The Ideal Gas Law may very well be the most powerful of the gas laws because it relates all four fundamental properties of a gas: its molar quantity, its temperature, its pressure, and the volume of the container it is in. Using the gas constant, \(R\), we see a fundamental relationship between these four quantities of a gas:
$$PV = nRT$$
The gas constant is a magnitude with some nasty-looking units: \(.0821 \; \frac{L \cdot atm}{mol \cdot K}\). While these units look intimidating, when the Ideal Gas Law is correctly manipulated, \(R\)'s units will cancel until you only have the unit that is expected when solving for one of the other four quantities.
One great problem-solving strategy is to solve the equation in terms of variables before substituting in any values.
Example 1: If we want to solve for volume, and the other four quantities are known, we can just solve the literal equation for \(V\) by dividing both sides of the Ideal Gas Equation by \(P\), giving
$$V = \frac{nRT}{P}$$
This is true wherever the Ideal Gas Law holds as the two equations are direct algebraic derivations of each other.
Example 2: If, by chance, we want to know the temperature of a gas when the other four values are known, we can solve the Ideal Gas Law for \(T\). In this case, we divide both sides of the equation by \(nR\):
$$T = \frac{PV}{nR}$$
Usually this type of manipulation, without any numbers, should be executed to make the calculations performed afterward much easier (and it will save you some writing!). In the next section we will begin to introduce specific values for these quantities to solve more specific, less generic problems.
With specific values given for any three of the four fundamental gas properties, we can solve for the fourth using the Ideal Gas Law, by the method described above. Be cautioned that temperatures must always be converted to Kelvins if they are not in that unit already!
Example 3: A \(3.2\)-mole sample of nitrogen trioxide gas has a temperature of \(655\) Kelvins. It is in a container of volume \(5.50 \; L\). What is the pressure being applied to the gas?
Solution: We are given \(n\), \(T\), and \(V\), so we can solve directly for \(P\) using the Ideal Gas Law. Divide both sides of the Ideal Gas Equation by \(V\):
$$P = \frac{nRT}{V}$$
All given quantities are in appropriate units, so we can substitute immediately and find \(P\):
$$P = \frac{3.2 \; mol \cdot .0821 \; \frac{L \cdot atm}{mol \cdot K} \cdot 655 \; K}{5.50 \; L} = 31 \; atm$$
Example 4: A \(5.9\)-mole sample of boron trifluoride gas has a temperature of \(23 \; C\) and is at SP (stands for "standard pressure," or \(1\) atmosphere). If the container the gas is in is a perfect sphere, what is its radius?
Solution: We are given \(n\), \(T\), and \(P\), so we can find \(V\), which is necessary to find the sphere's radius. We will use the Ideal Gas Law to solve for \(V\) and then the equation for the volume of a sphere to solve for \(r\) (the radius of the sphere).
Before proceeding, note that our temperature is given in Celsius and must be converted to Kelvins:
$$23 \; C = 23 + 273 \; K = 296 \; K$$
We just add \(273\) to the magnitude to convert from Celsius to Kelvins. According to the Ideal Gas Law,
$$PV = nRT$$
We can solve for \(V\) by dividing both sides of the equation by \(P\):
$$V = \frac{nRT}{P}$$
Furthermore, the equation for the volume of a sphere is
$$V = \frac{4}{3}\pi r^3$$
We do not even have to solve for \(V\) since we can combine the two equations together before any values are plugged in. Substitute the volume's expression in terms of the sphere's radius into the manipulated Ideal Gas Equation:
$$\frac{4}{3}\pi r^3 = \frac{nRT}{P}$$
Be sure you understand what we just did--we had a system of equations, and performed a substitution to get rid of an unneeded variable.
