Gases. Elements that exist as gases at 25 0 C and 1 atmosphere

Gases

Elements that exist as gases at 250C and 1 atmosphere

• 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.

Physical Characteristics of Gases

Kinetic Theory of Gases

• KMT Based on 5 Assumptions:

1. G. have large number of particles that are very far apart from each other

relative to their size.

2. Collisions between g. particles and between g. particles and container walls

are elastic. (No net loss of kinetic energy)

Kinetic Theory of Gases cont.

3. G. particles are in constant motion

4. There are no forces of attraction between G. particles.

5. The temperature of a gas depends on the average kinetic energy of the particles of the gas. KE= ½ (mv2). All gases have the same

KE at the same temperature.

Ideal vs. Real Gases

• Ideal gas: hypothetical gas that perfectly fits all of the assumptions of the KMT.

• Real gas: one that does not behave exactly according to the assumptions of the KMT.

• Real gas behavior more likely at high pressures and low temperatures.

Questions

1. What happens to gas particles when a gas is compressed?

2. What happens to gas particles when a gas is heated?

3. Describe the conditions under which a real gas is most likely to behave ideally.

Gases and Pressure

• Pressure: Force per unit area.

Pressure =

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

4/16/2012

Get out your notebooks and your books if you

have them.

Gases

Gases and Pressure

• Calculate the pressure of the following:

1. F=500N, A= 325cm2

2. F=500N, A= 13cm2

3. F=500N, A= 6.5cm2

Gases

Gases and Pressure• Convert the following

pressures:

1. P = 0.830atm to mmHg, and kPa

2. P = 1.75atm to mmHg and kPa

3. The critical pressure of carbon dioxide is 72.7atm, what is this value in Pa?

Gases

Gases and Pressure• Standard Temperature

and Pressure (STP)

• = 1atm and 0°C

Gases

Dalton’s Law of Partial Pressures• The pressure of each gas in a

mixture is the partial pressure of that gas.

• Dalton’s Law of partial pressures = the total pressure of a gas mixture is the sum of all the partial pressures of the

component gases.

• PT = P1 + P2 + P3 + ….

Gases

Dalton’s Law of Partial Pressures

• Water Displacement, gas collected over water.

• Always mixed with water vapor, which has specific

water pressure.

• Patm = Pgas + PH2O

Gases

Dalton’s Law of Partial Pressures• Three of the primary

components of air are CO2, O2, and N2. In the sample

containing a mixture of only these gases at exactly 1atm, the

partial pressures of PCO2 = 0.285 torr and PN2 = 593.525

torr. What is the partial pressure of oxygen?

Gases

Dalton’s Law of Partial Pressures• Oxygen gas from

decomposition of KClO3, was collected by water

displacement. The barometric pressure and temperature was 731.0 torr and 20.0°C. What

was the partial pressure of the oxygen collected?

4/19/2012

Get out your notebooks

Boyle’s Law

• Volume of a fixed mass of gas varies inversely with pressure at a constant temperature.

• PV = k , k is a constant

• Product always remains the same, so can be written:

• P1V1 = P2V2

Boyle’s Law

Constant temperatureConstant amount of gas

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

Charle’s Law• Volume of a fixed mass of gas at

constant pressure varies directly with the Kelvin Temperature.

• V = kT , k is a constant or

Charle’s Law• Kelvin Scale: -273.15ºC, or 0 K. Absolute

Zero at which all movement stops.

• K = 273.15 + ºC

• Convert 50ºC to K

• K = 273.15 +50

• K = 223.15K

Charle’s Law

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

Guy-Lussac’s Law• Pressure of a fixed mass of

gas at a constant temperature varies directly with the Kelvin Temperature.

• P = kT , k is a constant or

The gas in a container is at a pressure of 3.00atm at 25ºC. Directions on the container warn the user not to keep it in a place where the temp. exceeds 52ºC. What would the pressure in the container be at that temp?

P1 = 3.00 atm

T1 = 298.15 K

P2 = ?

T2 = 325.15K

P2 = T2 x P1

T1

325.15 K x 3.00atm298.15 K= = 3.27atm

P1/T1 = P2/T2

Combined Gas Law• Expression of the relationship between

pressure, volume and temperature of a fixed amount of gas.

Combined Gas LawA helium filled balloon has a volume of 50.0L at 25ºC and 1.08atm. What volume will it have at 0.855atm and 10.0ºC.

P1 = 1.08 atm

T1 = 298.15 KP2 = 0.855atm

T2 = 283.15 K

V2 = T2 x P1 x V1

P2 x T1

283.15 K x 1.08atm x 50.0L298.15 K x 0.855atm

=

= 60.0L

V1 = 50.0 L V2 = ?

Avogadro’s Law

V number of moles (n)

V = k x nConstant temperatureConstant pressure

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

Avogadro’s Law

Standard Molar Volume: Volume occupied by 1 mole at STP = 22.4L

1mol gas = 22.4L gas

What Volume does 0.0685mol of gas occupy at STP?

Multiply the amount in moles by the conversion factor, 1mol gas = 22.4L gas

What quantity of gas, in moles, is contained in 2.21L at STP?

Multiply the volume in liters by the conversion factor, 1mol gas = 22.4L gas

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

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

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

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

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

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

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

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

Gas effusion is the process by which gas particles pass through a tiny opening.