Homework - KOCWcontents.kocw.net/KOCW/document/2015/pusan/limmanho/1.pdfChap 1. The properties of gases (2015) Spring Physical Chemistry (I) by M Lim 2 • Gas: the simplest state

HomeworkChap 1. The properties of gases

(2015) Spring Physical Chemistry (I) by M Lim

1

Problems: 1A.1 1A.8 1A.10 1C.1 1C.2

1C.3 1C.4 1C.9 1C.12 1C.16

Chap 1. The properties of gases


2

• Gas: the simplest state of matter– A collection of molecules (or atoms) in continuous random motion

– Average speeds increases as T is raised

– The molecules of a gas are widely separated (negligible intermolecular forces)

• The perfect gas: an idealized version of a gas– Obey the perfect gas law: pV = nRT

• Real gas: do not obey the perfect gas law, (high p or low T)

– Van der Waals eqn

– Virial equation

1장수업목표 1: 기체의성질• Pressure: Standard pressure: 1 bar = 105 Pa

cf. 1 atm = 101325 Pa =760 Torr = 760 mmHg

• Temperature: the direction of the flow of energy

• Perfect gas law:

• Partial pressure:

o

m

nRTp

V

V RTV

n p

J Jp x p


3

A B Cp p p p

1A.1 Variables of states


4

The physical state of a sample of a substance (its physical condition):

defined by its physical properties (V, p, T, n)

(a) pressure: p = F/A (1 Pa = N/m2 = kgm−1s−2)

Standard pressure: 1 bar = 105 Pa


)( op


5



6

A (area)

h (height)

V=Ah (volume)

Barometer: measures the atmosphere

F = mg = ρAhg

Δp = F/A = ρgh

Example 1A.1

Mechanical equilibrium

1A.1 The variables of states


7

(b) Temperature (thermometer)

the property that indicates the direction of the flow of energy

Celsius scale (Ө/ºC ): the length of a column of a liquid

perfect gas temperature scale: the pressure of the perfect gas

= thermodynamic temperature scale (T/K): T/K = Ө/ºC + 273.15

1A.2 Equation of state


8

Boyle’s law: pV = constant (at cont. T)

Charles’s law: V = constant • T (at const p)

Avogadro’s principle: V = constant • n (at const. p, T)

The perfect gas law: pV = nRT

• The perfect gas (or ideal gas) law:

A real gas obeys in the limit of p→ 0

1A.2 Equations of states

V

nRTp

m

VV

n


9

2 3 3

1 22.414

1 273.15

101325 22.414 10

1 273.15

pV atm lR

nT mol K

Nm m

mol K

3 3 3 31 10 1 10 l cm dm m

• Standard ambient temperature and pressure (SATP)

T=298.15 K, p = 1 bar

Vmo = 24.789 l/mol

• Standard temperature and pressure (STP)

T= 0 ºC, p = 1 atm

Vmo = 22.414 l/mol

Molar volume:

(R=NAk)

o

m

RTV

p

1A.2 (b) Mixtures of gases

JJ total A B C J

total

nx n n n n n

n


10

• Partial pressure: pJ = xJ p

• Mole fraction: xJ

• Dalton’s law:

부분압 (partial pressure): 혼합기체에서 특정 한 기체의 압력

1A B C

A B C A B C

x x x

p p p x x x p p

A B Cp p p p

1A.2 (b) Mixtures of gases


11

1장수업목표 2: 기체운동론

• Maxwell-Boltzmann distribution of speed

• Root mean square (rms)speed:

• Mean speed:

• Collision frequency:

• Mean free path:


12

2

3

22 24

2

Mv

RTM

f v v eRT

11

22 2

3RTc v

M

1

28mean

RTv

M

1

relrel

rel

v pz v N

kT

v kT

z N p

1B. The kinetic model


13

1. The gas consists of molecules of mass m in ceaseless random

motion.

2. The size of the molecules is negligible, in the sense that their

diameters are much smaller than the average distance travelled

between collisions.

3. The molecules interact only through brief, infrequent, and

elastic collisions (탄성충돌).

In the kinetic theory of gases it is assumed that KE of the molecules is the only contribution to E of the gas.

1B.1 The model


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1B.1(a) Pressure and molecular speeds-1a

21

3pV nMc

p of a perfect gas according to the kinetic model

a molecule

2

2

2

Momentum change of a molecule after collision:

2

1Number of colliding molecule in :

2

= 2 =2

x

Ax

A x xtotal x

x

x

P mv

nNt Av t

V

nN Av t nMAv tP mv

V V

P nMAvF

t V

F nMvp

A V

2 2

2 2 2 2 2

3

where

x

x y z

nM v nMc

V V

c v v v v

1 22

AM mN

c v

2

2 3

3

nMcpV nR

RTc

M

T

반은 오른쪽 나

머지 반은 왼쪽.

