powerpoint to accompany chapter 9a gases. brown, lemay, bursten, murphy, langford, sagatys:...

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PowerPoint to accompany Chapter 9a Gases

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Page 1: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

PowerPoint to accompany

Chapter 9a

Gases

Page 2: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Characteristics of Gases Like liquids and solids, they:

Still have mass and momentum Have heat capacities & thermal conductivities Can be chemically reactive

Unlike liquids and solids, they: Expand to fill their containers Are highly compressible Have extremely low densities

Page 3: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Page 4: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Pressure

Pressure is the amount of force applied to an area.

Atmospheric pressure is the weight of air per unit of area.

P =FA

Page 5: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Units of Pressure

Pascals 1 Pa = 1 N/m2 = 1 Kg/ms2

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

Page 6: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Standard Pressure

Normal atmospheric pressure at sea level.

This is ambient pressure, equal all around us, so not particularly felt

It is equal to: 1.00 atm 760 torr (760 mm Hg) 101.325 kPa 14.696 psi

Page 7: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Manometer

Figure 9.2

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

Page 8: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Manometer calculation:

Patm= 102 kPa

Pgas> Patm

Pgas= Patm + h

Pgas= 102 kPa + (134.6mm – 103.8mm)Hg

Pgas= 102 kPa + (32.6mm Hg) Note 1mmHg = 133.322 Pa Pgas= 102 kPa + (32.6x133.322x10-3 kPa)

Pgas= 106 kPa = 1.06 bar =1.06/1.013 atm

Pgas= 1.046 atm

Page 9: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Page 10: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Boyle’s Law

The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure: V ~ 1/P

Figure 9.4

Page 11: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

As P and V are inversely proportional

A plot of V versus P results in a curve.

Since

V = k (1/P)

This means a plot of V versus 1/P will be a straight line.

PV = k

Figure 9.5

Page 12: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Charles’s Law: initialwarm balloon, T big, V big

Page 13: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Charles’s Law: finalcold balloon T & V little

Page 14: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Charles’s Law

The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

VT = k

A plot of V versus T will be a straight line.

Figure 9.7V = kT

Page 15: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Avogadro’s Law The volume of a gas at constant temperature

and pressure is directly proportional to the number of moles of the gas.

Mathematically, this means: V = kn

Figure 9.8

Page 16: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Page 17: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

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:

nTP V = R

i.e. PV = nRT (where R = 8.314 kPa L/mol K)

Page 18: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

V = k (1/P)

nTP V = R V

P

Page 19: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Ideal Gas Law Calculations

Heat Gas in Cylinder from 298 K to 360 K at fixed piston position. (T incr. so P increases)

Same volume, same moles but greater P P1 = (nR/V1)T1 or P1/T1 = P2 / T2

P2 = P1T2 / T1

P2 = P1(T2 /T1 ) = P1(360/298) = 1.2081P1

Page 20: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Ideal Gas Law Calculations

Move the piston to reduce the Volume from 1 dm3 to 0.5 dm3. (V decreases so P increases)

Same moles, same T, but greater P P1V1 = (nRT1) = P2V2

P2 = P1V1 / V2 = P1 (V1 / V2)

P2 = P1(1.0/0.5) = P1(2.0)

Page 21: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Ideal Gas Law Calculations

Inject more gas at fixed piston position & T. the n increases so P increases)

Same volume, more moles but greater P P1 = n1(RT1/V1) or P1/n1 = P2 / n2

P2 = P1 n2 / n1 for n2 = 2 n1

P2 = P1(2)

Page 22: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Ideal Gas Law Calculations

R = (PV / nT) When kPa and dm3 are used as is common R = 8.314 kPa dm3 / mol K or J/mol K It is useful to write the ideal gas law this way

so that R is the constant and other variables can change from initial to final conditions

This again reduces the gas law to a simple arithmetic ratio calculation

The only tricks are to ensure all units are consistent or converted as needed.

Page 23: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

V = (nRT/P) = 1mol (8.314kPa dm3 /molK)(273.15K)/100 kPaV/mol = 22.71 dm3 = 22.71L

Page 24: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Gas Densities and Molar Mass

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

nV

PRT=

Most of our gas law calculations involve proportionalities of these basic commodities.

Page 25: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Page 26: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Gas Densities and Molar Mass

We know that moles molar mass = mass

So multiplying both sides by the molar mass () gives:

n = m

PRT

mV = i.e. =

PRT

Page 27: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Molecular Mass

We can manipulate the density equation to enable us to find the molar mass of a gas:

Becomes

PRT =

RTP =

Page 28: PowerPoint to accompany Chapter 9a Gases. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Characteristics of Gases

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

End of Part 9a Ideal Gas Law