• April 24, 2013
Perseverance
• Aficionado: a devotee; a fan; an enthusiastic person about
a sport or hobby
• Do Now: Quad Card
• Topic:
Air pressure
Basic Gas Laws
(Boyle’s, Charles’s & Gay-Lussac’s)
What makes a hot air balloon
inflate?
HINT: look at
the name!
Part 1: What Is a Gas Law?
• the gas laws are simple, mathematical relationships between the
pressure (P), volume (V), temperature (T), and moles (n), of a gas.
Basic gas laws involve P, V, and T only.
Combined Gas Law
• the 5 basic gas laws:
PV = PV
Boyle’s Law
P1V1 = P2V2
Charles’s Law Gay-Lussac Law
V1 = V2
T1 T2
P1 = P2
T1 T2
1
T1
1
2
T2
2
Partial Pressures Law
PT = P1 + P2 + P3
• gas laws use the Kelvin temperature scale (K). Why? The Celsius
scale (C) has negative values and a zero value (the Kelvin scale does
not). If we used the Celsius scale, we might calculate a zero or
negative volume/pressure from it, which can’t exist!
• to convert a Celsius temp into Kelvin, just add 273 C + 273 = __K
• notice that Kelvin temps do not have a degree sign, just a “K”
• to convert a Celsius temp into Kelvin, just add 273 C + 273 = __K
• notice that Kelvin temps do not have a degree sign, just a “K”
• there is a certain temperature that is considered “standard,” as well as
a standard pressure. The values for standard temp and pressure (STP)
are 273 K and 1 atm.
• all the gas laws (except Charles’s law) involve pressure. Most people
are not familiar with the many units pressure can be measured in
(except maybe psi). So here they are:
NOTE: the STP values
Unit
Abbr. STP value shown here can be used
to convert one pressure
atmospheres
atm
1 atm
millimeters of mercury mmHg 760 mmHg unit to another, which
will need to be done
pounds per square inch
psi
14.7 psi
quite a bit in your
kilopascals
kPa 101.325 kPa calculations!
• remember: units of volume = milliliters (mL), liters (L), and cubic
centimeters (cm3).
Part 2: Boyle’s Law (1662)
Boyle’s Law
• Boyle’s Law states that the pressure of a fixed
P1V1 = P2V2
mass of gas varies inversely with the volume at
a constant temperature.
• this means if you compare the initial volume and
pressure of a gas with the new conditions of the gas,
P
you will get an inverse relationship every time
• P1 and V1 indicate initial (or starting) conditions
V
• P2 and V2 indicate new (or final) conditions
Steps for Solving ANY Gas Law Problem:
1. Write out a column of information down the left-hand side.
Make sure all of your variable’s units match (i.e. if P1 is in kPa, then
P2 must be in kPa as well). If one doesn’t match, convert it to match
the other, using a conversion table. Put a question mark in the space
for the variable you are trying to solve for (what you DON’T have).
2. Write the original equation for the gas law you will be using.
3. Rearrange the equation to solve for the variable you need.
Steps for Solving ANY Gas Law Problem:
1. Write out a column of information down the left-hand side.
Make sure all of your variable’s units match (i.e. if P1 is in kPa, then
P2 must be in kPa as well). If one doesn’t match, convert it to match
the other, using a conversion table. Put a question mark in the space
for the variable you are trying to solve for (what you DON’T have).
2. Write the original equation for the gas law you will be using.
3. Rearrange the equation to solve for the variable you need.
4. Plug in the values and units you have in to the rearranged equation,
and make sure all your units will cancel except for one. This will be
the unit for your answer.
5. Calculate, then box your answer!
Ex1: Using 14.3 L of N2 as the initial volume, calculate the volume that
would result if the pressure was raised from 150 kPa to 250 kPa.
Ex1: Using 14.3 L of N2 as the initial volume, calculate the volume that
would result if the pressure was raised from 150 kPa to 250 kPa.
P1 = __________
150 kPa
P1V1 = ____
P2V2
V2 = (150 kPa)(14.3 L) =
____
14.3 L
V1 = __________
250 kPa
P
P
2
2
250 kPa
P2 = __________
V2 = 150 14.3 ÷ 250 =
?
V2 = __________
V2 = P1V1
P2
V2 = 8.58 L
Charles’s Law
Part 3: Charles’s Law (1787)
V1 = V2
• Charles’s Law states that the volume of a fixed mass
T1 T2
of gas varies directly with the temperature at a
constant pressure.
• this means that as the volume of gas increases, so
V
does the temperature
Ex2: A sample of gas occupied a volume of 5.0L at a
temp of 37.0C. If the temp were to increase by 6C,
T
what would be the volume of the gas under this new condition?
Ex2: A sample of gas occupied a volume of 5.0L at a temp Charles’s Law
of 37.0C. If the temp were to increase by 6C, what would
V1 = V2
be the volume of the gas under this new condition?
T1 T2
5.0 L
V1 = __________
V1T2 =____
V2T1
____
V2 = (5.0 L)(316 K) =
310
37
T1 = ____C____K
T1
T1
V2 = __________
?
310 K
T2 = ____C____K
316
43
V = 5.0 316 ÷ 310 =
V2 = V1T2
T1
2
V2 = 5.10 L
Gay-Lussac Law
Part 4: Gay-Lussac’s Law (1802)
P1 = P2
• Gay-Lussac’s Law states that the pressure of a
T1 T2
fixed mass of gas varies directly with the
temperature at a constant volume.
• this means that as the pressure of gas increases,
P
so does the temperature
Ex3: The gas left in a used aerosol can is at a pressure
T
of 125.3 kPa at 17C. If the can is thrown into a fire, what
will the pressure be inside the can at 1045C?
Ex3: The gas left in a used aerosol can is at a pressure
Gay-Lussac Law
of 125.3 kPa at 17C. If the can is thrown into a fire, what
P1 = P2
will the pressure be inside the can at 1045C?
T1 T2
P1 = __________
125.3 kpa
P1T2 = P____
____
2T1
P2 = (125.3 kPa)(1318 K) =
290
T1 = ____C____K
17
T1
T1
P2 = __________
?
290 K
1318
1045
P2 = P1T2
T2 = ____C____K
P = 125.3 1318 ÷ 290 =
T1
2
P2 = 569.47 kPa