Oule-Thomson Effect (焦耳—湯普森擴散效應)

Example Problem 3.6

Example Problem 3.6

A sample of N2 gas undergoes a change from an initial statedescribed by T=200 K and Pi=5.00 bar to a final state described byT=400 K and Pf=20.0 bar. Treat N2 as a van der Waals gas with theparameters a=0.137 Pa m6 mol-2 and b=3.8710-5 m3 mol-1. We usethe path N2 (g, T=200 K, P=5.00 bar)N2 (g, T=200 K, P=20.0bar) N2 (g, T=400 K, P=20.0 bar), keeping in mind that all paths willgive the same answer for U of the overall process.

Example Problem 3.6

a. Calculate using the result of

Example Problem 3.5. Notice that Vm,i=3.2810-3 m3 mol-1 andVm,f = 7.8810-4 m3 mol-1 at 200 K, as calculated using the vander Waals equation of state.

Example Problem 3.6a

,

,,

m f

m i

Vm

T m mV T

UU dVV

,, ,

6 -33 3 -1 3 3 -1

1 1

1 10.137 Pa m mol3.28 10 m mol 7.88 10 m mol

132J

T mm i m f

U aV V

Solution

Example Problem 3.6aExample Problem 3.6

b. Calculate using the following relationship

for CV,m in this temperature range:

The ratio Tn/Kn ensure that Cv,m has the correct dimension.

, ,f

i

T

T m V mTU C dT

2 3, 2 5 8

1 1 2 322.50 1.187 10 2.398 10 1.0176 10V mC T T T

J K mol K K K

Solution

Example Problem 3.6

c. Compare the two contributions to Um. Can UT,m be neglectedrelative to UV,m ?

Example Problem 3.6a

UT,m is 3.1% of UV,m for this case. In this example, and for most processes, UT,m can be neglected relative to UV,m for real gas.

Solution

To a good approximation = 0

for real gases under most conditions. Therefore, it is sufficiently accurate to consider U as a function of T only [U = U(T) for real gases in processes that do not involve unusually high gas densities.

,

,,

m f

m i

Vm

T m mV T

UU dVV

3.4

The density of liquids and solids varies only slightly with the external pressure over the range in which these two forms of matter are stable. This conclusion is not valid for extremely high pressure conditions such as those in the interior of planets and stars. However, it is safe to say that dV for a solid or liquid is very small in most processes. Therefore,

because V 0. This result is valid even if (U/V)T is large.

3.4

For this process defined by P = Pexternal = constant,

Because P = Pi = Pf , this equation can be rewritten as

The Variation of Enthalpy with Temperature at Constant Pressure

Figure 3.4The initial and final states are shown foran undefined process that takes place atconstant pressure.

3.4

the value of H can be determined for an arbitrary process at constant P in a closed system in which only P-V work occurs by simply measuring qp , the heat transferred between the system and surroundings in a constant pressure process.

The Variation of Enthalpy with Temperature at Constant Pressure

,

,

f f

i i

P

P T

PP

P

T T

P P P mT T

P P P m

H q

H HdH dT dPT P

dq HCdT T

H C T dT n C T dT

H C T nC T

Example Problem 3.7

Example Problem 3.7

A 143.0-g sample of C(s) in the form of graphite is heated from 300to 600 K at a constant pressure. Over this temperature range, Cp,mhas been determined to be

Calculate H and qp. How large is the relative error in H if youneglect the temperature-dependent terms in Cp,m and assume thatCp,m maintains its value at 300K throughout the temperatureinterval?

4

411

3

37

2

24

1-1-,

10800.7

10919.110947.11126.019.12mol K J

KT

KT

KT

KTC mp

Example Problem 3.7

.qH (3.28),equation From

kJ85.461056.110789.4

1049.60563.019.12

mol g00.120.143

10800.710919.1

10947.11126.19.12

mol g00.120.143

p

600

3005

511

4

48

3

35

2

2

1

600

3004

411

3

37

2

24

1

,

J

KT

KT

KT

KT

KT

g

KTd

KT

KT

KT

KT

molJg

dTTCMmH

f

i

T

Tmp

Solution

Example Problem 3.7

If we had assumed Cp,m =8.617 J mol K, which is the calculatedvalue at 300 K,

=143.0 g / 12.00 g mol-18.617 J K mol [600K-300K]=30.81 kJThe relative error is (30.81kJ - 46.85kJ)/46.85 kJ=-34%. In this case,it is not reasonable to assume that Cp,m is independent of temperature.

3.5 How Are Cp and Cv Related?

PTPT

VU

TV

TPTCP

VU

TVC

TVP

TV

VUCC

dTdq

PdVdVVUdTCdq

dVVUdTCdV

VUdT

TUdVPdqdU

VT

PVV

PV

PPTVP

TV

TV

TVext

since

By first law:

(In a reversible process)

3.5 How are Cp and Cv Related?

TVCCor

TVCC mmVmPVP2

,,

2

Express abstract partial derivatives with experimentally available data and

Because and are positive for ideal and real gases,CP CV > 0

3.5 How are Cp and Cv Related?

For an ideal gas

nRCC

nRP

nRPTVPand

VU

VP

PT

0

For liquids and solids VP CC

The partial derivative (V/ T)p =V is much smaller for liquids and solids than for gases. Therefore, generally

3.6

3.6 The Variation of Enthalpy with Pressure at Constant Temperature

For isothermal Processes, T=0

Usince V

P VT T

VT

T T T

T V

H U PV dH dU PdV VdPH UC dT dP C dT dV PdV VdPP V

UC dT P dV VdPV

d

H U dVP VP V dP

dPTdT

T V T P

P

H dP dV dVT V V TP dT dP dT

Example Problem 3.8

Example Problem 3.8

Evaluate (H/P)T for an ideal gas.

