CHEMISTRY 115
THE GAS LAWS
I.
Introduction to this Handout:
In this handout, I will attempt to demonstrate that the only equation necessary to solve gas law
problems is the general equation:
PV = nRT
II.
(1)
Rearranging the Ideal Gas Law:
Be careful when you manipulate the equation algebraically!! Remember that you can perform any
operation you want to one side of an equation, as long as you perform the identical operation to the
other side.
A)
Solving for P:
If we want to solve for the pressure, P, given that we have known values for n, V, and T,
simply divide both sides of the equation by the volume, V:
PV nRT

V
V
This gives the following relationship for the pressure:
P
B)
nRT
.
V
(2)
Solving for V:
If we want to solve for the volume, V, given that we have known values for all the other
parameters, we simply divide both sides by P:
PV nRT

P
P
This gives the following relationship for the volume:
V
C)
nRT
P
(3)
Solving for n:
Remember my recommendations for solving problems: First, balance any chemical reaction
equations, if any. Next, always convert any information you can into moles!! If the
temperature, pressure and volume all have known values, then the only unknown in the Ideal
Gas Law will be the pressure, (or the partial pressure, in some cases). To solve for n, the
number of moles of gas present, simply divide both sides of the Ideal Gas Law by RT:
PV nRT

RT
RT
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This gives the following form for the number of moles, n:
PV
 n.
RT
I.e., n 
PV
.
RT
(4)
III) Using Partial Pressures:
The partial pressure of a gas is defined as the pressure that gas would exert if it were alone in the
container. Thus more than 1 gas occupies the same volume V, which is held at a temperature given
by T. The only things that distinguish one of the gases from the others are the partial pressure of
that gas and the number of moles of that gas that is present. From the definition, we can use Eq. (2)
to relate the partial pressure of gas A, PA to the number of moles of gas A that are in the container:
PA 
nA RT
.
V
(5)
Since all gases that are present in the same container will be in the same volume, V, and at the
same temperature, T, then we can rewrite Eq. (5) in order to separate the values that are identical
for all the gases, from those that can vary, (the partial pressures and numbers of moles for each):
 RT 
 RT 
 RT 
PA  nA 
 , PB  nB 
 , PC  nC 
,
 V 
 V 
 V 
(6)
Note that the factor given by (RT/V) will be the same for all the gases. We need only calculate it
once. Then, the partial pressures of each gas can be simply found by multiplying that factor by the
number of moles of that gas.
IV) Using the Ideal Gas Law to Determine MWs:
The book derives a special formula to use in order to determine the molecular weight of a gas from
the pressure a sample of known mass exerts in a particular volume and at a particular temperature.
Rather than try to remember another formula, simply follow the instructions below:
1)
As always, you should first calculate how many moles of a substance are present, when
you have enough data to do so. Since P, V, and T all have known values, simply use the
Ideal Gas Law to solve for n, the number of moles of gas present in the container, i.e.,
PV
use n 
.
RT
2)
Now we know how many moles are present. Since we were given the mass of the gas in
the container, m, we can now directly solve for the MW of the gas. [Recall always that
MW is a ratio of mass in grams to moles – i.e., it has units of g/mole]:
MW 
m
g

n mole
(7)
Note how easy this was!
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IV) Using the Ideal Gas Law to Determine Densities:
First, recall that the density of a substance is the ratio of the mass of a sample of that substance
divided by the volume it is contained in:
d
m
.
V
(8)
The book derives a special formula to use in order to determine the density of a gas from known
values of the following: P, T, and the MW of the gas. They derive a formula that involves all of
these values. Instead, let’s proceed as usual by looking at which values in the Ideal Gas Law are
known, and which are unknown. In the equation below, I have underlined all known parameters,
(including the gas constant, R):
PV  nRT .
(9)
Note that two parameters are unknown – n and V. There are two ways we can proceed from here:
1)
Whenever two parameters have unknown values, you can usually assume a value for one of
them and then solve for the other. [Be careful: you must use the same value you assumed
throughout the problem!]
In this case, let’s assume a value for V and solve for n, as usual. The easiest choice of value is
the value 1. I.e., let’s assume the volume is exactly 1 liter. [By assuming an exact value, we
are not affecting the number of sig figs in the problem. That is determined by the other data
that was given.]
We can now rearrange Eq. (9) and solve for the number of moles present:
n
P 1 liter 
RT
.
(10)
Since we know the molecular weight of the substance, we can now use it and the number of
moles to find the mass of gas in our 1 liter container:
m  n  MW .
(11)
[It is easiest to follow the units here, in order to avoid dividing when you should multiply,
etc. I.e., moles  g/mole  g.]
Now we have a known value for the mass of the gas and a known value for the volume it is
in, (we assumed a value of 1 liter), and can now determine the density of the gas in g/L, using
Eq. (8).
2)
We can also determine the density algebraically, by starting with the Ideal Gas Law, as
expressed in Eq. (9). Let’s rearrange Eq. (9) until all our unknowns are on the left-hand side
and all are known values are on the right-hand side, by dividing both sides by RT and also by
V. This gives the relationships shown below:
P
n

RT V

n
P

.
V RT
(12)
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Note that we are now able to determine a density in the form of moles/L. Multiplying our
result by the MW of the substance, (with units of g/mole), will give us the ratio we are
seeking, with units of g/L. I.e.,
n
d  MW   
V 
(13)
[I.e., g/mole  mole/L yields g/L – the density of the gas.]
V.
Determining MW from the Density of a Gas:
The book goes one step further and derives the relationship between MW and known values for the
density of a gas, the pressure it exerts and the temperature of the gas. Once again, it is not
necessary to memorize another formula! Let’s again note which parameters are known and which
unknown in the Ideal Gas Law, keeping in mind that our goal is to determine the MW of the gas:
PV  nRT .
(14)
Rearranging this equation to place all unknowns on one side and known values on the other, we get
Eq. (12), which I will repeat here:
n
P

