GasesObjective

I will know the properties of gasses and be able to calculate the quantifiable properties

of gassesSuccess Criteria

I will be able to calculate each of the four quantifiable properties using the

appropriate equation

Video Links

• Liquid Nitrogen and Balloon Demo Link• Charles Law Balloon Demo Link• Boyle’s Law Demo Link

Collecting Background

• 1. What are the physical properties of a gas? 2. What are the quantitative (measurable) properties of gasses?

• 3. What are the units of pressure? How do you convert between the units?

• 4. Describe the different ways that the measurable properties can be related (the laws)?

Characteristics

What are the physical properties of a gas?• Gases are fluid– In essence, gases can flow

• Gases have low density• Gases are highly compressible• Gases completely fill their container

Pressure• Pressure is defined as the amount of force

exerted per unit area of surface– How hard the gas is pushing on the container

• There are several units of pressure– Atmosphere (atm)– Kilopascal (kPa)– Millimeters of Mercury (mm Hg)– Torr (torr)– Pounds per square inch (psi)– http://www.youtube.com/watch?v=XfFdNNiIAJw– http://www.youtube.com/watch?v=O9p8SRmQlPo

Pressure

• Converting Units of Pressure• 1.00 atm=101.325 kPa=760 torr=760 mm Hg• Practice Problems– The vapor pressure of water at 50.0oC is 12.33kPa.

What is the value in atmospheres?• (12.33kPa)(1.00atm/101.325kPa)=0.122atm

– In thermodynamics, the standard pressure is 100.0kPa. What is this value in mm Hg?• (100.0kPa)(760mm Hg/101.325kPa)=750.1mm Hg

Quantifiable Properties

• There are four measurable properties of gases– P is pressure– T is temperature (must be measured in Kelvin)• K = oC + 273

– V is the volume– n is the number of moles of the gas

STP

• Standard Temperature and Pressure (STP)• Standard Temperature is 0oC • Standard Pressure is 1.00atm• ΔG° = ΔH° – T ΔS°• The superscript indicates at STP

Boyle’s Law• Boyle’s Law discusses the relationship of pressure and

volume• This shows that as the pressure increases, the

volume decreases.– This is an inverse relationship

• PV = k1

• P = k1/V where k1 is the pressure-volume constant• And if we keep Temperature and the number of

particles constant, but change either pressure or volume we

then have: P1V1=P2V2

Boyle’s Law

• A sample of oxygen gas has a volume of 150.0mL at a pressure of 0.947atm. What will the volume be at a pressure of 1.000atm if the temperature remains constant?– P1V1=P2V2

– (0.947atm)(150.0mL)=(1.000atm)V2

– V2=142mL

– Be sure your answers are in Significant Figures!

Boyle’s Law

• A balloon has a volume of 456mL at a pressure of 761torr. It is taken under water in a submarine to a depth where the air pressure in the submarine is 3.3atm. What is volume of the balloon if the temperature remains constant?

– Be sure all variables are in the same units• (761torr)(1.00atm/760torr)=1.00atm

– P1V1=P2V2

– (1.00atm)(456mL)=(3.3atm)V2

– V2=138mL=140mL

Charles’ Law• Charles’ Law discusses the relationship

between Temperature and Volume• This shows that as the temperature increases,

the volume increases– This is a direct relationship

• V/T=k2

• At a constant pressure and number of particles, if either temperature

or volume change we have: V1/T1=V2/T2

Charles’ Law

• A sample of neon gas has a volume of 752mL at 25.0oC. What is the volume at 50.0oC if the pressure remains constant?– Change the degrees in to Kelvin• 25.0oC + 273 = 298 K; 50.0oC + 273 = 323 K

– V1/T1=V2/T2

– (752 mL)/(298 K)=V2/(323K)

– V2 = 815 mL

Gay-Lussac’s Law• Gay-Lussac’s Law discusses the relationship

between temperature and pressure• This shows that as the temperature increases,

the pressure increases.• P = k3T where k3 is the temperature-pressure

constant• P/T = k3

• At a constant volume and number of particles, if either the temperature or

pressure changes, we use this equation: P1/T1=P2/T2

Gay-Lussac’s Law

• At 122oC, the pressure of a sample of nitrogen is 1.07atm. What will be the pressure at 205oC, assuming constant volume?– First change temperature to Kelvin• 122oC + 273 = 395 K; 205oC + 273 = 478 K

– P1/T1=P2/T2

– (1.07 atm)/(395 K)=P2/(478 K)

– P2= 1.29 atm

Combined Gas Law

• When the three different laws just discussed are analyzed, what would they be if they are all combined?

• PV = k1

• V/T=k2

• P/T = k3

Combined Gas Law

• PV/T = k4

• If the number of moles remains constant, we can solve for any of the other variables

• P1V1/T1= P2V2/T2

• Also used is Standard Temperature and Pressure (STP)

Combined Gas Law

• At a pressure of 780.0 torr and 24.2oC, Xenon has a volume of 0.350L. What will the volume be at STP?

• First change temperature to Kelvin– 24.2oC + 273 = 297.2 K

• Change the pressure to atmospheres– (780.0 torr)(1.00 atm/760 torr) = 1.03 atm

– P1V1/T1= P2V2/T2

– (1.03atm)(0.350 L)/(297.2K) = (1.00atm)V2/(273K)– V2 = 0.330 L

Diffusion and Effusion

• Diffusion is the movement of particles from high density to low density– Popping a balloon

• Effusion is the passage of gas molecules under pressure through a tiny opening– Nail in a tire

Graham’s Law of Diffusion

• In Graham’s Law of Diffusion, we will analyze the kinetic energy of two different gases– Kinetic energy formula is 1/2mv2

• m is the mass (in this case molar mass)• v is the velocity of the molecule

– Graham’s Law states that the kinetic energy of the first gas will equal the kinetic energy of the second gas

m1v12= m2v2

2

Graham’s Law of Diffusion

• Oxygen molecules have an average speed of 480 m/s at room temperature. At the same temperature, what is the average speed of molecules of sulfur hexafluoride?– Molar mass of Oxygen is 32.00 g/mol (O2)– Molar mass of SF6 is 146.07 g/mol– m1v1

2= m2v22

– (32.00 g/mol)(480 m/s)2= (146.07 g/mol) v2

– v = 220 m/s

Dalton’s Law of Partial Pressures• Dalton’s law of partial pressures states that the total

pressure of a mixture of gases is equal to the sum of the partial pressures of each gas

• PT = P1 + P2 + P3 + ……• A 42.0 mL sample of hydrogen is collected over water

at 30.0oC. What is the total pressure if the pressure of the dry gas 845.0 torr and the vapor pressure of water at 30.0oC is 18.6 torr?

• PT = P1 + P2 + P3 + ……• PT = 845.0 torr + 18.6 torr• PT = 863.6 torr

Essential Information• 1.00 atm=101.325 kPa=760 torr=760 mm Hg• STP: 273 K and 1.00 atm• Dalton’s Law: PT = P1 + P2 + P3 + ……

• Combined Gas Law: P1V1/T1= P2V2/T2

• Gay-Lussac’s Law: P1/T1=P2/T2

• Charles’ Law: V1/T1=V2/T2

• Gramham’s Law: m1v12= m2v2

2

• Boyle’s Law: P1V1=P2V2

• Ideal Gas Law: PV=nRT where R is the gas constant

Essential