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# Pressure of a gas

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### Kinetic Model of Gases

Section 1.9, 1.11

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### Assumptions

z A gas consists of molecules in ceaseless random motion

z The size of the molecules is

negligible in the sense that their diameters are much smaller than the average distance traveled

between collisions

z The molecules do not interact, except during collisions

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### =

2

z M molecular weight; V volume

z c root-mean-square speed (rms speed)

2 / 2 1 2

3 2

2 2

1

N

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z r.m.s. speed

z mean speed

2 / 2 1 2

3 2

2 2

1

N

1 2 3

N

c c

c 0.921

3

8 1/2

 

= 

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2

3

1 nMc pV

nRT pV

=

=

2

## McRT

•Effect of Molecular Weight

•Temperature Effect

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### Kinetic Energy of Molecules

z Ek = 3/2 RT

Ek kinetic energy; T temperature; R gas constant

z The average kinetic energy per molecule

B

k

### ε

kB Boltzmann constant = R/6.02×1023

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### Partial Pressure

zDalton’s Law

The total pressure observed for a mixture of gases is equal to the sum of the pressures that each individual component gas would exert

Ptotal = P1+P2+P3+…+PJ PJ = xJ Ptotal

Ptotal total pressure; PJ partial pressure of component J;

χJ molar fraction of component J.

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### Diffusion & Effusion

z Diffusion

Molecule of different substances mingle with each other.

z Effusion

Escape of a gas through a small hole.

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### Diffusion & Effusion

z Rates of diffusion and effusion of gases increase with increasing temperature.

z For effusion the rate decreases with increasing molar mass.

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### Diffusion & Effusion

z Graham’s Law

At a given pressure and temperature, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

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### Diffusion & Effusion

z The rate at which hydrogen and carbon dioxide effuse under the same conditions of pressure and temperature are in the ratio

2

2

1/ 2 1/ 2

2 2

Rate of effusion of H 44.01

4.672

Rate of effusion of CO 2.016

CO H

M M

= = =

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### Diffusion & Effusion

z Separation of uranium-235 from uranium-238, in the form of volatile solids UF6

http://www.columbia.edu/itc/chemistry/chem-c1403/text_chapters/nukes.html http://www.uic.com.au/uicchem.htm

http://www.uilondon.org/index.htm

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### Effusion as a separation technique

Use porous membranes to separate light gases from heavy ones

z average speed of gas molecules depends on the masses of their molecules

z heavy molecules in a mixture move slower on average than light ones

z gases made of light molecules diffuse through pores in membranes faster than heavy molecules

Differences from dialysis

z membrane is permeable, not semipermeable: all gas molecules in the mixture can pass through it

z size of molecules isn't usually important: pores in membrane are much larger than gas molecules

z ...molecular velocity (and so, molecular mass) is the basis for separation, not size

Examples

z separating helium from oxygen

z separating uranium isotopes as volatile UF6

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### Molecular Collisions

z C = 平均自由路徑 / 飛行時間 = λ / [1/ z] = λ z

λ = RΤ / [21/2NAσ p]

Z = [21/2NAσ c p] / RT

λ: 平均自由路徑;

Z : 碰狀頻率 collision frequency

σ : 碰狀截面積; σ = πd2 p : 壓力; T: 溫度;

NA: 亞佛加厥常數

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### Molecule Collisions

z λ ∝ 1/p 平均路徑隨壓力減少而增加

z λ ∝ 1/σ 分子之碰狀截面積越大, 平均自由路徑隨之減短

z z ∝ p 碰撞頻率隨壓力增加而增大

z z ∝ c ∝ 1/ Μ1/2 分子量越大的分子其碰撞頻率會低於分 子量小的分子

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### Maxwell distribution of speeds

The Maxwell distribution of speeds f = F (s) ∆S

F (s) = 4π[m / 2πkBT](3/2) s2 e-(ms2/2kBT)

## f : 運動速率在某個範圍內的分子之比例s: 分子運動速率 speed;

∆s : 速率的範圍 interval of speed KB : Boltzmann Constant

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### Maxwell distribution of speeds

f = F (s) ∆S

F (s) = 4π[m/ 2πkBT](3/2) s2 e-(ms2/2kBT) f ∝ ∆S

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### Maxwell distribution of speeds

f = F (s) ∆S

F (s) = 4π[m / 2πkBT](3/2) s2 e-(ms2/2kBT)

z s2

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### Maxwell distribution of speeds

f = F (s) ∆S

F (s) = 4π[m / 2πkBT](3/2) s2 e-(ms2/2kBT)

e -x (x =ms2/2kBT)

z 這是一個"衰減"函數.當速率 (s) 非常大的時候指數值就 相當小.也就是說具有極高運動速率的氣體分子比例是非 常小的.

z 分子量(M)越大, 指數值就越小. 大分子具有高運動速率 的比例較小.

z 溫度(T)升高, 指數值越大. 溫度越高具有較快運動速率的 分子比例也越大.

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### Maxwell distribution of speeds

f = F (s) ∆S

F (s) = 4π[m / 2πkBT](3/2) s2 e-(ms2/2kBT) 4π[m / 2πkBT](3/2)

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### Distribution of Translational Energy

z Maxwell-Boltzmann Distribution Law

z ε kinetic energy (=1/2 mu2) f = F (ε) ∆ε

F (ε) = 2π / (πkBT)(3/2) ε1/2 e-(ε/kBT)

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