Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Gases Chapter 5 Become familiar...

Copyright©2000 by Houghton Mifflin Company. All rights reserved.

1

GasesChapter 5

Become familiar with the definition and measurement of gas pressure.

Learn the gas law and ideal gas equation.Understand the concept of the partial pressures.Study the kinetic molecular theory and gas

Effusion/DiffusionMain differences between ideal and real gases.


2

Elements that exist as gases at 250C and 1 atmosphere


3

• Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids.

• Exerts pressure on its surroundings.

Physical Characteristics of Gases


4

Figure 5.1a: The pressure exerted by the gases in the atmosphere can be demonstrated by boiling water in a large metal can (a) and then turning off the heat and sealing the can.


5

Figure 5.01b: As the can cools, the water vapor condenses, lowering the gas pressure inside the can. This causes the can to crumple.


6

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Pressure = ForceArea


7

Sea level 1 atm

4 miles 0.5 atm

10 miles 0.2 atm


8


9

As P (h) increases V decreases


10

Boyle’s Law


11

P 1/VP x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas


12

Boyle’s Law*

Pressure Volume = Constant (T = constant)

P1V1 = P2V2 (T = constant)

V 1/P (T = constant)

(*Holds precisely only at very low pressures.)


13

As pressure increases, the volume of SO2 decrease


14

A gas that strictly obeys Boyle’s Law is called an ideal gas.


15

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg


16

As T increases V increases

Volume-Temperature Relationship


17

Charles's Law

Figure 5.9: Plots of V versus T as in Fig. 5.8 except here the Kelvin scale is used for temperature.


18

Charles’s Law

The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.

V = bT (P = constant)

b = a proportionality constant


19

Charles’s Law

VT

VT

P1

1

2

2 ( constant)


20

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L

= = 192 K

V1/T1 = V2/T2


21

Avogadro’s Law

V number of moles (n)

V = constant x n

V1/n1 = V2/n2

Constant temperatureConstant pressure


22

Figure 5.18: The effects of increasing the number of moles of gas

particles at constant temperature and pressure.


23

Ideal Gas Equation

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: V (at constant n and T)1P

V nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT


24

Ideal Gas Law

PV = nRT R = proportionality constant

= 0.08206 L atm mol

P = pressure in atm

V = volume in liters

n = moles

T = temperature in Kelvins

Holds closely at P < 1 atm


25

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.42L)

(1 mol)(273.15 K)

R = 0.082067 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.42 L.


26


27

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.6 L


28

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constant

nRV

= PT

= constant

P1

T1

P2

T2

=

P1 = 1.20 atm

T1 = 291 K

P2 = ?

T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm


29

Ex 9

• P1 = 345 torr P2 = 468 torr

• T1 = -15 oC T2 = 36 oC

• V1 = 3.48 L V2 = ?


30

Ex 10

• n1 = 0.35 n2 = 0.35

• P1 = 568 torr P2 = 1.18 atm

• T1 = 13 oC = 286 K T2 = 56 oC =329


31

Gas Stoichiometry

What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L

5.5


32

S.Ex 5.13• CH4 + O2 → CO2 + H2O• 2.80 L 35.0 L V?• 25oC 31oC 125oC• 1.65 atm 1.25 atm 2.50 atm• 0.189 mol 1.75 mol ?• Balance eq.

• CH4 + 2 O2 → CO2 + 2 H2O

• CH4 is limiting !• Mol of CO2 = 0.189 > V = 0.189 x 0.0821 x398

2.50 atm= 2.47 L


33

Density (d) Calculations

d = mV =

PMRT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTP

M = d is the density of the gas in g/L


34

S. Ex 5.14

• P = 1.50 atm; 27oC ; d = 1.95 g/L; MM=?

• MM = dRT = 1.95 g/Lx0.0821L.atm/(K.mol)x300K

P 1.50 atm

= 32.0 g/mole


35

Dalton’s Law of Partial Pressures

For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone

PTotal = P1 + P2 + P3 + . . .


36

Dalton’s Law of Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2


37

S.Ex 5.15

• At 25 oC

He46L, 1 atm

+O2

12L, 1 atm

He+O2

5.0L, P = ?

↓

P1V1= P2V2

PHe = 46Lx1atm/5.0L= 9.2 atm

PO2 = 12Lx1 atm/5.0L=2.4 atm

PT = 9.2atm + 2.4 atm=11.7 atm


38

Mole Fraction – Χ

The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture.

