Combined Gas Law
The Combined Gas Law – does just that; it
combines Boyle’s Law (P1V1 = P2V2) and
Charles’ Law (V1/T1 = V2/T2).
Used when a pressure and temperature
change occur
Combined Gas Law
new volume = old volume x pressure ratio x K temperature ratio
Each ratio is considered independently…
– Ask each question separately
– Use “Old-New” table
Old
Pressure (P)
Volume (V)
Temp. (T) in
Kelvins
New
What happens to
the gas volume?
Combined Gas Law - Example
Calculate the volume of a gas at STP if
500 cm3 of the gas are collected at 27 oC
and 96.0 kPa.
STP = Standard temperature & pressure:
• 273 K (0 oC)
• 101.3 kPa
Combined Gas Law - Example
If 400 cm3 of oxygen is collected over
water at 20oC, and the atmospheric
pressure is 97,000 Pa, what is the volume
of the dry oxygen at STP?
STP = Standard temperature & pressure:
• 273 K (0 oC)
• 101.3 kPa
Gas Density
Density = mass ÷ volume
Since the volume of gas varies with pressure
(inversely) and temperature (directly), these
affect the density of a gas.
What happens to the density as the pressure on
a gas increases? Decreases?
What happens to the density as the temperature
on a gas increases? Decreases?
Gas Density
THEREFORE: The density varies directly
with the pressure, and it varies inversely
with the temperature.
VERY important that you know these
relationships!
Units for gas density
grams/dm3 or
grams/liter (1 dm3 = 1 liter)
Grams/cm3 would give very small numbers
for most gases.
Gas Density - Examples
What is the density of a gas which has a mass
of 4.50 g and occupies 2.50 dm3?
Solution:
4.50 g
= 1.80g/dm3
2.50 dm3
Gas Density Formula
P1
= P2
T1D1
T2D2
Gas Density - Examples
If the density of helium is 0.179 g/dm3 at STP (273 K
and 101.3 kPa), what is its density at 99.0 kPa and 27
oC?
How to solve…
New density = old density x pressure ratio (new/old) x temp. ratio (old/new)
(direct relationship)
Old
Pressure (P)
Temp (T)
Density (D)
New
(inverse relationship)
What happens to
the gas density?
Gas Density - Examples
Solution
New density = (0.179 g/dm3)x(99
kPa/101.3 kPa)x(273 K/300 K)
= 0.159 g/dm3
Graham’s Law
Only a few physical properties of gases depend on the
identity of the gas.
The rate of effusion for a gas is inversely
proportional to the square root of its molar
mass
Diffusion – Movement of one material
through another (high conc.
low conc.)
• Ex: cooking breakfast in the morning
Effusion – Flow of a gas through a small
opening.
• Ex: gas leaving a balloon, balloon deflates
Graham’s Law of Diffusion
Mathematically rate of effusion looks like
this:
Rate of effusion =
1
√molar mass
So, when comparing the rate of one gas to
another…
Graham’s Law of Diffusion restated
Rate Gas A =
Rate Gas B
Molar mass gas B
Molar Mass gas A
Graham’s Law of Diffusion Example
What is the molar mass of a gas that
diffuses 3 times faster than O2 under
similar conditions?
Molar
Mass
Rate
Gas A
???
3 times faster = 3m/s
Gas B = O2
32g/mol
1 times fast =
Rate Gas A
=
Molar mass gas B
Rate Gas B
Molar Mass gas A
1m/s
Graham’s Law of Diffusion Example
What is the molar mass of a gas that
diffuses 3 times faster than O2 under
similar conditions?