gases are easily compressed gases can expand large amount of space between particles
TRANSCRIPT
Properties of Gases and Gas Laws
Ch. 14.1 & 14.2
Properties of Gases
Gases are easily compressed
Gases can expand› Large amount of space between
particles
Properties of Gases
Compressing a gas causes pressure to increase
3 Factors Affect Gas Pressure› Volume of the container (V)
Measured in Liters› Amount of gas (n)
Measured in the number of moles› Temperature
Measured in Kelvins (T)
Properties of Gases
Amount of gas affects on pressure› Increase in number of particle› Increases in number of collisions› Increase in Pressure
As the amount of gas changes, the pressure changes directly› Ex. When the number of moles is doubled
the pressure double.› Also works when the moles decreases
Properties of Gases
Gases will flow to the area with lower pressure.› Ex:
Deflating balloon Gas will leave the balloon into the surroundings
Vacuum sealed container Gas will rush into the container
Properties of Gases
Volume’s effects on pressure› Decrease in volume of the container› Increase in the number of collisions› Increase in pressure
Volume inversely affects pressure A volume decrease causes a pressure
increase If the volume is halved, the pressure is
doubled. Squeezing a pack of Ketchup
Properties of Gases
Temperature’s affects on pressure› Increase in Temperature› Increase in the speed of the gas particles› Increase in the number of collisions› Increase in pressure
Temperature directly affects pressure› If the temperature double, so will the
pressure› Ex: Frozen balloon
The Gas Laws
Boyle’s Law› For a given mass of gas at constant
temperature, the volume of a gas varies inversely with pressure.
The Gas Laws
Boyle’s Law
›P ×V =P ×V₁ ₁ ₂ ₂
› Pressure in units of kilopascals (kPa)› Volume in units of liters (L)
The Gas Laws
Boyle’s Law Ex:› The volume of a raft has an initial volume
of 1.2 liters and an initial pressure of 87kPa. If the final volume was 2.9 liters what was the final pressure?
› P =87kPa V =1.2L P =? V =2.9₁ ₁ ₂ ₂› P ×V =P ×V (₁ ₁ ₂ ₂ P ×V )/ V =P₁ ₁ ₂ ₂› (87x1.2)/2.9= 36 kPa
The Gas Laws Charles Law: As the temperature of an
enclosed gas increases, the volume increases, if the pressure is constant.
The Gas Laws
Charles Law:
› V₁/T₁=V₂/T₂
› Volume in units of Liters (L)› Temperature in units of Kelvins (K)
The Gas Laws
Charles Law Ex:› The volume of an inflated balloon at 24˚C
has a volume of 4 liters. The balloon is then moved to a room with a temperature of 58˚C. What is the Volume? T = 24˚C+ 273 = 292 K T = 58˚C + 273 = 331 K₁ ₂ V₁/T₁=V₂/T₂ V = (V xT )/ T₂ ₁ ₂ ₁ (4.00L x 331 K)/ 297 K= 4.46L
The Gas Laws
Gay-Lussac’s Law› As the temperature of an enclosed gas
increases, the pressure increases, if the volume is constant.
› P /T =P /T₁ ₁ ₂ ₂
› Pressure: kPa› Temperature: K
The Gas Laws
Gay-Lussac’s Law Ex:› An aerosol can is stored at 25˚C and has a
pressure of 103 kPa. If the f=can is heated to 928˚C, what is the final pressure?
› T =25˚C + 273= 298 K₁› T = 928˚C + 273= 1201 K₂› P /T =P /T T (P /T )=P₁ ₁ ₂ ₂ ₂ ₁ ₁ ₂› 103kPa x (1201 K / 298 K)= 415 kPa› Or 4.15 x 10² kPa
The Gas Laws
The Combined Gas Law› Calculates for situations where the amount
of gas is constant.
› (P x V )/T = (P x V )/ T₁ ₁ ₁ ₂ ₂ ₂
› P: kPa› V: L› T: K
The Gas Laws
Combined Gas Law Ex:› The volume of a balloon is 30.0 L at 313 K
and 153 kPa pressure. What would the volume be at standard temperature and pressure (STP)
› STP: 273K and 101.3 kPa› (P x V )/T = (P x V )/ T₁ ₁ ₁ ₂ ₂ ₂› V = (V x P x T )/ (P x T )₂ ₁ ₁ ₂ ₂ ₁› (30L x 53kPa x 273K)/(101.3kPa x 313K)= 39.5 L
Ideal Gas Law14.3
Ideal Gas Law
Ideal Gas Law was created to calculate the number of moles of a contained gas.
Symbol for number of moles is “n” Number of moles is directly
proportional to volume
Ideal Gas Law
1mol of every gas is 22.4 L at STP
The Ideal Gas Constant R=(P x V)/(T x n)
At STP :› (101.3 x 22.4)/(273 x 1)=
8.31(L·kPa)/(K·mol)› R= 8.31
Ideal Gas Law
Common Equation
PV= nRT P: kPa V: L n: mol R: (L·kPa)/(K·mol) T: K
Review
Converting mass to moles. 12 grams of CO₂ how many moles?
› 12grams/(MMof C + 2 x the MMof O)› 12/(12+2·16)= .2727 mols
Ex.
What is the pressure of 113g of Xenon gas at 187˚C, held in a 1.2L container?
nRT/V=P 113g/131.3MM of Xe= .8606 mols T= 187˚C+ 273= 460k (.8606·8.31·460)/1.2= 2741.44 kPa
Gases, Part 3
Real Gases vs. Ideal Gases: Ideal gases follow the gas laws at all
pressures and temperatures.
Real gases can not be described by the gas laws at certain temps or pressures.
An ideal gas has particles with NO volume, and there are no attractions between the particles.
A real gas has particles that have volume, and there are interactions between particles.
Real gases differ most from ideal gases at high pressures and low temperatures.
Partial pressure: The pressure that one gas contributes to the total pressure.
Dalton’s Law: Ptotal = P1 + P2 + P3 + …
101.3 kPa
On top of Mt. Everest, air pressure is 33.73 kPa. Since oxygen is 21% of air, the pressure of oxygen is 7 kPa.
You need 10.67 kPa Oxygen to live, so mustHave compressed O2.
Diffusion: The tendency of molecules to spread out evenly.
Effusion: A gas escapes through a tiny hole.
Lower the molar mass of a gas = Faster Effusion and Diffusion.
Graham’s Law: RateA / Rate B = (Molar mass B / Molar mass A)^.5