AP CHEM NOTES: GAS LAWS
gas and
H2O(g)
When Gases Are Collected Over Water:

Whenever gases are collected over water, there will be water vapor
in the gas sample as well as the desired gas. Therefore, the total
pressure inside the bottle will include the partial pressure of the gas
and the partial pressure of the water vapor. When doing gas law
problems, if you are asked to determine something for the “dry” gas,
you must correct the pressure at which it was collected to be just the
gas and not the water. The temperature and the volume collected
would stay the same for the gas since if we could physically remove
the water vapor they wouldn’t change for the gas itself.

Problem: A 425.0 mL sample of a gas is collected over water at 730.0 torr and 23°C. What would the volume of
the “dry” gas be at STP?
o Since this gas is collected over water, we must subtract the pressure of the water vapor at the
temperature it was collected from P1—see the table of variables below.
V1 = 425.0 mL
P1 = 730.0 torr – 21.07 torr = 708.9 torr
T1 = 23°C = 296 K
V2 = ?
P2 = 760 torr
T2 = 273.15 K
P1V1 P2 V2

T1
T2
V2 
P1V1T2 (708 .9 torr)(425 .0 mL)(273 .15 K)
=
= 366 mL
P2 T1
(760 torr)(296 K)
Avogadro’s Law:


States that equal volumes of gases at the same temperature and pressure contain the same number of particles. In
other words: for a gas at constant temperature and pressure, the volume is directly proportional to the number of
moles of the gas.
V1
V
V1 V2
V

V  n (n = moles) 
=k

= k and 2 = k

n1
n2
n1 n 2
n
Molar Volume of Gases:


molar volume → the volume of 1 mol of a gas
Using PV = nRT and solving for V at STP, the molar volume would be as follows: (remember: n = 1 mol)
L  atm

nRT (1 mol)(0.08206 K  mol )(273 .15 K)
V=
=
= 22.41 L
1 atm
P
It is generally accepted that the molar volume of all “ideal” gases is 22.4 L at STP. Of course, this will change with
temperature and pressure changes. And of course, no gas is truly “ideal” so there will be some discrepancy.
Density of Gases:

Problem: Calculate the density of CO2 at STP.
o To solve this, you know the molar volume (22.4 L) and molar mass (44.01 g) for CO 2 at STP.
density =

44.01 g
mass
=
= 1.96 g/L
22.4 L
volume
Problem: Calculate the density of CH4 at 25°C and 1.020 atm.
o To solve this, you know the molar mass (16.05 g) for CH4, but we must determine the molar volume at
this temperature and pressure before finding the density.
L  atm
nRT (1 mol)(0.08206 K  mol )(298 K)
V=
=
= 24.0 L
1.020 atm
P
d=
16.05 g
= 0.669 g/L
24.0 L
Another method to find gas density when not at STP, and probably the easier way is as follows:

Problem: The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar
mass of the gas.
gRT
g
dRT
gRT
o Since PV =
, then P =
. And since
= density, then P =
. Thus, the final equation:
MM
V
MM
V(MM)
MM =
MM =
(1.95 Lg )(0.08206
dRT
P
L  atm
K  mol
)(300 . K)
1.50 atm
= 32.0 g/mol
Pressure vs. Moles



Pressure and moles are directly proportional to each other when all other variables remain constant—doesn’t work
in a balloon since adding more gas increases the volume instead of the pressure (the pressure inside a balloon will
always be the same as atmospheric pressure as long as the balloon can still stretch).
Equation:
P1
P2
=
n1
n2
Since it doesn’t matter what gas is involved, you can find the mole ratio between two gases in a container if you
know their partial pressures by rearranging the equation to:
P1
n1
=
P2
n2
The following is a typical problem using this concept.

Problem: If a balloon has 15.0 g O2 and 15.0 g CO2 in it when the atmospheric pressure is 750.0 torr, what is the
partial pressure of the O2 in the balloon?
o Since you are given grams, you need to change it to moles.
15.0 g O2 = 0.469 mol O2
15.0 g CO2 = 0.341 mol CO2
total mol = 0.810 mol
PO2
Ptotal
=
nO2
ntotal
PO2 =
(Ptotal )(nO2 )
(750.0 torr)(0.469 mol O2 )
=
= 434 torr O2
ntotal
0.810 mol