We can now solve for \(r\). Multiply both sides of the equation by \(\frac{3}{4\pi}\):
$$r^3 = \frac{3nRT}{4\pi P}$$
Now take the cube root of both sides of the equation:
$$r = \sqrt{\frac{3nRT}{4\pi P}}$$
Having \(n\), \(T\), and \(P\) in acceptable units, we substitute in every variable (and \(R\)) that we know, letting us solve for the radius of the sphere:
$$r = \sqrt{\frac{3 \cdot 5.9 \; mol \cdot .0821 \; \frac{L \cdot atm}{mol \cdot K} \cdot 296 \; K}{4\pi \cdot 1.0 \; atm}} = 7.0 \; L^{\frac{1}{3}}$$
We are technically done, but these units are confusing. It would be best to convert them to centimeters. To do that, note that
$$1000 \; mL = 1 \; L$$
Taking the cube root of this unit conversion gives
$$10 \; mL^{\frac{1}{3}} = 1 \; L^{\frac{1}{3}}$$
Furthermore, we know that \(1 \; mL = 1 \; cm^3\), so substitute that into the left-hand side:
$$10 \; (cm^3)^{\frac{1}{3}} = 1 \; L^{\frac{1}{3}} \Rightarrow$$
$$10 \; cm = 1 \; L^{\frac{1}{3}}$$
This is our new unit conversion, which we can apply to our radius:
$$7.0 \; L^{\frac{1}{3}} \cdot \frac{10 \; cm}{1 \; L^{\frac{1}{3}}} = 70. \; cm$$
Example 5: We have a sample of ozone gas in a \(45.0 \; cm^3\) container at STP (stands for "standard temperature and pressure," \(273 \; K\) and \(1 \; atm\)). What is the mass of the gas sample?
Solution: We are given \(V\), \(T\), and \(P\). We are asked for the mass of the gas, which can be found with a combination of knowing \(n\) and ozone's molar mass. Let's find the molar mass now of this chemical, with a chemical formula of \(O_3\) (I suggest memorizing this one):
$$3(16.0) = 48.0 \; \frac{g}{mol}$$
The gas constant's volume unit is liters, so we should also convert our given container volume to liters:
$$45.0 \; cm^3 \cdot \frac{1 \; mL}{1 \; cm^3} \cdot \frac{1 \; L}{1000 \; mL} = .0450 \; L$$
Now we can use the Ideal Gas Law to solve for \(n\), the moles of gas:
$$PV = nRT \Rightarrow n = \frac{PV}{RT}$$
We know all of the quantities on the right-hand side of the equation, so begin substituting:
$$n = \frac{1.0 \; atm \cdot .0450 \; L}{.0821 \; \frac{L \cdot atm}{mol \cdot K} \cdot 273 \; K} = .0020 \; mol \; O_3$$
We already calculated the molar mass of ozone, so we can use that as our final conversion factor from moles of gas to grams of gas:
$$.0020 \; mol O_3 \cdot \frac{48.0 \; g \; O_3}{1 \; mol \; O_3} = .096 \; g \; O_3$$
These three problems give a comprehensive overview of how the Ideal Gas Law can be used to solve quantitative problems.
The other gas laws can also be derived from the Ideal Gas Law, but it requires a bit of creativity. Let's return to our equation:
$$PV = nRT$$
We can divide both sides of the equation by \(nT\) to get all variables on one side of the equation:
$$\frac{PV}{nT} = R$$
Now, let's say the four fundamental properties of gases' values refer to some specific point in time, so label all of them with a \(1\) subscript:
$$\frac{P_1V_1}{n_1T_1} = R$$
Let's say we are working with the gas at some other point in time. To denote the fundamental quantities at this new point in time, give all of the variables subscripts of \(2\). By the Ideal Gas Law (which applies regardless of the time elapsed):
$$P_2V_2 = n_2RT_2$$
We can solve this equation for \(R\) as well:
$$R = \frac{P_2V_2}{n_2T_2}$$
Note that \(R\) is a constant so the subscripts are irrelevant. In fact, since we are working with the same \(R\), we know that for any two points in time:
$$\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$$
Here is where the derivation comes into play. All of the other gas laws assume that one or more of these quantities is a constant, so it remains unchanged over time.
Example 6: The Combined Gas Law holds when the moles of gas is kept constant, so \(n_1 = n_2\), and we can cancel them out in the above equation:
$$\frac{P_1V_1}{nT_1} = \frac{P_2V_2}{nT_2} \Rightarrow$$
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
If you know the other Gas Laws at this point, see if you can derive them from \(\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}\). If you do not know them yet, experiment with holding other combinations of \(P\), \(V\), \(n\), and \(T\) constant to see what you can come up with. You can check your answers with the other resources provided on the website.