Combined Gas Law
 The good news is that you don’t have to
remember all three gas laws! Since they are
all related to each other, we can combine
them into a single equation. BE SURE
YOU KNOW THIS EQUATION!
P1 V1
P2 V2
=
T1
No, it’s not related to R2D2
T2
Combined Gas Law
If you should only need one of the
other gas laws, you can cover up
the item that is constant and you
will get that gas law!
P1 V1
T1
P2 V2
T2
=
Boyle’s Law
Charles’ Law
Gay-Lussac’s
Law
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a
pressure of 0.800 atm and a temperature of 29°C.
What is the new temperature(°C) of the gas at a
volume of 90.0 mL and a pressure of 3.20 atm?
Set up Data Table
P1 = 0.800 atm V1 = 180 mL T1 = 302 K
P2 = 3.20 atm V2= 90 mL T2 = ??
*Make volume units the same!
Calculation
P1 = 0.800 atm
V1 = 180 mL
P2 = 3.20 atm
V2= 90 mL
P1 V1
T1 = 302 K
T2 = ??
P2 V2
=
T1
T2
T2 = P2 V2 T1
P1 V1
T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
T2 = 604 K - 273 = 331 °C
= 604 K
Learning Check
A gas has a volume of 675 mL at
35°C and 0.850 atm pressure. What
is the temperature in °C when the
gas has a volume of 0.315 L and a
pressure of 802 mm Hg?
Note: Volumes must be the same
unit and pressures must be the
1. List variables:
V1 = 675 mL = 0.675 L
T1 = 35°C = 308 K
P1 = 0.850 atm
V2 = .315 L
P2 = 802 mmHg = 1.06 atm
2.Decide on the appropriate gas law:
Everything’s changing, so Combined!
3. Rearrange to solve for unknown:
T2 =
(P2) (V2) (T1)
(P1) (V1)
= 179.2 K =
(178.4K)
Ideal gas law
 IF
WE COMBINE ALL OF THE
LAWS we’ve looked at
TOGETHER - INCLUDING
AVOGADRO’S LAW - WE GET:
PV=nRT
Ideal gas constant(R)
R
IS A CONSTANT THAT CONNECTS
THE 4 VARIABLES
R
IS DEPENDENT ON THE UNITS OF
THE VARIABLE FOR PRESSURE
– TEMP IS ALWAYS IN KELVIN
– VOLUME IS ALWAYS IN LITERS
– PRESSURE IS IN EITHER atm OR
mmHg OR kPa
 Because
of the different pressure
units we use there are 3 different
values for “R””
– IF PRESSURE IS
L•atm
R=.0821
GIVEN IN atm
mol•K
L•mmHg
R=62.4
mol•K
– IF PRESSURE R=8.314 L•kPa
IS GIVEN IN
mol•K
– IF PRESSURE
IS GIVEN IN
mmHg
kPa
Learning Check
Dinitrogen monoxide (N2O), laughing
gas, is used by dentists as an anesthetic.
If 2.86 mol of gas occupies a 20.0 L tank
at 23°C, what is the pressure (in atm) in
the tank in the dentist office?
Using Ideal gas law
1. List variables:
n = 2.86 moles
L•atm
R  0.0821
P ?
mol•K
 296K
V  20.0 L
T
2. Rearrange to solve for unknown:
PV = nRT
P = nRT
(2.86 mol)(.0821)(296K)
V
3, Plug & Chug:
20.0 L
=3.48 atm
GAS DIFFUSION AND
EFFUSION
 Diffusion is the
movement of
molecules to fill a
container
 Effusion is the
movement of
molecules through
a small hole into an
empty container.
Graham’s Law
Rates of effusion of
gases are inversely
proportional to the
square root of their
molar masses, at
constant temp. &
pressure.
Rate for A
Rate for B
M of B
M of A
Thomas Graham, 1805-1869.
Professor in Glasgow and London.
M = molar mass & Gas B is the heavier gas!
Graham’s
Law
Molecules effuse thru holes in a rubber
balloon – that’s the main reason they
get ‘whimpy’ after awhile!
They do this at a rate that is inversely
proportional to molar mass.
Therefore, He (4 g/mol) effuses more
rapidly than O2 (32 g/mol) at same T.
(It’s lighter!)
He
Lighter gases effuse faster than heavier ones
Graham’s
Law
 We can use the entire equation to calculate the
actual speed of gas particles, however…
 We will just use the square root side to
COMPARE rates of effusion (speeds)
 Ex. Compare the rates of effusion of oxygen gas
& hydrogen gas.
 1st – find their molar masses!
 O2 = 32.0 g/mol H2 = 2.0 g/mol
2nd -Put the heavier gas (Gas B) in the numerator!
Rate for A
4
Rate for B
M32ofgB
M
2.0ofgA
You’re not done yet!
Graham’s Law
 The number 4 is not much of a “comparison”!
 You must put your answer in sentence form!
 Try this:
 “_______ gas travels (or effuses) at a rate
___(Lighter
times gas)
faster than _________ gas.”
So, the
(#)answer is…
(heavier)
 “Hydrogen gas travels (or effuses) at a rate
4 times faster than oxygen gas.”
Graham’s Law
 You try it!
 Compare the rates of effusion of Ar and nitrogen gas (N2)
28.0 g/mol
Rate for A
1.19
Rate for B
39.9 g/mol
39.9
M
of B
M
of A
28.0
“Nitrogen gas travels (or effuses) at a
rate 1.19 times faster than argon gas.”
All of these gas laws work just ducky
assuming the gases are ‘ideal”
(Points with “no volume” & “no mutual
attraction”
Most of the time, gases conform to
ideal conditions.
So, when are gases not “ideal”?
Under conditions of low temperature
& high pressure
(force molecules close enough to
affect each other!)
Deviations from Ideal Gas Law
Real Molecules have
volumes and
attractive forces
between them.
Gases are not “Ideal”
under conditions of
high pressure & low
temperature which
bring particles close
enough together to
affect each other!
STOP HERE