2 Pressure What is pressure? Pressure is the amount of force - - PowerPoint PPT Presentation

Gases: Chapter 9

• Characteristics: Unlike liquids and solids, they:
• have large spaces between the gas molecules

→ not tightly packed

• expand to fill the container → take the

shape of the container

• diffuse into one another and mix
• 4 properties that affect its physical

behaviour:

• amount of gas (in moles)
• volume
• temperature
• pressure

1

2

Pressure

What is pressure? Pressure is the amount of force applied to an area.

• it is the force per unit area – (force divided

by the area over which the force is applied)

• The unit is N.m-2 which is one pascal (Pa)

) ( ) ( ) (

2

m A N F Pa P 

3

Units of Pressure

• Pascals

1 Pa = 1 N/m2

• Bar

1 bar = 105 Pa = 100 kPa

• mm Hg or torr

The difference in the heights in mm (h) of two connected columns of mercury

• Atmosphere

1.00 atm = 760 torr

4

Barometric Pressure

• The simplest device

used to measure pressure is the Toricelli barometer

• Standard atmosphere

(atm) – is the pressure exerted by a mercury column of 760 mm (density of Hg = 13.5951 g.cm-3 (0

0C and gravity =

9.80665 m.s-2)

5

1 atm = 760 mmHg 1 atm = 760 Torr Example: What is the height of a column of water that exerts the same pressure as a column of mercury 76.0 cm (760 mm) high? PH2O = PHg Using equation (g x h x d) g x hH2O x 1.00 g/cm3 = g x 76.0 cm x 13.6 g/cm3

m cm cm g cm g cm h

O H

3 . 10 10 03 . 1 / 00 . 1 / 6 . 13 . 76

3 3 3

2

    

6

Common Units of Pressure

Atmosphere atm Millimeter of mercury mmHg 1 atm = 760 mmHg Torr Torr = 760 Torr Pascal Pa = 101 325 Pa Kilopascal kPa = 101.325 kPa Bar bar = 1.01325 bar Millibar mb = 1013.25 mb

7

Manometer

Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Figure 9.2

Standard Pressure

• Normal atmospheric pressure at sea level.
• It is equal to:

–1.00 atm –760 torr (760 mm Hg) –101.325 kPa

Manometers

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Example: When a manometer is filled with liquid mercury (d = 13.6 g/cm3), the barometric pressure is 748.2 mmHg, and the difference in mercury levels is 8.6 mmHg. What is the gas pressure Pgas? Pgas = Pbar + ∆P = 748.2 mmHg - 8.6 mmHg = 739.6 mmHg

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Gas Laws

Boyle’s Law: The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

V P 1 

12

Charles’s Law: - The volume of a fixed amount

• f gas at constant pressure is directly

proportional to its temperature. A plot of V versus T will be a straight line.

T V 

13

Standard Temperature and Pressure

STP conditions:

• standard temperature for gases = 0 0C =

273.15 K

• standard pressure, 1 atm = 760 mmHg

Avogadro’s Law: - at a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas

n V 

) ( 4 . 22 1 STP at gas L gas mol 

14

Example: What is the mass of 1.00 L of cyclopropane gas, C3H6, when measured at STP? 1L =1dm3

mol L mol L x L x STP at gas L mol 0446 . 4 . 22 1 00 . 1 00 . 1 ) ( 4 . 22 1     

g mol g mol m m n H C

• f

mass 88 . 1 . 08 . 42 0446 . .

