVRP

WUWQUTTTT RR

00

3/1

23

827

Ideal gas so dTncdTCdU VV

0000 21

21

23

TncWTncTTncU VVV

(b) From above 023TTR

32

(c)

OLLLLLL VPnR

TnRTVPVVPP914

8271

914

827

000

025.5 TTL

(d) For the left side the gas expands so, from part (a) 02

1TncW VL

000 25.4)125.5()( TncUTncTTncU VLVLVL

00 5.025.4 TncTncWUQWQU VVLLLLLL

075.4 TncQ VL

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Example: Problem 4-4

Consider u=u(T,P) dPP

udT

T

udu

TP

dT

dP

P

u

T

u

dT

du

TP

VTPV T

P

P

u

T

u

T

u

)1(VTP

V T

P

P

u

T

uc

1P

v

v

TPV

T

T

P

v-

v

PvTv

T

P

VT

P

In equation (1) )2(

TPV P

u

T

uc

?

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From problem 4-3: vPcT

uP

P

In equation (2)

TPV P

uPcc

v

TVP P

uPcc

vT

VP P

uPcc

v)(

)(-v VPT

ccPP

u

Hence we have shown that this partial can be calculated in terms of easily measurable or tabulated quantities.

{To derive, start with definition of cP and use h=u+Pv}

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Recapitulation:The key concept is that adiabatic work is independent of path.U=internal energy function or adiabatic work function.

Q and W are methods for changing the internal energyW=mechanical method of energy transferQ=non-mechanical method of energy transfer

When the transfer is over, heat and work are no longer useful or meaningful.

Heat is not a noun! It is wrong to make a statement such as “the ocean holds a lot of heat.”Using heat as a verb can be confusing:

to heat : add energy to or raise the temperature of

36

Avoid using the term “thermal energy”. This is obscure or ambiguous. Sometimes it is used to mean “heat” and at other timesinternal energy. Often it is not clear what is meant.

Consider the adiabatic compression of a gas. It is often said that there is no flow of thermal energy into the gas, but then we are told that the “thermal energy” has increased.

In Chapter 2 we introduced the work PdvWIn Chapter 3 we introduced the 1st Law Pdvduq In this Chapter we introduced the specific heat (often tabulated)which permits us to now calculate đq and so du can be obtained.

The specific heat is a very important property of systems andso extensive tabulations exist and much theoretical work has also been done. It will be considered throughout the course.

đ

đ

37

For an adiabatic process undergone by an ideal gas,

constantPv

Along the way, we introduced another state variable, the enthalpy.This is very convenient when the pressure is held constant and also processes involving a change of phase. h=u+Pv

Also, for any reversible process, VV T

UC

For an ideal gas, we added the condition that U=U(T) only!

38

Comments on a thin wire.

The linear expansivity α and Young’s Modulus Y are introduced.Young’s Modulus is always positive.The linear expansivity is almost always positive. (The few exceptions include rubber.)

In an assignment, you showed:

dYA

1dL

L

1dT

If there is no change in the tension, a slight increase in L will result in an increase of T. On the other hand, if the length is held fixed and the tension increases, T will decrease. If both the length and tension change, the T change can be positive or negative.