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 (1)
II. 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:
This gives the following relationship for the pressure:
. (2)
B) 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:
This gives the following relationship for the volume:
(3)
C) 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:
This gives the following form for the number of moles, n:
.
I.e., . (4)
III. 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 its density, the pressure it exerts and the temperature. However, we are often given the mass of the gas and the volume of the container it is in, instead of the density. Of course, we can calculate the density of the gas from the mass and volume, using d = m / V. However, rather than memorize specific formulas for specific circumstances, I will demonstrate that you can always use the Ideal Gas Law in its normal form, regardless of what information is given to us. For example, given the volume of the container and the mass of the gas contained within that volume, as well as the pressure and temperature, we can determine the molar mass of the gas by following the steps below:
1) As always, whenever possible, the first step in any problem should be to solve for the number of moles present of a substance. Since P, V, and T all have known values, we can simply use the Ideal Gas Law to solve for n, the number of moles of gas present in the container, i.e., using .
2) Now we know how many moles are present. Since we were told 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]:
(5)
Note how easy this was!
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:
. (6)
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):
. (7)
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 choose a convenient value for one of them and then use this value in order to solve for the other. [Be careful: you must use the same assumed value throughout the problem!]
In this case, let’s assume a value for V and solve for n, as usual. The most convenient choice is to use V = 1 liter, (exactly). [By assuming an exact value, we are not affecting the number of sig figs in the problem. The sig figs will be determined by the other data that was given.]
We can now rearrange Eq. (7) in order to solve for the number of moles present:
. (8)
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:
. (9)
[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, m, and a known value for the volume it is in, (we assumed a value of V = 1 liter), so we can now determine the density of the gas in g/L, using Eq. (6) – i.e., using d = m/V.
2) We can also determine the density algebraically, by starting with the Ideal Gas Law, as expressed in Eq. (7). Let’s rearrange Eq. (7) 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:
. (10)
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.,
(11)
[Don’t memorize this formula – use unit analysis. 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:
. (12)
Rearranging this equation to place all unknowns on one side and known values on the other, we get Eq. (10), [which is repeated below as Eq. (13)]:
. (13)
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. (13) 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. (13) 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:
. (14)
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:
(15)
2) We can also solve the problem algebraically. Eq. (13) 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:
(16)
The following example from the textbook shows how you can solve such problems without having to use anything other than the Ideal Gas Law in its normal form.
Example: [Example 5.9 in text]
A chemist synthesized a gaseous compound of chlorine and oxygen and finds its density is 7.71 g/L at 36 °C and 2.88 atm. Calculate its molar mass and determine its molecular formula.
The book uses a special formula to solve this problem. I will show below how we can use the above described methods to solve it using PV = nRT to determine the molar mass:
Method 1:
Since density is uniform, we can choose any size sample we want for the volume and use it to solve for the mass of gas that must be present. Let’s choose a sample with a volume given by V = 1 L, exactly. [Since this is an exact value, it has an infinite number of sig figs. Thus the other numerical data in the problem will determine how many sig figs are in our answer.] We can now use the density to convert our known volume of the gas into a mass of this gas:
V = 1 L ´ 7.71 g /1 L = 7.71 g.
Now consider PV = nRT. We have numerical values for all data except for the number of moles of gas in our 1 L sample. [As always, solve for the number of moles present as soon as possible!]
n = PV / RT = (2.88 atm) (1 L) / (0.0821 L-atm/mol-K) (309K)
n = 0.1135 moles
We now know the mass in our 1 L sample and the number of moles present in that same sample. Since the molar mass is the ratio of mass divided by moles, (its units are g/mol), we can now calculate the molar mass:
MW (units = g/mol) ® MW = mass/moles = 7.71 g / 0.1135 moles = 67.9 g/mol
Method 2:
Let’s use the Ideal Gas Law to determine the ratio of moles to volume – i.e., the ratio of the two unknowns in the Ideal Gas Law. [We are given values for P and T, but not for n or V.]
Next, we can use unit analysis to convert the density into units of g/mol, using the result given above for mol/L. [It tells us that 0.1135 mol of the gas occupy 1 Liter.]
Final Step: Determining Molecular Formula from Molar Mass:
Once we know the molar mass, we can use it to help us determine the molecular formula. Usually, we are given (or have determined) an empirical formula for the substance. However, in this case we are told only that the molecule contains only 2 elements – chlorine and oxygen. Thus, the molecular formula must have the form ClxOy, where x and y have integer values. Thus the molar mass must have the following form relative to the atomic masses of chlorine, (35.45), and oxygen, (16.00):
MW = 67.9 = (35.45) x + (16.00) y
First, x cannot be larger than 1. [If it were equal to 2, the contribution from chlorine to the molar mass would be 70.9, which is larger than the entire molar mass.] Thus, x = 1 is the only possible outcome. We now have only one integer to determine – the value of y:
67.9 = (35.45) (1) + (16.00) y → 32.4 = 16.00 y → y = 2
Finally, we can deduce that the molecular formula must be ClO2.
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):
(17)
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:
(18)
This means that the product of P times V will be the same for both experiments. I.e.,
(19)
This leads directly to Boyle’s Law:
. (20)
B) 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.
(21)
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:
(22)
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
(23)
This leads directly to the desired relationship between the two experiments:
. (24)
Another algebraic approach is to write down the form for the Ideal Gas Law for each experiment. In the example above, we would begin by writing the following:
and (25)
The underlined data are those which remain unchanged during experiments 1 and 2.
Next, divide the 2nd equation by the 1st one and then cancel out any data that appears identically in numerator and denominator:
. (26)
C) An Alternative Approach:
Since two of the parameters are unknown and are held constant during these experiments, one approach that can be used is to choose a convenient value for one of the two parameters and then use its value and the values given in Experiment 1 for the other data to determine a value for the other unknown parameter. The values for these constant parameters are then used with the data given for Experiment 2 to determine the required answer.