The Properties of Gases

박영동 교수자연과학대학 화학과

기체는 왜 다루는가 ?역사적인 배경과 의미1. Boyle’s Law – First Scientific Experiment, 16612. Charles’s Law – Definition of Temperature, 1780s3. Avogadro’s Hypothesis – 4. Combined Ideal Gas Law and Kinetic Theories–

First Successful Scientific Law derived from purely mathematical approach.

실용적인 배경과 의미1. 기체는 열역학의 기초 , 열역학 제 1, 2 법칙을 설명2. 화학 평형 , 화학적 퍼텐셜의 기초3. 화학반응의 근본적 이해를 도움 - 반응동력학4. 기체를 다루는 공정 이해에 중요한 기초를 줌

The perfect gas equation of state

Isotherms of an ideal gas and a real gas.

Boyle’s law in 1660.

an ideal gas and a real gas

/p F Apressure is the force F acting on an area AN/m2 = Pa(pascal)1 atm = 1.013×105 Pa = 760 torr1 bar = 1×105 Pa

The perfect gas equation of state

pV nRT

Charles’s law in 1780s.

Charles’s law and a perfect gas

Avogadro’s principle in 1811.The molar volume of various

gases at SATP(RT, 1bar), dm3 ∙ mol-1 Perfect gas 24.7896

He 24.8Ar 24.4H2 24.8N2 24.6O2 24.8

NH3 24.8The gas constant, R

8.31447 J∙K-1∙mol-18.20574×10-2

L∙atm∙K-1∙mol-1

1.98721 cal∙K-1∙mol-1

Pressure and partial pres-sure

Partial pressure of idea gases

and A BA B

n RT n RTp pV V

,

( )A BA B A B

n RT n RT RTp p p n nV V V

and A A B Bp x p p x p ,

and A BA B

A B A B

n nx xn n n n

,

= mole fraction of ix i

Dalton’s concept of partial pressure in 1801.

Kinetic model of an ideal gasMaxwell distribution of speeds

2

3/2 2

12( ) with ( ) 4 ( / 2 ) exp( )

Msf F s s F s M RT s

RT

an ideal gas in a container of side l

Basic assumptions:

very small particles, all with non-zero mass. in constant, random motion.perfectly elastic collisions with the walls.negligible interactions among molecules except collisions.The total volume of the gas molecules is negligible.The molecules are perfectly spherical in shape, and elastic.no relativistic effects. no quantum-mechanical effects. instant collision with the wall.The equations of motion of the molecules are time-re-versible.

01 ( )F s ds

Kinetic model of an ideal gas for basics

Click here to see KineticTheory_of_Gas.pdf

Maxwell distribution of speeds

1/2

0

8( ) RTc sF s dsM

1/22

0

3( ) RTc s F s dsM

1/22mp

RTcM

Maxwell distribution of speeds

1/2

0

8( ) RTc sF s dsM

1/21/2

2

0

3( ) RTc s F s dsM

1/22mp

RTcM

c(mp) 422 c 476

c(rms) 517

Diffusion, Effusion

(a) Diffusion, (b) Effusion

Effusion when diameter is smaller than the mean free path λ.

Effusion rate is propor-tional to speed of mole-cules, and area of a small hole.

Effusion rate ∝c ∝(T/M)1/2

Molecular Collisions

cross sectional area σ= πd2

sweep volume= ‘c’ σ

‘c’

collision rate z =‘c’ σ × number density of mole-cule

= 21/2NAσcp/RTmean free path λ= RT/(21/2NAσp)

number density of molecules = N/V = nNA/V = pNA/RT

collision cross-section/nm2

Ar 0.36C2H4 0.64C6H6 0.88CH4 0.46Cl2 0.93CO2 0.52H2 0.27He 0.21N2 0.43Ne 0.24O2 0.40

SO2 0.58

Real Gases

potential energy variation

Interaction between molecules

Interactionsphase transitionfinite sizes of molecules

Isotherms of CO2

Phase transition and the critical tempera-ture

Critical Temperature

CO2 at the pressure of 75 atm

31 ℃

http://www.youtube.com/watch?v=8ZbZVikZP9w&feature=related

Compression Factor, Z

Compression factor at 0 ℃

0 /m m m

m

V V pVZV RT p RT

Z 실제기체의 몰부피이상기체의 몰부피

21 ...m

m m

pV B CZRT V V

The van der Waals equation of state

volume effect

size effectaccessible volume is smaller than the physical volume due to the molecular volume.

interaction effectpressure measured is smaller than the ‘ideal’ pressure due to the attrac-tions between molecules.

0 0p V nRTvan der Waals correction to

V nb0V

2np aV

0p

2

2 ( )anp V nb nRTV

The van der Waals equation

van der Waals isotherms interpretation of the van der Waals equation

2

20, and 0.p pV V

at critical point,

2m m

RT apV b V

2

83 , , 27 27c c c

a aV b T pRb b

van der Waals coefficients　 a/(atm dm6 mol- 2) b/(10-2 dm3 mol-1)Ar 1.337 3.20

CO2 3.610 4.29He 0.0341 2.38Xe 4.137 5.16

The liquefaction of gases

The principle of the Linde refrigerator.

The van der Waals gas and Boyle Temperature

Z=0 at Boyle temperature TB,

B = b - a/RT = 0

2m m

RT apV b V

van der Waals coefficients TB (K)　 a/(atm dm6 mol-

2)b/(10-2 dm3 mol-

1) vdW(a/bR) exp

Ar 1.337 3.20 509 411.5CO2 3.610 4.29 1025 714.8He 0.0341 2.38 17.5 22.64Xe 4.137 5.16 977 405.9

21 ...m

m m

pV B CZRT V V

2(1 / ( / ) ...)(1 / ) m m

m m m m

RT RT RT b V b VV b V b V V

22

2

(1 ( ) ...)

11 ( ) ( ) ...

m m m m

m

m m

RT b b apV V V V

pV a bbRT RT V V

BaT

bR

( )aB bRT