1 6 3

    



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Ideal-Gas Equation

• So far we’ve seen that:

V  1/P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law)

• Combining these, we get, if we call

the proportionality constant, R:

nT P V = R

i.e. PV = nRT

16

The Ideal Gas Equation

R = 8.3145 m3.Pa.mol-1.K-1 (SI unit-use this) = 8.3145 J.mol-1.K-1(SI unit)(1m3 Pa = 1J) = 0.082057 L.atm.mol-1.K-1 = 8.314 kPa. dm3 mol-1.K-1

nRT PV 

Where R = gas constant

17

J s m kg Pa m s m kg Pa m s kgm Pa m s m kg Pa s m kg N m A N F P 1 2 2 3 2 1 2 1 2 2 2 2 1 ) 2 ( ) (           

Units (1m3 Pa =1J)

R = 8.3145 m3 Pa mol-1 K-1

R= 8.3145 J mol-1 K-1

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19

Volume conversion (from milliliters to litres and then to cubic meters)

3 4 3

10 05 . 3 1000 1 1000 1 305 m L m mL L mL V



    

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Pressure conversion

• 1 atm → 101325 Pa
• 1.37 atm → x Pa
• X = (1.37 atm x 101325 Pa) / 1 atm

= 138815 Pa = 1.38 x 105 Pa

21

Example: What is the pressure, in kPa, exerted by 1.00 x 1020 molecules of N2 in a 305 mL flask at 175 0C?

2 23 20 20 23

000166 . 10 022 . 6 1 10 00 . 1 10 00 . 1 10 02 . 6 1 N mol molecules mol molecules x molecules x molecules mol         

kPa Pa m K K mol Pa m mol V nRT P 03 . 2 10 03 . 2 10 05 . 3 448 . . . 3145 . 8 000166 .

3 3 4 1 1 3

       

  

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General Gas Equation

For a fixed mass of gas,

f f f i i i

T V P T V P 

f f f f i i i i

T n V P T n V P 

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Example: If a fixed amount of gas held at constant volume goes from STP conditions to boiling at 100 0C, what is the new pressure?

f f f f i i i i

T n V P T n V P 

atm K K atm T T P P

i f i f

37 . 1 273 373 00 . 1     

f f i i

T P T P 

24

Applications of the Ideal Gas Equation – Molar mass

Molar mass determination PV = nRT

M m n 

M mRT PV 

25

Example: Propylene is used in the production of

• plastics. A glass vessel weighs 40.1305 g when

clean, dry and evacuated; it weighs 138.2410 g when filled with water at 25 0C(density of water = 0.9970 g/mL) and 40.2959 g when filled with propylene gas at 740.3 mmHg and 24.0 0C. What is the molar mass of propylene? Mass of water to fill vessel = 138.2410 g – 40.1305 g = 98.1105 g Density = mass/volume Volume = mass/density Volume of water (vol. of vessel) = 98.1105 g/0.9970 g/ml = 98.41 mL = 0.09841 L

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Mass of gas = 40.2959 g – 40.1305 g = 0.1654 g Temperature (K) = (24 0C + 273.15 0C) x 1K/10C = 297.15 K R = 8.3145 m3 Pa mol-1 K-1 Pressure = 740.3 mmHg/760 mmHg = 0.9741 atm (convert to Pa) = 98700 Pa

mol g m Pa K K mol Pa m g PV mRT M / 08 . 42 10 841 . 9 98700 15 . 297 3145 . 8 1654 .

3 5 1 1 3

      

  

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Gas Densities and Molar Mass

If we divide both sides of the ideal-gas equation by V and by RT, we get: PV = nRT

n V P RT =

28

We can then replace n/V with P/RT The density of a gas at STP is calculated by dividing its molar mass by the molar volume (22.4 L/mol)

M V n V M n V m d     

RT MP V m d  

Applications of the Ideal Gas Equation - Gas Densities

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Example: What is the density of O2 at 298 K and 0.947 atm? (Convert 0.947 atm to Pa) R = 8.3145 m3 Pa mol-1 K-1

L g L m m g K K mol Pa m Pa mol g RT MP V m d / 24 . 1 1000 1 5 . 1239 298 3145 . 8 95954 00 . 32

3 3 1 1 3 1

       

  

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Gases in Chemical Reactions

Air bags use the decomposition of sodium azide, NaN3 to produce N2 (g) that inflates the air bag Example: From the following equation determine

what volume of N2 is produced when 70.0 g of NaN2 is decomposed at 735 mmHg and 26 0C?

