More Gas Laws
FUNDAMENTALS
TEMPERATURE
TEMPERATURE
  Temperature
is a measure of the average
kinetic energy of the molecules of a
substance
  Kinetic Energy = ½ m × v2
•  Where m = mass and v = velocity
  Temperature
of a material is related to the
velocity of its particles
HEAT
  Heat
is energy transferred due to
differences in temperature
  The energy always flows from higher to
lower temperature
TEMPERATURE SCALES
  Celsius or Centigrade (0°C):
•  Based on the boiling and freezing points of water
•  Designed so that there would be 100 degrees
between them
  Kelvin
(K):
•  Based on absolute zero; the temperature where
all molecular motion stops
•  0 Kelvin = -273 oCelsius
Temperature Conversions
K = oC + 273
oC = K -273
STP
  Standard
temperature is 273K or 0 °C
  Standard pressure (1 atm)
  Together standard temperature and
pressure are abbreviated STP
•  Standard set of conditions for experimental
measurements
•  Allows comparisons to be made between sets of
data when working with gases
TEMP SAMPLE CONVERSIONS:
TRY THESE:
SOLUTIONS:
1. 
86 K to oC
1. 
-187 oC
2. 
6.23 oC to K
2. 
279.23 K
3. 
191 K to oC
3. 
-82 oC
4. 
9.18 oC to K
4. 
282.18 oC
5. 
894 K to oC
5. 
621 oC
NOW YOU TRY!!
 
If the sentence uses Celsius, rewrite it using
Kelvin, or vice versa.
1. 
It has to be 0°C or colder before it will
snow outside. 0 + 273 = 273 K
2. 
If your body temperature is 313K, you
probably have the flu. 313 - 273 = 40°C
3. 
I prefer my classroom to be 296K.
296 - 273 = 23°C
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
=
T 1 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 313 K occupies a volume
of 2.32 L. If the temperature is raised to
348 K what will the volume be assuming
the pressure remains constant?
GIVEN:
V1 = 2.3 L
T1 = 313K
V2 = ??
T2 = 348K
WORK:
Cross
Multiply
and Divide
Ex. Problem #2: Charles’s Law
Ø  A
gas at 348 K occupies a volume
of .45L. At what temperature will the
volume be .95L?
GIVEN:
V1 =
T1 =
V2 =
T2 =
WORK
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 295 K. If the temperature rises to
333 K what will be the gas pressure in the
tank?
GIVEN:
P1 = 3.20 atm
T1 = 295K
P2 = ??
T2 = 333K
WORK:
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 303 K. If the
pressure is increased to 150 KPa what is
the new temperature?
GIVEN:
WORK:
Cross
Multiply
and Divide