2

2

nRT / /V

/ / /

0

V V

TT

T V T

P T nR V andT

V RT d nRT P P nRT PP

H P V nR nRT nRT nRTT V T V VP T P V P P nRT

Solution

Example Problem 3.8

Example Problem 3.9

Calculate the change in enthalpy when 124 g if liquid methanolinitially at 1.00 bar and 298 K undergoes a change of state to 2.50bar and 425K. The density of liquid methanol under theseconditions is 0.791 g/cm3, and Cp,m for liquid methanol is 81.1 J K-1mol-1.

, ,

-1 -1-1

6 3 5

-3 3

3

124g81.1 J K mol 425 K-298 K32.04 g mol

124g 10 m 10 Pa 2.50 bar-1.00 bar0.791 g cm cm bar39.9 10 J+23.5 J=39.9 kJ

f f

i i

T P

P m P m f i f iT P

H n C dT CdP nC T T V P P

Solution

Joule-Thomson effect ()Joule-Thomson effect (throttling process)1852-62James P. Joule and William Thomson(porous plate )

3.7 The Joule-Thompson Experiment

47

Chap 2 The First Law

Joule-Thomson coefficient, enthalpy

49

Joule-Thomson expansion


.

HPT

0d , 0 if 0,d , 0 if 0d expansion, For

TTp

51


:, ( P2 < P1 ). (V2 > V1 ),:

w = P1V1 P2V2 = U2 - U1 (q = 0 )(U2 + P2V2 ) = (U1 + P1V1 ) or H2 = H1 (isenthalpic)

: T P. Joule-Thomson coefficient: J-T

HpTJ P

TPPTT

12

120

lim

P

T

PTHTJ C

PHTH

PH

PT

PPHPHT

CTH

pT

HT

pT

PH


2

1

0

1 2 2 2 1 10

2 1 2 2 1 1

2 2 2 1 1 1 2 1

0

or

V

left rightV

w w w PdV P dV PV PV

qU U U w PV PV

U PV U PV H H

Joule-Thompson Experiment is an isenthalpic process


Joule-Thompson Experiment

Figure 3.5

In Joule-Thompson experiment, a gas is forced through a porous plug using apiston and cylinder mechanism. The pistons move to maintain a constant pressurein each region. There is an appreciable pressure drop across the plug, and thetemperature change of the gas is measured. The upper and lower figures show theinitial and final states, respectively. As shown in the text, if the piston and thecylinder assembly forms an adiabatic wall between the system (the gases on bothsides of the plug) and the surroundings, the expansion is isenthalpic.

55


The modern method of measuring m is indirect, and involves measuring the isothermal Joule-Thomson coefficient, the quantity:

Which is the slope of a plot of enthalpy against pressure at constant temperature.

PTJT

CpH

TCpCTTHpP

HH ppTJPT

ddddd

57

TCpCTTHpP

HH pTJPPT

ddddd

TJPC

Example Problem 3.8Example Problem 3.10

Using equation (3.43), (H/P)T=[(U/V)T+P](V/P)T+V, andEquation (3.19), (U/V)T=T(P/T)V-P, derive an expressiongiving (H/P)T entirely in terms of measurable quantities for a gas.

1

T T T

V T

P

dH U VP VdP V P

P VT P P VT P

VT V TV V V TT

Solution


0

Joule-Thompson coefficient

limJ T PH H

T TP P

for an isenthalpic process dH=0

0PT

P J TT

HdH C dT dPP

H CP

TCpC

TTHpP

HH

ppTJ

PT

dd

ddd

3.7 The Joule-Thompson ExperimentFigure 3.6

Figure 3.6All along the curves in the figure, J-T and J-T is positive to the left ofthe curves and negative to the right.To experience cooling uponexpansion at 100 atm, T must liebetween 50 and 150 K for H2. Thecorresponding temperatures for N2are 199 and 650 K.

Example Problem 3.11

Example Problem 3.11

Using equation (3.43), (H/P)T=[(U/V)T+P](V/P)T+V showthat J-T = 0 for an ideal gas.

J-T1 1

1 0

/1

1 0

T T T TP P

TP

P T

P

H U V VP VC P C V P P

VP VC P

nRT PP V

C P

nRT VC P

Solution

62

, >0, . , H2, He. 0, 0..,

3.8 Liquefying Gases Using an isenthalpic Expansion (Figure 3.7)

Figure 3.7Schematic depiction of the liquefactionof a gas using an isenthalpic Joule-Thompson expansion. Heat is extractedfrom the gas exiting from thecompressor. It is further cooled in thecountercurrent heat exchanger beforeexpanding through a nozzle. Becauseits temperature is sufficiently low,liquefaction occurs.

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Oule-Thomson Effect (焦耳—湯普森擴散效應)