반만 벽과 충돌


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1B.1(b) The Maxwell-Boltzmann distribution of speed

2

2

2

2

1

22

1direction

2

Probability having ,

2

x

x

x x

x x

mvE

kT

m

k

k

v

Tx

Tx

mf v

v KE mv

v v

f v e e

ekT

: Maxwell-Boltzmann

velocity distribution

2

3

22 24

2

Mv

RTM

f v v eRT

Maxwell-Boltzmann distribution of speeds

molar mass,

gas constant,

A

A

M mN

R kN

2

2

2 1

21

2

x

ax

x x

mv

kTx

e dxa

f v dv

Ne dv

kTN

m

ma

kT


16

2

3

22 24

2

Mv

RTM

f v v eRT

2 2 2

2

~

~

~ ,

si

x x x

y y y

z z z

x y z

x y z

v v dv

v v dv

v v dv

v v v v

dv dv dv v

22 2

2

2

3

22 2 2

3

2222

0 0

3

22

n

2

sin2

42

x x y y z z

x y z

yx zx x y y z z

x y z

v dv v dv v dv

x x y y z zv v v

mvmv mvv dv v dv v dv

kT kT kTx y z

v v v

mvv dv

kT

v

mv

k

dvd d

f v dv f v dv f v dv

me e e dv dv dv

kT

me v dvd d

kT

me

kT

2

3

22 2

2

42

mv

v dv v dvT

v v

kTm

f v v ekT

v dv f v dv

molar mass,

gas constant,

A

A

M mN

R kN

2

1

22

direction

2

xmv

kTx

x

mf v e

v

kT



17

2

3

22 24

2

Mv

RTM

f v v eRT



18

1B.1(c) Mean values

0

n n f vv v dv

2

3

22 24

2

Mv

RTM

f v v eRT


19

2

3

22 24 0

2

Mv

RTM

f v v e vRT

2

2

2

2

2

0

0

2 3 2

0

3

20

4 5 2

0

1

2

1

2

4

1

2

3

8

ax

ax

ax

ax

ax

e dxa

xe dxa

x e dx a

x e dxa

x e dx a

1B.1(c) Mean values

0

2 2 2

0

meanv vf v dv

c v v f v dv

Root mean square (RMS) speed

Mean speed

Most probable speed


20

11

22 2

1 1

2 2

1 1

2 2

3

8 8

3

2 2

3

mean rms

mp rms

RTc v

M

RTv v

M

RTv v

M

2

3

22 24 0

2

Mv

RTM

f v v e vRT

1B.1(c) Mean values

Ex 1B.1 vmean of N2 molecules in air at 298 K.

1 11

3 1

8 8 8.314 298475

28.02 10 mean

RT JK mol Kv ms

M kg mol


21

1

2

1

2

82 2

8 where

rel mean

A B

A B

RTv v

M

kT m m

m m

Mean relative speed

1 11

3 1

8 8 8.314 2982 2 728

28.02 10

RT JK mol Kc ms

M kg mol

1B.1(c) Mean values

Bi 1B.2 vrel of N2 molecules in air at 298 K.


22

1

relrel

rel

v pz v N

kT

v kT

z N p

freeze the position of all the molecules except one

number density,

/

A

rel rel

pN

V kT

pV nRT nN kT kT

z v t t N vV

1B.2 (a) The collision frequency, z(b) The mean free path,

For 1 atm of N2 molecules at 298 K.

18 2 1 2

23 1

9 1

1

9 1

0.43 10 728 101325

1.381 10 298

7.7 10

728 95 (~ 1000 times of )

7.7 10

rel

rel

v pz

kT

m ms Nm

JK K

s

v msnm d

z s

2

collision crosssection,

d

1장 수업목표 3: 실제기체

• Compression factor:(압축인자)

• Van der Waals equation:

• Virial equation: (라틴어의 “힘”)

m m

o

m

V pVZ

V RT

2

mm V

a

bV

RTp

2

2

( ) ( )1

1 '( ) '( )

m

m m

pV B T C TZ

RT V V

B T p C T p

23(2015) Spring Physical Chemistry (I)

by M Lim

1C.1 Deviations from perfect behavior


24

Real gases interact with one another (high p low T)