2NaN3 (s) →2Na(l) + 3N2 (g)

mol mol N n mol mol g g NaN n 62 . 1 2 3 08 . 1 ) ( 08 . 1 / 01 . 65 . 70 ) (

2 3

    

31

P = 735 mmHg/760 mmHg = 0.967 atm (convert to Pa) n = 1.62 mol R = 8.3145 m3 Pa mol-1 K-1 T = (26 0C + 273 0C) x 1K/10C = 299 K

L m L m Pa K K mol Pa m mol P nRT V 1 . 41 1 1000 0411 . 97981 299 3145 . 8 62 . 1

3 3 1 1 3

      

 

32

Law of Combining Volumes

Example: For the following reaction, determine what volume of SO2 is produced per liter of O2 consumed? Both gases are measured at 25 0C and 745 mmHg.

) ( 2 ) ( 2 3 ) ( 2

2 2

g SO s ZnO O s ZnS  

 

) ( 667 . 3 1 2 ) ( 2 ) ( 3 1

2 2 2

g SO L mol L mol x g SO mol x g O mol L     

33

Mixtures of Gases

34

Dalton’s law of partial pressures:

• The total pressure of a mixture of gases is

the sum of the partial pressures of the components of the mixture Ptot = PA + PB + ……

tot A A

P RT n V 

tot B B

P RT n V 

B A tot

V V V  

35

Where XA is the mole fraction (fraction

• f all the molecules in a mixture

contributed by that component)

tot A tot tot tot A tot A tot A tot tot tot A tot A

n n P RT n P RT n V V and n n V RT n V RT n P P     ) / ( ) / ( ) / ( ) / (

A tot A tot A tot A

x V V P P n n   

36

Example: What are the partial pressures of H2 and He in a gaseous mixture containing 1.0 g of H2 and 5.00 g He that are in a confined volume of 5.0 L at 20 0C? (convert volume)

mol mol mol He mol g g H mol g g ntot 75 . 1 25 . 1 50 . / 003 . 4 00 . 5 / 02 . 2 00 . 1

2

                    

Pa m K K mol Pa m mol V RT n P

tot 5 3 1 1 3

10 5 . 8 0050 . 293 3145 . 8 75 . 1      

 

37

Partial pressure of H2 and He

atm atm n n P P atm atm n n P P P P n n

tot He tot He tot H tot H tot H tot H

. 6 75 . 1 4 . 8 25 . 1 4 . 2 75 . 1 4 . 8 50 .

2 2 2 2

          

38

Partial Pressures and Mole Fraction

• So, when one collects a gas over water, there is

water vapour mixed in with the gas.

• To find only the pressure of the desired gas, one

must subtract the vapour pressure of water from the total pressure. Figure 9.13

Collecting a gas over water Gas collected over water is “wet” – made up of the desired gas and water vapour Ptot = Patm = Pgas + PH2O

• r

Pgas = Patm – PH2O

40

Kinetic-Molecular Theory

1.Gases consist of large numbers of molecules that are in continuous, random motion. 2.The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained. 3.Attractive and repulsive forces between gas molecules are negligible.

Kinetic-Molecular Theory

4.Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. 5.The average kinetic energy of the molecules is proportional to the absolute temperature.

1. Gas molecules contain translational kinetic energy (energy of objects moving through space). The faster the molecules move, the greater is their translational energies →greater forces exerted when they collide. 2. Frequency of molecular collisions – no. of collisions per second. The higher the frequency, the greater the total force of the collisions (increases with no. of molecules per unit volume and with molecular speed). 3. When a molecule hits the walls of a container, momentum is transferred as the molecule reverses direction→impulse. The magnitude of the impulse is directly proportional to the mass and velocity of a molecule.

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The Kelvin temperature (T) of a gas is directly proportional to the average translational kinetic energy of its molecules. Absolute zero of temperature – the temperature at which the translational molecular motion should cease.