.
V RT
(12)
As before, there are two different ways we can choose to solve this problem.
1)
First, we can choose any value we want for one of the unknown parameters in Eq. (12) and
then use the chosen value to determine the other one. The simplest choice is to choose V
identical to 1 liter. Then we can use Eq. (12) to solve for the number of moles of this gas that
would be present in a 1 liter sample of the gas.
Next, we can apply the same value we chose to determine the mass present in our 1 liter
sample, using the value we were given for the density of the gas:
d
m
m

 m.
V 1 liter 
(15)
Now we have determined the mass of our sample and the number of moles in the same
sample and can use those values to determine the MW of the gas:
MW 
2)
m
n
g 



 moles 
(16)
We can also solve the problem algebraically. Eq. (12) gives us the ratio of moles to the
volume, (in liters), using the known values of P and T. Next, we note that a numerical value
for the density has also been given to us, (in g/L). We can now use the value we determined
for the ratio of moles to liters with the density to solve for the MW, (with units of g/mole).
This is most easily accomplished using unit analysis:
MW (g/mole) 
d
L 
g
 
 n / V   L mole 
–4–
(17)
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VI. Boyle’s Law Problems, Charles’ Law Problems, etc.:
It is not necessary to memorize Boyle’s Law or Charles’ Law, etc., in order to solve problems that
involve them. The easiest way to proceed in any of these types of problems is to first isolate all
parameters that vary on one side of the equation and all those that remain constant on the other.
The easiest method is to underline all parameters being held constant first, and then rearrange the
equation, if necessary. I will illustrate a couple of types of problems each below:
A)
Boyle’s Law Problems:
These are problems where the pressure and volume of a gas is varying from one experiment
to the next, but the temperature and sample size remain unchanged. [Note: since the sample
of gas is the same in both experiments, the number of moles of gas is the same for both
experiments.] In the Ideal Gas equation written below, I have underlined the termperature and
numbers of moles of gas, (as well as the gas constant, R):
PV  n RT
(18)
The next step is to rearrange the equation in order to place all parameters that vary on one
side and all the parameters that remain constant on the other. Note that we already have the
desired form in this case.
Since everything on the right-hand side is held constant, we can rewrite the equation as
follows:
PV  constant
(19)
This means that the product of P times V will be the same for both experiments. I.e.,
PV
1 1  constant and PV
2 2  constant
(20)
This leads directly to Boyle’s Law:
PV
1 1  PV
2 2.
B)
(21)
Charles’ Law Problems or Any Other Similar Problem:
Alternatively, you might be told that the pressure is held constant while the volume and
temperature change or that the volume is held constant while the pressure and temperature
change. In either of these cases, the easiest way to solve them is to follow the same approach
we used for Boyle’s Law problems.
For example, suppose you are told that the pressure is held constant as the volume and
temperature change. The easiest way to proceed is to first write down the Ideal Gas Law,
underlining all parameters that remain constant – the pressure and number of moles of gas, in
this example.
PV  n RT
(22)
Next, we need to rearrange the equation in order to place all constants on one side and all
variables on the other. I.e., we will divide both sides by P, and will also divide both sides by
T, as shown below:
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V nR

 constant
T
P
(23)
Since everything on the right-hand side is held constant, the entire right-hand side will be
equal to a constant value, for both experiments. Thus we can use Eq. (22) for each of the
experiments, giving
V1
V2
 constant and
 constant
T1
T2
(24)
This leads directly to the desired relationship between the two experiments:
V1 V2
 .
T1 T2
C)
(25)
An Alternative Approach:
Since two parameters are held constant during these experiments, one approach that can be
used is to simply choose a convenient value for one of the two parameters and use the data
given for Experiment 1 to determine a value for the other parameter. These two values would
then be used with the data for Experiment 2.
For example, suppose you were told that the sample size, (i.e., n), and the volume were held
constant as the pressure and temperature were changed. As before, I have underlined the
parameters that are held constant, but I have identified that the pressure and temperature from
Expt. 1 are being used:
P1 V  n RT1
(26)
The value for R is of course known, so the only unknowns in this equation are the volume
and the number of moles of gas. Choose a value for either of these parameters. Then solve for
the value the other must have, using the data from Expt. 1. Let’s choose V = 1 Liter, exactly.
We can now rearrange Eq. (26) to solve for the value of n:
nn
P1 V P1 1 L 

RT1
RT1
(27)
Now we can rewrite Eq. (26) using the data from Expt. 2, with our chosen value for V and
our now known value for n:
P2 V  n RT2
 P2 1 L  n RT2
(28)
One of the values for Expt. 2 will be given to us, the other can now be solved for. To
illustrate, suppose the question was phrased as follows: “A sample of gas was placed in a
container at 300 K. The pressure of the gas was found to be 1.00 atm. If the same sample in
the same container was heated to a temperature of 400 K, what would be its pressure at this
new pressure?” Let’s assume the volume of the container was 1 L, exactly. We can now
determine how many moles of gas must be present in the container, using the pressure and
temperature data from the initial experiment:
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n
1 atm 1 L 
 0.0821 L-atm/mol-K  300K 
 0.0406 moles
Now, we can use our data with the data for the 2nd experiment, and determine the pressure
when the temperature is raised to 400 K:
PV  nRT
 P2 1 L   0.0406 moles  0.0821 L-atm/mol-K  400K 
This gives the following value for the new pressure:
P2  1.33 atm
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