chapter 13 gases. make a flip book boyle’s gas law how are pressure and volume related? as the...
TRANSCRIPT
Chapter 13
Gases
Make a Flip Book
Boyle’s Gas Law
How are pressure and volume related?
As the pressure increases, the volume decreases proportionally.
Boyle’s Gas Law
Identify and label 2 situations
A diver blows a 0.75 liter air bubble 10m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the volume of the air in the bubble at the surface?
Identify and label 2 situations
A diver blows a 0.75 liter air bubble 10m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the volume of the air in the bubble at the surface?
10m underwater Surface
V1 = 0.75 L V2 = ?
P1 = 2.25 atm P2 = 1.03 atm
Solve using Boyle’s Law
P1V1 = P2 V2
2.25 0.75 = 1.03 V2
Use magic Algebra both sides by 1.03
V2 = 2.25 0.75 1.03
V2 = 1.6 L
Check Answer
P1V1 = P2 V2
2.25 0.75 = 1.03 V2
V2 = 1.6 L
The pressure decreased by about half, 2.25 to 1.03.
So, the volume should double. Is 1.6 about twice 0.75?
STP
•Standard Pressure
•1 atm
•760 torr
•760 mmHg
Standard Temperature
• 273º K
• 0º C
Charles’s Law
How are temperature and volume related?
The volume of a given amount of gas is directly proportional to its kelvin temperature at constant pressure.
Temperature in kelvin units
Convert celsius degrees to kelvins
Tk = 273 + Tc
Charles’s Law
A helium balloon in a closed car occupies a volume of 2.23 L at 40.0C. If the car is parked on a hot day and the temperature inside rises to 75.0C, what is the new volume of the balloon, assuming the pressure remains constant.
Set up two situations
Car with balloon
T1 = 40C+ 273 = 313
V1 = 2.32 L
Car parked in the heat
T2 = 75.0C + 273 = 348
V2 = unknown
2.32 2
313 348348 2.32 313 2
2 2.58
V
V
V L
Assume the pressure and amount of gas remain constant
What volume will the gas in the balloon occupy at 250 K?
A gas at 89C occupies a volume of 0.67L. At what celsius temperature will the volume increase to 1.12L?
The celsius temperature of a 3.00L sample of gas is lowered from 80.0C to 30.0C. What will be the resulting volume of the gas?
Assume the pressure and amount of gas remain constant
What volume will the gas in the balloon occupy at 250 K?1.8LA gas at 89C occupies a volume of 0.67L. At what celsius temperature will the volume increase to 1.12L?330CThe celsius temperature of a 3.00L sample of gas is lowered from 80.0C to 30.0C. What will be the resulting volume of the gas?2.58L
Gay-Lussac’s Law
How are temperature and pressure of gas related?
Remember the temperature must be in kelvins.
Example
The pressure of the oxygen gas inside a canister is 5.00 atm at 25.0 C. The canister is located at a camp high on Mount Everest. If the temperature there falls to -10.0 C , what is the new pressure inside the canister?
Set up 2 situations
Warm
T1 = 25.0 C +273
P1 = 5.00 atm
Cold
T2 = -10.0 C + 273
P2 = unknown
5.00 2
298 2635
263 2298
2 4.41
P
P
P atm
Analyze the answer
The temperature decreased, so the pressure should decrease. The unit is atm matches the pressure unit given.
Examples
The pressure in an automobile tire is 1.88atm at 25.0 C. What will be the pressure if the temperature increases to 37.0 C.
Helium gas in a 2.00L cylinder is under 1.12 atm pressure. At 36.5 C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder?
Examples
The pressure in an automobile tire is 1.88atm at 25.0 C. What will be the pressure if the temperature increases to 37.0 C.
1.96 atm
Helium gas in a 2.00L cylinder is under 1.12 atm pressure. At 36.5 C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder?
-138.0 C
The Combined Gas Law
Temperature measured in kelvins.
Example
A gas at 110 kPa and 30C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0C and the pressure increases to 440kPa, what is the new volume?
Set up 2 situations
Initial
P1 = 110 kPa
T1 = 30.0 C +273
V1 = 2.00L
Raised Temp
P2 = 440 kPa
T2 = 80.0 C + 273
V2 = unknownLV
V
V
58.02
2303
2110
440
353353
2440
303
00.2110