More Gas Laws
March 13, 2015
CHARLES’S LAW
Charles’s Law
Ø  French
physicist
(1746-1823)
Ø  First
person to fill a
balloon with
hydrogen gas
Ø  Made
the first solo
balloon flight
Charles’s Law
Ø 
Charles’s Law states at constant pressure the
volume of a fixed amount of gas is directly
proportional to its absolute temperature
Ø 
If volume INCREASES, temperature INCREASES
Ø 
If volume DECREASES, temperature DECREASES
V1 V2
=
T1 T2
Ø 
If the temperature is given in Celsius it must
first be converted to Kelvin
Ex. Problem #1: Charles’s Law
Ø  A
gas sample at 40oC occupies a volume
of 2.32 L. If the temperature is raised to
75oC what will the volume be assuming
the pressure remains constant?
 
GIVEN:
V1 = 2.32 L
T1 = 40+273= 313K
V2 = ??
T2 = 75+273= 348K
WORK:
V1 = V2
T1 T2
(2.32 L) = (x L)
313 K
348 K
Cross
Multiply
and Divide
Ex. Problem #2: Charles’s Law
Ø  A
gas at 75oC occupies a volume
of .45L. At what temperature will the
volume be .95L?
GIVEN:
V1 = .45 L
T1 = 75+273= 348K
V2 = .95 L
T2 = ???
WORK:
V1 = V2
T1 T2
(.45 L) = (.95 L)
348K
xK
Cross
Multiply
and Divide
GAY-LUSSAC’S LAW
Gay-Lussac’s Law
Ø  If
the number of moles
and volume are constant,
then pressure is directly
proportional to temp (K)
Ø 
If pressure INCREASES, then
temperature INCREASES
Ø 
If pressure DECREASES, then
temperature DECREASES
P1 P2
=
T1 T2
Ex. Problem #1: Gay-Lussac’s Law
Ø 
The pressure of a gas in a tank is 3.20
atm at 22oC. If the temperature rises to
60oC what will be the gas pressure in the
tank?
GIVEN:
WORK:
P1 = P2
P1 = 3.20 atm
T
T
1
2
T1 = 22+273= 295K
(3.20) = (?? atm)
P2 = ??
295 K
333 K
T2 = 60+273= 333K
Cross
Multiply
and Divide
Ex. Problem #2: Gay-Lussac’s Law
Ø  A
gas in a container has a pressure of 85
KPa at a temperature of 30oC. If the
pressure is increased to 150 KPa what is
the new temperature?
Cross
GIVEN:
WORK:
Multiply
P1 = P2
P1 = 85 KPa
and Divide
T
T
1
2
T1 = 30+273= 303K
(85KPa) = (150KPa)
P2 = 150 KPa
303 K
?? K
T2 = ???
COMBINED GAS LAW
The Combined Gas Law
Ø  Derived
from Boyle’s, Charles’s and
Gay-Lussac’s work
•  As pressure increased, volume decreased
•  As temperature increased, volume increased
•  As temperature increased, pressure increased
Ø 
Resulted in a combined formula to
calculate changes observed in a gas
due to changes in either temperature,
pressure or both
Combined Gas Law Equation
Ø  By
combining the equations for Boyle’s
Charles’s and Gay-Lussac’s Laws. We
derive the Combined Gas Law Equation
where:
P1V1 P2V2
=
T1
T2
Ex. Prob #1: Combined Gas Law
Ø  A
sample of a gas at 15°C and 2.0 atm
has a volume of 2 mL. What volume will
the gas occupy at 38°C and 1 atm?
Ø  Need to use the Combined Gas Law!
P1V1 P2V2
=
T1
T2
Ex. Prob #1: Combined Gas Law
Don’t forget to convert Celsius to Kelvin!!
P1= 2 atm
V1=2 mL
T1=15°C=288K
P2= 1 atm
V2=?
T2=38°C=311K
P1V1 P2V2
=
T1
T2
€
(2atm) (2mL) = (1atm) (V2)
288K
311K
Cross Multiply!
2×2×311 = 1×288 V2
1244 = 288 V2
V2 = 4.32 mL
Ex. Prob #2: Combined Gas Law
Ø 
A sample of nitrogen gas, N2, occupies 750 mL
at 75.00C under a pressure of 810 torr. What
volume would it occupy at STP?
(810 torr) (750mL) = (760 torr) (V2)
348K
273K
Cross Multiply!
810×750×273 = 760×348 V2
165847500 = 264480 V2
V2 = 627.1 mL