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Pressure

The force per unit area of a surface.Units: N/cm2

N, Newton: SI unit of forceAs area of contact changes, force changes500 N = 1.7 N 500 N = 83.3 N300 cm2 cm2 6.0 cm2 cm2

Christina,

Will you go to the prom with me?

Steve

Barometer

• Used to measure the pressure of gases.

barometerplanet.com

Units of Pressure for Gases

• Millimeters of mercury (mm Hg)• Torricelli or 1 torr = 1mm Hg • 760 mm Hg = 1 atmosphere at sea level

when temp is 0oC• Pascal = pressure exerted by a force of one

newton acting on an area of one square meter. Pa = N/m2

• 1.013 x 105 kPa = 1 atmosphere= 10.1N/cm2

Pressure conversion sample:

Express 0.725 atm in a) mm Hg and b) kilopascals (kPa)

A) 0.725 atm x 760 mm Hg = 551 mm Hg 1 atm

B) 0.725 atm x 101.325 kPa = 73.5 kPa 1 atm

Standard Conditions (STP)

Standard Temperature is 0oC or 273K

Standard Pressure is 1 atm or 760 mm of Hg

You must have done by next meeting:

A list of the units of pressure (see page 364) andA list of the gas laws:Dalton’s Law of Partial Pressures: PT = P1+P2+…

Boyle’s: P1V1 = P2V2 T constant

Charles’s: V1 = V2 P constant

T1 T2

Gay-Lussac’s: P1 = P2 V constant

T1 T2

Combined Gas: P1V1 = P2V2

T1 T2

Ideal Gas: PV = nRT

Dalton’s Law of Partial Pressures

The total pressure of a gas mixture is the sum of the partial pressures of the component gases.

Gas collected through water picks up water vapor, so allow inside and outside water levels in a gas collection device to stabilize and:

Patm = Pgas + PH20

Dalton sample problem:

Oxygen gas is collected by water displacement. The barometric pressure and the temperature during the experiment are 731.0 torr and 25.0oC. What was the partial pressure of the oxygen collected?

PT = Patm = 731.0 torr

PH20 = 23.8 torr (see vapor pressure of water at 25.0 oC from table in handout or book – Table A-8)

PT = Patm = 731.0 torr

PH20 = 23.8 torr (see vapor pressure of water at 25.0 oC from table in handout or book – Table A-8)

Patm = PO2 + PH20

So PO2 = Patm – PH20

PO2 = 731.0 torr – 23.8 torr = 707.2 torr

Boyle’s Law – at constant temperature, volume of a fixed gas varies inversely with the pressure.

If 100.0 mL of a gas, originally at 760 torr, is compressed to a pressure of 800 torr, at a constant temperature, what would be its final volume?

P1V1 = P2V2 --> V2 = P1V1

P2

V2 = 100mL(760 torr) = 95.0 mL

800 torr

Charles’ Law – the volume of a fixed mass of gas at constant pressure varies directly with the

Kelvin temperature.

A sample of neon gas has a volume of 752 mL at 25.0oC. What will the volume at 100.0oC be if pressure is constant?

V1 = V2 --> V2 = V1T2

T1 T2 T1

V2 = 752 mL (100.0oC) = 300.8 mL

25.0oC

Gay-Lussac’s Law – the pressure of a fixed mass of gas at constant volume varies directly with

the Kelvin temperature.

At 122oC the pressure of a sample of nitrogen gas is 1.07 atm. What will the pressure be at 205oC, assuming constant volume?

P1/T1 = P2/T2 --> P2 = P1T2

T1

P2 = 1.07 atm(205&273) = 1.29 atm

122+273

Gay-Lussac’s Law of combining volumes

At constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.

H2 + Cl2 --> 2HCl

1L 1L 2L H:Cl:HCl = 1:1:2

Formulas must be written correctly and chemical equation balanced.

Combined Gas Law – expresses the relationship between pressure, volume, and temperature of

a fixed amount of a gas. PV = k T

P1V1 = P2V2 --> To find V2: V2 = P1V1T2

T1 T2 P2T1

Problem: The volume of a gas is 27.5mL at 22.0oC and 0.974 atm. What will be the volume at 15.OoC and 0.993 atm?Temps to K: 22+273 = 295K and 15+273=288K

The volume of a gas is 27.5mL at 22.0oC and 0.974 atm. What will be the volume at 15.OoC and 0.993 atm?

V2 = P1V1T2

P2T1

V2 = 0.974atm(27.5ml)(288K)

0.993atm(295K)V2 = 26.3 mL

Hint: in solving gas law problems, use the combined gas law and quantities that don’t change will cancel out.

V1 = 27.5mlT1 = 295KP1 = 0.974atmV2 = ?T2 = 288KP2 = 0.993atm

Avogadro’s Law – equal volumes of gases at the same temperature and pressure contain equal

numbers of molecules.Ratios apply here also.2H2 + O2 --> 2H2O

2molecules 1molecule 2molecules2mol 1mol 2mol2volumes 1volume 2volumesStandard molar volume of a gas is the volume

occupied by one mole of a gas at STP.Standard molar volume = 22.4 L/mol

Steve,

Yes, I will go to the prom with you.

Christina

At STP, what is the volume of 7.08 mol of nitrogen gas?

7.08 mol (22.4L) = 158 L 1 mol

A sample of gas occupies 11.9 L at STP. How many moles of the gas are present?

11.9L (1 mol) = 0.531 mol 22.4L

Assuming all volume measurements are made at the same temperature and pressure, what volume of

hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor?

Write and balance the equation.

Label known and unknown.

Do unit analysis.

Gas Stoichiometry – dealing with proportional relationships between reactants and products in

a chemical reaction.2CO2(g) + O2(g) --> 2CO2(g)

2molecules 1 molecule 2 molecules2 mol 1 mol 2 mol2 volumes 1 volume 2 volumes

Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor?

Write the correct chemical reaction first:

2H2 + O2 --> 2H2O

Indicate known and unknown:Solve:

Ideal Gas Law – the mathematical relationship among pressure, volume, temperature, and the

number of moles of a gas.PV = nRT --> R = PV R is the ideal gas

nT constant.At STP, R = 1 atm(22.414 L) = 0.0821 L atm

1 mol(273.15K) mol K

R is the ideal gas constant.

What pressure, in atmospheres, is exerted by 0.325 mol of hydrogen gas in a 4.08 L container

at 35oC?PV = nRT --> P = nRT VT = 35 + 273 = 308KP = 0.325 mol 0.0821L atm) (308K) 4.08 L mol KP = 2.01 atm

Gases spread out in a container – diffusion

Gases can randomly pass through a tiny opening in a container (leak out) - effusion