Χi = ni = ____ ni ______ ntotal n1 + n2 +n3 + …


39

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT


40

S.E. 5.16

• The partial pressure of oxygen is 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of oxygen in air.

• ΧO2 = PO2 = 156 torr = 0.210

Ptotal 743 torr


41

S.Ex 5.17

• The mole fraction of nitrogen in air is 0.7808. Calculate the partial pressure of nitrogen in air when the atm pressure is 760 torr.

• PN2 = ΧN2 x Ptotal = 0.7808x760 torr

= 593 torr


42

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm


43

2KClO3 (s) 2KCl (s) + 3O2 (g)

Bottle full of oxygen gas and water vapor

PT = PO + PH O2 2


44

S.Ex 5.182KClO3 (s) → 2KCl (s) + 3O2 (g)

O2 collected at 22oC, at Ptotal = 754 torr; V=0.650 L ; v.p. of water at 22oC = 21 torr

Ptotal = PO2 + PH2O = PO2 + 21 torr

Thus, PO2 = 754 torr – 21 torr = 733 torr

Ideal gas eq. to find moles of O2

Moles O2 → moles KClO3 → g KClO3


45

Kinetic Molecular Theory of Gases

1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points that is, they possess mass but have negligible volume (V zero).

2. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. The collision of particles with the wall of the container are the cause of the pressure exerted by the gas

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy


46

Kinetic theory of gases and …

• Compressibility of Gases

• Boyle’s Law

P collision rate with wall

Collision rate number densityNumber density 1/VP 1/V

• Charles’ LawP collision rate with wall

Collision rate average kinetic energy of gas molecules

Average kinetic energy T

P T


47

Kinetic theory of gases and …

• Avogadro’s Law

P collision rate with wall

Collision rate number densityNumber density nP n

• Dalton’s Law of Partial Pressures

Molecules do not attract or repel one another

P exerted by one type of molecule is unaffected by the presence of another gas

Ptotal = Pi


48

Figure 5.19: Path of one particle in a gas. Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of the container: The concept of Average Speed


49

The Meaning of Temperature

Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

(KE)32avg RT


50

Root Mean Square Velocity

2uurms

MMRT

rms

Aeave

u

RTmuNKE

3

232

21

arg )()(


51

The distribution of speedsfor nitrogen gas molecules

at three different temperatures

The distribution of speedsof three different gases

at the same temperature

urms = 3RTM


52

S.Ex. 5.19• Calculate the root mean square velocity for the

atoms in a sample of He gas at 25oC.

• MM = 4.00 g/mol x1kg/1000g • =4.00x10-3kg/mol

molkgx

KmolKJx

MM

RTurms /1000.4

298)(./(314.8333

smx

smkgJkg

Jx

/1036.1

/1086.1

3

226


53

Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.


54

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

NH3

17 g/molHCl

36 g/mol

NH4Cl


55

Effusion: describes the passage of gas into an evacuated chamber.


56

Rate of effusion for gas 1Rate of effusion for gas 2

2

1

MM

Distance traveled by gas 1Distance traveled by gas 2

2

1

MM

Effusion:Effusion:

Diffusion:Diffusion:


57

S.Ex. 5.20• Calculate the ratio of the effusion rates of H2 gas and UF6

2.13016.202.352

2

6

6

2 H

UF

MM

MM

UFforeffusionofRateHforeffusionRate


58

2

1

21

1

MMMM

gasofeffusionoftimegasofeffusionoftime

timerate


59

Real Gases

Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).


60

Figure 5.25: Plots of PV/nRT versus P for several gases (200 K).


61

Figure 5.26: Plots of PV/nRT versus P for

nitrogen gas at three temperatures.


62

Real gases tend to behave like ideal gases, when

• P is low (large volume of container relative to volume of molecule )

• T is high ( at high T attractive and repulsive forces are negligible, relative to the high speed of molecules)


63

Figure 5.27: (a) Gas at low concentration— relatively few interactions between particles. (b) Gas at high concentration—many more interactions between particles.


64

Real Gases

[ ]P a V nb nRTobs2( / ) n V

corrected pressurecorrected pressure corrected volumecorrected volume

PPidealideal VVidealideal


65

Date post:	20-Dec-2015
Category:	Documents
View:	214 times
Download:	2 times

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Gases Chapter 5 Become familiar...

Documents