ratio method of solving basic gas laws. boyle’s law don’t hate me just because i’m beautiful...
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Ratio Method of SolvingBasic Gas Laws
Boyle’s Law
Don’t hate me just because I’m beautiful
Robert Boyle
Boyle’s Law
Demonstrations that illustrate Boyle’s Law• Marshmallow or balloon in the bell jar (marshmallow or balloon
gets big when pressure gets low)
Boyle’s Law
Volume is inversely related to PressureIf volume decreases, pressure increasesIf volume increases, pressure decreases
(they move in opposite directions)
You can see this relationship when temperature and moles of gas are held constant
Boyle’s Law
Mathematically, PV = kWhere “k” is a constant
Boyle’s Law
Another way to write Boyle’s law is
P1V1 = P2V2
Temperature is constant!! If it is not, then you can not use Boyle’s Law
Boyle’s Law Problem #1
A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be
the new volume?Start with a “data table” to organize what you know and what you don’t know
• V1 = 0.030L
• V2 = ?
• P1 = 1.1 atm
• P2 = 0.20 atm
Boyle’s Law Problem #1
A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be
the new volume?
• V1 = 0.030L
• V2 = ?
• P1 = 1.1 atm
• P2 = 0.20 atm
V2 = 0.030L x 1.1 atm
0.20 atm= 0.17 L
Bigger number on top
makes Volume get larger
Pressuregot smallerso Volumemust getlarger
Boyle’s Law Problem #2
A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the
new volume?
Start with your data table
• V1 = 0.025L
• V2 = ?
• P1 = 0.60 atm
• P2 = 0.95 atm
Boyle’s Law Problem #2
A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the
new volume?
• V1 = 0.025L
• V2 = ?
• P1 = 0.60 atm
• P2 = 0.95 atm
V2 = 0.025L x 0.60 atm
0.95 atm= 0.016 L
Smaller number on top
Makes volume get smaller
Pressuregot larger, so Volumemust get smaller
Charles’ Law
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Charles’ Law
Demonstrations that illustrate Charles’ Law• Balloon on flask (heat the flask then the balloon expands)
Charles’ Law
Volume is directly related to TemperatureIf temperature increases, volume increases
If temperature decreases, volume decreases(they move in the same direction)
You can see this relationship when pressure and moles of gas are held constant
Charles’ Law
V = kTWhere “k” is a constant of proportionalityNote that V does NOT have to equal T, but they are
directly proportional to each other
Charles’ Law
Another way to write this is:(V1/ T1) = (V2/T2)
OrV1T2 = V2T1
Pressure must remain constant or you can not use this law!
Charles’ Law Problem #1
A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume?
Start with a data table to organize what you know and what you don’t know
• V1 = 2.0 L
• V2 = ?
• T1 = 298 K
• T2 = 278 K
Charles’ Law Problem #1
A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume?
• V1 = 2.0 L
• V2 = ?
• T1 = 298 K
• T2 = 278 K
V2 = 2.0 L x 278 K
298 K= 1.9 L
Smaller number on top
makes volume get smaller
Temperaturedecreasedso volumemustdecrease
Charles’ Law Problem #2
A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen?
Start with your data table
• V1 = 3.25 L
• V2 = 2.50 L
• T1 = 298 K
• T2 = ?
Charles’ Law Problem #2
A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen?
• V1 = 3.25 L
• V2 = 2.50 L
• T1 = 298 K
• T2 = ?
T2 = 298 K x 2.50 L
3.25 L= 229 K
Smaller number on top
makes temperature sm
aller
Volume gotsmaller so
temperaturemust getsmaller
Gay-Lussac Law
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we may continue.
Gay-Lussac Law
Demonstrations that illustrate Gay-Lussac’s Law
Aerosol can in the campfire (As temperature increases, pressure increases until the can explodes)
Gay-Lussac Law
Pressure is directly related to TemperatureIf temperature increases, pressure increases
If temperature decreases, pressure decreases(they move in the same direction)
You can see this relationship when volume and moles of gas are held constant
Gay-Lussac Law
P = kTWhere “k” is a constant of proportionality
Gay-Lussac Law
Another way to write it:(P1 / T1) = (P2 / T2)
OrT1P2 = P1T2
Volume stays constant!
Gay-Lussac Law Problem #1
A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if
the can is heated to 500.K?
Start with a data table
• T1 = 298K
• T2 = 500.K
• P1 = 1.50 atm
• P2 = ?
Gay-Lussac Law Problem #1
A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if
the can is heated to 500.K?
• T1 = 298K
• T2 = 500.K
• P1 = 1.50 atm
• P2 = ?
P2 = 1.50 atm x 500 K
298 K= 2.52 atm
Big number on top makes
pressure get bigger
Temperature
increased so
pressuremustincrease
Gay-Lussac Law Problem #2
A can of spraypaint at 298K experiences an increase in pressure from 101.3 kPa to 425 kPa. What temperature would cause such a change?
Start with a data table
• T1 = 298 K
• T2 = ?
• P1 = 101.3 kPa
• P2 = 425 kPa
Gay-Lussac Law Problem #2
A can of spraypaint at 298K experiences an increase in pressure from 101.3 kPa to 425 kPa. What temperature would cause such a change?
• T1 = 298 K
• T2 = ?
• P1 = 101.3 kPa
• P2 = 425 kPa
T2 = 298 K x 425 kPa
101.3 kPa= 1250 K
Big number on top makes
Temperature get biggerPressureincreasedsotemperaturemustincrease
Avogadro’s Law
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Avogadro’s Law
DemonstrationBlowing up a balloon
As you add moles of gas, the volume increases
Avogadro’s Law
Volume is directly related to MolesIf moles of gas increases, volume increases
If moles of gas decreases, volume decreases(they move in the same direction)
You can see this relationship when temperature and pressure of gas are held constant
Avogadro’s Law
V = knWhere “k” is a constant of proportionality
Avogadro’s Law
Another way to write this:V1/n1 = V2/n2
Or V1n2 = V2n1
Avogadro’s Law Problem #1
A 12.2 L sample of gas contains 0.500 mol of O2 at 1.00 atm and 25.0oC. If 3.00 mol of O2 are
added, what will be the new volume?
Start with a data table
• V1 = 12.2 L
• V2 = ?
• n1 = 0.500 mol
• n2 = 3.500 mol
Avogadro’s Law Problem #1
A 12.2 L sample of gas contains 0.500 mol of O2 at 1.00 atm and 25.0oC. If 3.00 mol of O2 are
added, what will be the new volume?
• V1 = 12.2 L
• V2 = ?
• n1 = 0.500 mol
• n2 = 3.500 mol
V2 = 12.2 L x 3.500 mol
0.500 mol= 85.4 L
Bigger number on top
Makes volume get biggerMolesincreasedso volumemustincrease
Avogadro’s Law Problem #2
A 2.25 L balloon contains 0.475 mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain?
Start with a data table
• V1 = 2.25 L
• V2 = 6.50 L
• n1 = 0.475 mol
• n2 = ?
Avogadro’s Law Problem #2
A 2.25 L balloon contains 0.475 mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain?
• V1 = 2.25 L
• V2 = 6.50 L
• n1 = 0.475 mol
• n2 = ?
n2 = 0.475 mol x 6.50 L
2.25 L= 1.37 mol
Bigger number on top makes
Moles get bigger
Volumeincreased somoles must
increase