44

Nonideal (Real) Gases

Compressibility factor: For an ideal gas, PV/nRT = 1 For nonideal gas behaviour, at high pressures the volume

• f a gas becomes very small

→ 0 but this is not so as gas molecules occupy a volume themselves.

nRT PV

45

Real Gases

In the real world, the behaviour of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

Figure 9.21

Deviations from Ideal Behaviour

The assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature. Figure 9.22 Figure 9.23

Corrections for Non-ideal Behaviour

• The ideal-gas equation can be adjusted to

take these deviations from ideal behaviour into account.

• The corrected ideal-gas equation is known

as the van der Waals equation.

The van der Waals Equation

) (V − nb) = nRT n2a V2 (P +

Table 9.2

• Since molecules are

attracted to each other, the force of collisions of gas molecules with the walls of the container are less than expected for an ideal gas

• Gases behave ideally at high temperatures and low

pressures.

• Gases tend to behave nonideally at low

temperatures and high pressures.

 

nRT nb V V a n P           

2 2

50

Example: Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl2(g) confined to a volume of 2.00 L at 273 K. The volume of a = 0.657 (m3)2Pa .mol-2, and that of b = 0.0000562 m3 mol-1.

R = 8.3145 m3 Pa mol-1 K-1

 

Pa Pa m mol Pa m mol mol m mol m K K mol Pa m mol V a n nb V nRT P

6 2 3 2 2 2 3 2 1 3 3 1 1 3 2 2

10 00 . 1 1003343 ) ( ) 002 . ( ) ( 657 . 00 . 1 0000562 . x 1 002 . 273 . 3145 . 8 00 . 1            

   

51

To simplify calculation :

• R = 8.3145 m3 Pa mol-1 K-1
• nRT =
• n2a =
• V-nb =

K K mol Pa m mol 273 . 3145 . 8 00 . 1

1 1 3

 

 

 

1 3 3

0000562 . x 1 002 .



 mol m mol m

2 2 3 2

) ( 657 . 00 . 1



 mol Pa m mol

52

• EX. A 1.65 g sample of an unknown metal M reacts with excess

HCl and the liberated hydrogen is collected over water at 25 oC at a barometric pressure of 744 mmHg. The gas collected has a volume

• f 2367 mL. The equation for the reaction is:
• 2M(s) + 6HCl (aq) 2MCl3(aq) + 3H2(g)
• The gas constant: R = 8.3145 m3 Pa mol-1K-1
• Vapour pressure of H2O at 25 oC is 23.76 mmHg
• 1 atm = 760 mmHg
• 1 atm = 101325 Pa
• Calculate the following:
• (i)

partial pressure of dry hydrogen gas in SI units

• (ii)

moles of dry hydrogen gas in SI units

• (iii)

moles of M

• (iv)

molar mass of M

• (v)

density of hydrogen gas, in g L-1 at STP.

• 53

(i) partial pressure of dry hydrogen gas using SI units PH2 = (744 – 23.76) mmHg = 720 mmHg/760 mmHg atm-1 = 0.947 atm (convert to Pa) PH2 = 0.947 atm x 101325 Pa atm-1 = 9.60 x 104 Pa (ii) moles of dry hydrogen gas using SI units n = PV/RT

= (9.60 x 104 Pa x 2.367 x 10-3 m3 )/8.3145 m3 Pa mol-1K-1 x 298K )

= 0.0917 mol

54

(iii) moles of M From the balanced equation: moles of M = 2/3 mol of H2 = 2/3 x 0.0917 = 0.0611 mol of M (iv) molar mass of M molar mass of M = mass/mol = 1.65 g/0.0611 mol = 27.0 g mol-1 (M = Al) (v) density of hydrogen gas, in g L-1 at STP. Density = PM/RT = (101325 Pa x 2.02 g mol-1)/8.3145 m3 Pa mol-1 K-1 x 273 K. = 90.2 g m-3 ( convert volume from cubic metres to litres ) = 0.0902 g L-1

55