The Ideal Gas Law was first written in 1834 by Emil Clapeyron.

This is just one way to "derive" the Ideal Gas Law:

For a static sample of gas, we can write each of the six gas laws as follows:

PV = k_{1}

V / T = k_{2}

P / T = k_{3}

V / n = k_{4}

P / n = k_{5}

1 / nT = 1 / k_{6}

Note that the last law is written in reciprocal form. The subscripts on k indicate that six different values would be obtained.

When you multiply them all together, you get:

P^{3}V^{3}/ n^{3}T^{3}= k_{1}k_{2}k_{3}k_{4}k_{5}/ k_{6}

Let the cube root of k_{1}k_{2}k_{3}k_{4}k_{5} / k_{6} be called R.

The units work out:

k_{1}= L-atm

k_{2}= L / K

k_{3}= atm / K

k_{4}= L / mol

k_{5}= atm / mol

1 / k_{6}= 1 / mol-K

Each unit occurs three times and the cube root yields L-atm / mol-K, the correct units for R when used in a gas law context.

Consequently, we have:

PV / nT = R

or, more commonly:

PV = nRT

R is called the gas constant. Sometimes it is referred to as the universal gas constant. If you wind up taking enough chemistry, you will see it showing up over and over and over.

R's value can be determined many ways. This is just one way:

We will assume we have 1.000 mol of a gas at STP. The volume of this amount of gas under the conditions of STP is known to a high degree of precision. We will use the value of 22.414 L.

By the way, 22.414 L at STP has a name. It is called molar volume. It is the volume of ANY ideal gas at standard temperature and pressure.

Let's plug our numbers into the equation:

(1.000 atm) (22.414 L) = (1.000 mol) (R) (273.15 K)

Notice how atmospheres were used as well as the exact value for standard temperature.

Solving for R gives 0.08206 L atm / mol K, when rounded to four significant figures. This is usually enough. Remember the value. You'll need it for problem solving.

Notice the weird unit on R: say out loud "liter atmospheres per mole Kelvin."

This is not the only value of R that can exist. (Here's a whole bunch.) It depends on which units you select. Those of you that take more chemistry than high school level will meet up with 8.31447 Joules per mole Kelvin, but that's for another time. The ChemTeam will use the 0.08206 value in gas-related problems almost every time.

**Example #1:** A sample of dry gas weighing 2.1025 grams is found to occupy 2.850 L at 22.00 °C and 740.0 mmHg. How many moles of the gas are present?

**Solution:**

1) Notice that the units for pressure MUST be in atm., so the 740.0 mm Hg must be converted first.

740.0 mm Hg ÷ 760.0 mm Hg/atm = 0.973684 atm

2) Now, plug into the equation and solve for n:

(0.973684 atm) (2.850 L) = (n) (0.08206 L atm / mol K) (295.0 K) n = 0.1146329 mol

I will leave the answer with some extra digits and round off in the next example. It this was a problem where you would report the answer to the teacher, you would round off to four sig figs, 0.1146 mol.

**Example #2:** Using the problem above, what is the molar mass of the gas?

This is a very common use of this law and the odds are very good you will see this type of question on a test.

The key is to remember the units on molar mass: grams per mole.

We know from the problem statement that 2.1025 grams of the gas is involved and we also know how many moles that is.

We know that from doing the calculation above and getting 0.1146329 mol.

So all we have to do is divide the grams of gas by how many moles it is:

2.1025 g ÷ 0.1146329 mol = 18.34 g/mol

Let's go over those steps for using the Ideal Gas Law to calculate the molar mass of the gas:

1) You have to know the grams of gas involved. Usually the problem will just give you the value, but not always. You might have to calculate it.

2) You are going to have to calculate the moles of gas. Use PV = nRT and solve for n. Make sure to use L, atm and K.

3) Divide grams by moles and there's your answer.

**Example #3:** At STP, a 5.00 L flask filled with air has a mass of 543.251 g. The air in the flask is replaced with another gas and the mass of the flask is 566.107 g. The density of air is 1.29 g/L. What is the gas that replaced the air?

**Solution:**

1) Calculate mass of air in flask:

1.29 g/L times 5.00 L = 6.45 g

2) Calculate mass of flask:

543.251 g minus 6.45 g = 536.801 g

3) Calculate mass of unknown gas:

566.107 g minus 536.801 g = 29.306 g

4) Calculate moles of unknown gas:

PV = nRT(1.00 atm) (5.00 L) = (n) (0.08206) (273 K)

n = 0.22319 mol

5) Calculate molar mass of unknown gas:

29.306 g / 0.22319 mol = 131.3 g/molThe unknown gas could be xenon.

**Example #4:** (a) A gas has a temperature of 300. K and a pressure of 104 kPa. Find the volume occupied by 1.05 mol of this gas, assuming it is ideal. (b) Assuming the gas molecules can be approximated as small spheres of diameter 3.0 x 10¯^{10} m , determine the fraction of the volume found in part (a) that is occupied by the molecules.

**Solution:**

1) PV = nRT:

(104 kPa / 101.325 kPa/atm) (V) = (1.05 mol) (0.08206 L atm / mol K) (300. K)V = 25.184 L

To three sig figs, the answer is 25.2 L

2) PV = nRT with slightly different numbers:

PV = nRT whereP = 104000 Pa, n = 1.05 mol, R = 8.314 Pa m

^{3}/mol K, T = 300. K.V = nRT/p = (315) (8.31447 m

^{3}) / 104000 = 0.025183 m^{3}<--- note the 315, which comes from 300. x 1.05Note how only the final unit that survives in the answer is shown.

V = 0.0252 m

^{3}

3) The number of molecules in 1.05 moles:

(1.05 mol) (6.022 x 10^{23}molecules mol¯^{1}) = 6.3231 x 10^{23}molecules

4) The volume of an individual molecule is:

(4/3) (3.14159) (1.5 x 10¯^{10}m)^{3}= 1.4137 x 10¯^{29}m^{3}

5) The total volume of the molecules in m^{3}:

(6.3231 x 10^{23}molecules) (1.4137 x 10¯^{29}m^{3}/ molecule) = 0.00000894 m^{3}

6) The fraction of the gas volume that is occupied by matter is:

0.00000894 m^{3}/ 0.025183 m^{3}= 0.000355 or about 1/2817 of the total volume

**Example #5:** Of the following gases, which has density of 0.906 g/L at 315 K and 1.16 atm.

A) Ne; B) Ar; C) Kr; D) Xe; E) He

**Solution #1:**

PV = nRT(1.16 atm) (1.00 L) = (n) (0.08206 L atm / mol K) (315 K)

n = 0.044876 mol

0.906 g / 0.044876 mol = 20.19 g/mol <--- the molar mass of neon

**Solution #2:**

PV = nRTPV = (m/M)RT <--- where m = mass, M = molar mass

Since density = m/V, we have this (after substituting and rearranging):

PVM = mRT

M = mRT / PV

M = [(0.906 g) (0.08206 L atm / mol K) (315 K)] / [(1.00 L) (1.16 atm)] = 20.19 g/mol

Neon.

**Solution #3 (without PV = nRT):**

At STP , 1 mol of gas has a volume of 22.414 LUsing the Combined Gas Law, convert the volume to 315 K and 1.16 atm:

22.414 * (315/273) * (1.00/1.16) = 22.295 L

Now, you need to calculate densities until you get the correct answer:

Ne---> 20.180 g/22.295 L = 0.905 g/L <--- the correct answer

For comparison's sake, here is argon:

Ar ---> 39.95 g/22.295 L = 1.79 g/L <--- this is a wrong answer