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can be compressed exert pressure on whatever surrounds them expand into whatever volume is available easily diffuse into one another can be described

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Page 1: can be compressed exert pressure on whatever surrounds them expand into whatever volume is available easily diffuse into one another can be described
Page 2: can be compressed exert pressure on whatever surrounds them expand into whatever volume is available easily diffuse into one another can be described
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• can be compressed

• exert pressure on whatever surrounds them

• expand into whatever volume is available

• easily diffuse into one another

• can be described in terms of temperature, pressure, volume and amount

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• gases exert a pressure on the walls of their container.

PF (N)

A (m2)

SI unit: 1 N/m2 = 1 Pascal (Pa)

• pressure is defined as force per unit area:

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The atmospheric pressure can be measured using a barometer.

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760 mm Hg = 1 atm= 101 kPa

Standard atmospheric pressure supports a column of mercury about 760 mm high.

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Page 10: can be compressed exert pressure on whatever surrounds them expand into whatever volume is available easily diffuse into one another can be described

If the barometer reads 753.3 mm Hg, what is the atmospheric pressure in atm and kPa?

= 0.9912 atm753.3 mm Hg760 mm Hg

1 atm

= 100.1 kPa101 kPa

753.3 mm Hg760 mm Hg

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• Lord Kelvin proposed an absolute temperature scale defined by:

T(K) = T(C) + 273.15

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Page 13: can be compressed exert pressure on whatever surrounds them expand into whatever volume is available easily diffuse into one another can be described

Convert the following temperatures into Kelvin or celcius:

45 oC → 318.15 K

207 oC → 480.15 K-48 oC → 225.15 K

100 K → -173.15 oC

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Pressure Conversions:

1. 0.875 atm to mmHg

2. 745.0 mmHg to atm

3. 0.955 atm to kPa

4. 98.35 kPa to atm

5. 740.0 mmHg to kPa

6. 99.25 kPa to mmHg

665 mm Hg

0.9803 atm

96.5 kPa

0.9738 atm

98.34 kPa

746.8 mm Hg

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• In order to compare two gases, we choose a standard temperature and pressure:

STP: standard temperature and pressure → 273.15 K and 101.325 (data booklet) → one mole of gas has a volume of 22.4 L

SATP: standard ambient temperature and pressure → 298.15 K and 100.00 kPa → one mole of gas has a volume of 24.8 L

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Graph the following information

Pressure (KPa) Volume (L)

100 5.00

110 4.55

120 4.16

130 3.85

140 3.57

K (VP)

500

500.5

499.2

500.5

499.8

P1V1 = K = P2V2

Is there a relationship

between pressure

and volume?

volumepr

essu

re

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For a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure.

P1V1 = P2V2

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Graph the following data

Temperature (K) Volume (L)

298 5.00

323 5.42

348 5.84

373 6.26

398 6.68

K (V/T)

0.0167

0.0168

0.0168

0.0168

0.0168

V1/T1 = K = V2/T2

temperaturevo

lum

e

Is there a relationship

between pressure

and volume?

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Charles discovered that volume is directly proportional to its Kelvin temperature if a fixed mass remains under constant pressure.

V1 V2

T1 T2

=

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V versus T for different gases

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• The pressure of a gas is directly proportional to the Kelvin temperature, providing the volume and mass remain constant.

P1 P2=T1 T2

Joseph Louis Gay-Lussac and Jean-Baptistse Biot in their balloon on 24 August 1804

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V1P1 = V2P2

T1 T2

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• In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

• It follows that the volume of a gas at constant temperature and pressure is proportional to number of moles.

V1 V2

n1 n2=

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The relationship between volume V and number of moles n.

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The Ideal Gas EquationCombining the gas laws gives:

atmR Lmol K

0 08206.

where R is called the Gas constant:

PV = nRT

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• In order to compare two gases, we choose a standard temperature and pressure:

STP: 0C and 1 atm

• What is the volume of a mole of gas at STP?

VnRT

P

1 mol

0 08206.L atmmol K

273.15 K

1 atm=

= 22.4 L

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• If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?

2 H2O(g) 2 H2(g) + O2(g)

6.45 g H2O 18.02 g H2O1 mol H2O

2 mol H2O1 mol O2

1 mol O2

22.4 L O2

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• How many liters of CH4 at STP are required

to completely react with 17.5 L of O2 ?

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

17.5 L O2 22.4 L O2 1 mol O2

2 mol O2 1 mol CH4

1 mol CH4 22.4 L CH4

= 8.75 L CH4

22.4 L O2 1 mol O2

1 mol CH4 22.4 L CH4

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• Equal volumes of gas, at the same temperature and pressure contain the same number of particles.

• Moles are numbers of particles

• You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

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• How many liters of H2O at STP are produced

by completely burning 17.5 L of CH4 ?

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

17.5 L CH4 1 mol CH4 2 mol H2O

= 35.0 L H2O

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2 NaN3(s) 2Na (s) + 3N2(g)

Secondary reaction:

10 Na + 2 KNO3 K2O + 5 Na2O + N2(g)

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To measure the amount of gas produced in a reaction, it is often collected over water.

MgO(s) Mg(s) + O2(g)

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