Gas Problems?
How about gas LAW
problems?
Charles’ Law
Constants are
– # of particles
– Pressure
Variables are
– Temperature
– Volume
Direct relationship
Equation is: V1 =
T1
V2
T2
Try some problems
A gas has an initial temperature of 25°C and a
volume of 1.2 L. The gas is held under constant
pressure and allowed to expand to 2.5 L. What
is the temperature of the gas?
FIRST – convert your temperatures into K.
– K = 273 + °C
NEXT – Substitute your numbers into the
equation
V1 = 1.2 L
T1 = 298 K
V2 = 2.5 L
T2 = ?
1.2L = 2.5L
298K T2
Boyle’s Law
Constants are
– # of particles
– Temperature
Variables are
– Pressure
– Volume
Direct relationship
Equation is: P1V1 = P2V2
Try some problems
A gas has an initial pressure of 700 torr
and a volume of 1.9 L. The gas is held at
a constant temperature and allowed to
expand to 3.0 L. What is the final
pressure of the gas?
Substitute your numbers into the equation
P1 = 700 torr
V1 = 1.9 L
P2 = ?
V2 = 3.0 L
700 x 1.9 = P2 x 3.0
KINETIC MOLECULAR THEORY:
1. A gas is composed of particles considered to be hard
spheres with negligible volume.
2. The particles in a gas move rapidly in constant, random
motion. They travel in straight lines and move independently
of each other. The particles change direction when they
rebound from collisions with one another or with other
objects.
3. All collisions are perfectly elastic. Perfectly elastic means
that energy is transferred from one particle to another during
collisions, but the total energy remains constant
The average speed of oxygen molecules in air at 20 degrees
Is about 1660 km/hr or about 1000 miles/ hr!!!
Standard Temperature & Pressure
(STP)
There are standard values for temperature and
pressure
– The reason is that it creates consistent conditions for
scientists to compare their results.
Standard Temperature
– 0°C (K = 273 + °C  273K)
Standard Pressure: The following are equivalent
– 1 atm = 760 torr = 760 mm Hg = 101.3 kPa = 14.7 psi
Try some problems
A gas has an initial volume of 5.35 L at standard
temperature. The gas is held under constant
pressure and heated to 32°C. What is the final
volume of the gas?
FIRST – convert your temperatures into K.
– K = 273 + °C
NEXT – Substitute your numbers into the
equation
V1 = 5.35 L
T1 = 273 K
V2 = ?
T2 = 305 K
5.35L = V2
273K 305
MIXED PROBLEM PRACTICE
Do the following steps for EACH problem
– Identify the law needed to solve the problem
(Charle’s, Boyle’s, Gay-Lussac’s)
– Identify whether it’s a direct or inverse
relationship
– Solve the problem
A gas at 45°C and 1.2 atm is
cooled to a temperature of 5°C.
What is the new pressure of the
gas?
1.05 atm
45°C + 273 = 318 K
5°C + 273 = 278 K
1.2 atm = P2
318
278
Gay-Lussac’s Law
Direct Relationship
A 2.3L sample of gas under 500
torr of pressure is allowed to
expand with a final pressure of
2.3 atm. What is the final
volume?
.658L
2.3 atm x 760 torr = 1748 torr
1 atm
500 x 2.3 = 1748 x V2
Boyle’s Law
Inverse Relationship
A gas at a pressure of 700 torr is
pressurized to 1200 torr and
50°C. What was the initial
temperature of the gas in degrees
celcius?
-84.6°C
50°C + 273 = 323
500 = 1200
T1
323
T1 = 188.4 K – 273 = -84.6°C
Gay-Lussac’s Law
Direct Relationship
A syringe containing 5.5 mL of
gas at standard temperature is
allowed to expand to twice its
volume. What is the final
temperature?
546K
0°C + 273 = 273 K
V2 = 2 x 5.5 = 11 mL
5.5 =
273
11
T2
Charles’ Law
Direct Relationship
15L of a gas at standard pressure
is shrunk to a volume of 12.5L.
What is the new pressure, in torr?
912 torr
Standard pressure in torr = 760
760 x 15 = P2 x 12.5
Boyle’s Law
Inverse Relationship
The combined gas law:
If all three of these variables P, V and T
are changed in a new set of conditions
(for instance STP) and 5 of the six
variables are known, the sixth can be
calculated using….THE COMBINED
GAS LAW
P1V1 = P2V2
T1
T2