Ideal Gas Law Example – Unknown Gas Problem

Unknown Gas
Use the Ideal Gas Law equation to find the identity of an unknown gas.

The ideal gas law can be used to determine the identity of an unknown gas. If you are given the pressure, volume and temperature of the unknown gas, you can determine what the gas is likely to be made up of. This ideal gas law example problem shows the steps necessary to accomplish this task.

Gas Law Problem

A 276.58-g sample of X2(g) has a volume of 30.0 L at 3.2 atm and 27°C. What is element X?


The ideal gas law is expressed by the formula

PV = nRT

P = Pressure
V = Volume
n = number of moles of gas particles
T = Absolute Temperature in Kelvin
R is the gas constant.

The Gas Constant, R, while a constant, depends on the units used to measure pressure and volume. Here are a few values of R depending on the units.

R = 0.0821 liter·atm/mol·K
R = 8.3145 J/mol·K
R = 8.2057 m3·atm/mol·K
R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

The first step of any ideal gas law problem is to make sure your temperature is in the absolute temperature scale. This is a common source of error for a lot of students and it is best to get it out of the way as soon as you begin.

Our example has the temperature of 27°C. To convert this to Kelvin, use the formula

K = °C + 273

K = 27°C + 273
K = 300 K

Now let’s choose the value of the gas constant suitable for our example. The example uses Liters and atmospheres so the value of R we should use is

R = 0.0821 liter·atm/mol·K

Now we have all we need to use the ideal gas law to find the number of moles of our gas sample. Solving the equation for n yields

Ideal Gas Law math step 1

Plug in our values

Ideal Gas Law Math Step 2

n = 3.9 moles

We now know there are 3.9 moles of the unknown gas in the system. We also know these 3.9 moles has a mass of 276.58 grams. Now find how much one mole of the gas weighs.

Ideal Gas Law Math Step 4

molar mass of X2 = 70.9 grams/mol

X2 means our gas is diatomic or composted of two atoms of the element X. This means the atomic weight of X will be half the value of X2‘s molar mass.

atomic weight of X = ½(70.9 grams/mol)
atomic weight of X = 35.45 grams/mole

Looking on a periodic table, the element with atomic weight closest to 35.45 grams/mol is chlorine.


The identity of element X is chlorine.

The key points to watch for with this type of problem are the absolute temperature, units of the gas constant R, and the atomic mass of one atom of the unknown gas. Ideal gas law problems should always work with absolute scale temperatures, not relative temperatures like celsius or fahrenheit. The units on the gas constant should match the units you are working with, or else they will not cancel out. This is a easy error to avoid if you pay attention. This problem had a diatomic gas as the unknown. If we hadn’t remembered this step, we would have thought one mole of the gas had a mass of 70.9 grams and decided our gas was gallium (69.72 g/mol).