1C.1 Deviations from perfect behavior


25

Consequences of molecular interactions

Real gases interact with one another (high p low T)

CO2

1C.1(a) The compression factor, Z

o

m m

V V RTV V

n n p


26

m m

o

m

V pVZ

V RT

ZRTpVm

Even a real gas Z ≈ 1 at very low p

• Z > 1 repulsive F dominates• Z < 1 attractive F dominates

Taylor expansion


27

Exponential function:

http://en.wikipedia.org/wiki/Taylor_expansion

( )f x

Geometric series:2 3

0

11 for 1

1

n

n

x x x x xx

2 3

ln 12 3

x xx x Natural log:

http://en.wikipedia.org/wiki/Exponential_function

http://en.wikipedia.org/wiki/Taylor_expansion

http://en.wikipedia.org/wiki/Geometric_series


28

Taylor expansion

http://upload.wikimedia.org/wikipedia/commons/2/21/Sintay.svg

1C.1(b) Virial coefficients


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1 for ideal gasmpVZ

RT

21 ' 'p pZ Z pa b

2

11 1

m

m m

Z Z VV V

a b

2

2

1 ' '

1

m

m m

pVB p C p

RT

B C

V V

•Virial equation of state

, ,' ' '( )

( ) , ,

'(

)

)

(

B C B T C T

B C B T C T

Virial coefficients depend on Temperature.



30

2

2'( )

( ) () 1'(

)1m

m m

B T C TBpV

p pT

R

C T

T V V

2

21

( ) ( )' ) '( )1 (m

m m

B T C TB T C T

pVZ p p

RT V V


31

'( ) 2 '( )

( )( ) 2

1 m m

dZB T C T p

dp

dZ C TB T

d V V

At Boyle temperature TB B(T) = 0

('(

))

BB

T

TT

R


At TB the gas behaves

perfectly over a wider range

of conditions than at other

temperatures.

For large Vm and high T, the real-gas isotherms do not differ greatly from perfect-gas isotherms

1C.1(c) Critical constants


32

Critical constants(임계상수 ) (Tc, pc, Vc)liquids phase of substances does not form above Tc

Supercritical fluids (초임계유체): the single phase at T > Tc and much denser than typical of gases

Vapor pressure

(A, B, C) (C, D, E) (E-F)

CO2


33

1C.1(c) Critical constants

• First assume “hard sphere” molecules (repulsive force)

becomes

1C.2(a) The van der Waals equation


34

Only two parameters, derived from molecular concepts

RTpVm RTbVp m )(

bV

RTp

m 2

m m

RT ap

V b V

• Now put in attraction

So becomes

1.C.2(a) The van der Waals equation


35

Ex 1C.1 van der Waals Vm of CO2 at 500 K and 100 atm.

2 2

3 2 0

m m m m

m m m

V b V p RTV V b a

RT a abV b V V

p p p

6 2

2 3 1

3.592

4.267 10

a dm atm mol

b dm mol

3 1

22 3 1

33 3 1

0.453

3.61 10

1.55 10

RTb dm mol

p

adm mol

p

abdm mol

p

3 2 2 3

3 1

3 1

0.453 3.61 10 1.55 10 0

0.366

0.410

m m m

m

o

m

V V V

V dm mol

V dm mol

공식보다는 계산기나 컴퓨터 사용하여 푼다.

1C.2(b) The features of the equation


36

Van der Waals loop

06

)(

2

02

)(

432

2

32

mmm

mmm

V

a

bV

RT

dV

pd

V

a

bV

RT

dV

dp

,3

mV

,

3( ) 2

3

m m

m c

V b V

V b

2 2

3 3

2 2

2 2 8( ) 4

27 27

8

27

2 9 27

m

m

c

c

a a aRT V b b

V b b

aT

Rb

RT a ap

b b b


37

… ①

… ②

① 06

)(

342

mmm V

a

bVV

RT… ③

② + ③, results in0)32

()( 2

mmm VbVbV

RT

from ①,

8

3

8

273

27 2

,

a

bb

b

a

RT

VpZ

c

cmc

c

Critical compression factor

1C.2(b) The features of the equation


38

m m mr r r

c c c

V p TV p T

V p T

1C.2(c) The principle of corresponding states

Reduced variables (dimensionless)


39

Documents

Homework - KOCWcontents.kocw.net/KOCW/document/2015/pusan/limmanho/1.pdfChap 1. The properties of gases (2015) Spring Physical Chemistry (I) by M Lim 2 • Gas: the simplest state