Using the Ideal Gas Law: Calculate Pressure, Volume, Temperature, or Quantity of a Gas

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  • 0:07 The Ideal Gas Law
  • 0:59 Using the Ideal Gas Law
  • 3:02 Lesson Summary
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Lesson Transcript
Sarah Friedl

Sarah has two Master's, one in Zoology and one in GIS, a Bachelor's in Biology, and has taught college level Physical Science and Biology.

Expert Contributor
Dawn Mills

Dawn has taught chemistry and forensic courses at the college level for 9 years. She has a PhD in Chemistry and is an author of peer reviewed publications in chemistry.

In another lesson, you learned that the ideal gas law is expressed as PV = nRT. In this video lesson, we'll go one step further, examining how to rearrange the equation to solve for a missing variable when the others are known.

The Ideal Gas Law

In another lesson, you learned about ideal gases and the ideal gas equation. Ideal gases are just what they sound like - ideal. But since real gases behave similarly to ideal gases at normal temperatures and pressures, we can use the ideal gas equation to predict the behavior of real gases under these conditions.

First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volume in liters, 'n' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmospheres per moles Kelvin). Just like any equation, if we know three of those four variables (other than R, which we already know because it is a constant), we can rearrange the equation to calculate the unknown.

Using the Ideal Gas Law

Let's start with a very simple example to see how this works. Say we want to calculate the volume of 1 mole of gas at 273 K (which is the same as 0 °C) and 1 atmosphere of pressure. Here's what our equation looks like when we fill in the variables we do know:

1 atm * V = 1 mol * 0.0821 atm L / mol K * 273 K

If we want to find the volume (V), we simply rearrange the equation to get this variable by itself. We do this by dividing by the pressure, 1 atm (atmosphere). So, now our equation looks like this:

V = (1 mol * 0.0821 atm L / mol K * 273 K) / 1 atm

The moles cancel out, as do atmospheres and Kelvin. All we're left with in terms of units is liters, and then to get our volume, we simply do the math. Our final answer is 22.4 L. Make sense?

Let's try another example, this time solving a real-life example. Suppose you want to calculate the temperature of the gas in your bike tire. As long as you know the other variables, you can do this quite easily! In this case, the pressure is 1.14 atm, the volume of the tire is 5.00 L, and we have 0.225 moles of gas. So, our original equation looks like this:

1.14 atm * 5.00 L = 0.225 mol * 0.0821 atm L / mol K * T

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Additional Activities


You are exploring a new planet with a group of scientists to investigate the air and ground composition. All of the scientists are collecting a variety of samples to take back to the laboratory to study further. One of the samples you collected is an ideal gas that is unknown to exist on this planet. When you weighed the gas in your lab, it had a mass of 2 grams. At the time of collection, the temperature was 49 degrees C and the atmospheric pressure was 1006.24 mmHg. You were also able to determine that the sample of unknown gas occupies a volume of 0.5 L by using techniques in the laboratory. (The value for R is 0.082 L atm/K mol)


1. Convert the temperature to Kelvin.

2. Convert the pressure to atm.

3. Determine the moles (n) of the gas using the ideal gas law equation.


1. How many moles of ideal gas were recovered?

2. What is the molar mass of the gas? (Hint: molar mass is equal to the grams of ideal gas recovered divided by the moles of the ideal gas)

3. What is the likely identity of the gas (Hint: use the periodic table and molar mass to find the identity of the closest gas)


1. PV ÷ RT = n

P = 1006.24 * 0.00131 = 1.32 atm

V = 0.5 L

n = _____

R = 0.082 L atm/K mol

T = 49 + 273 = 322 K

Plug in the values from above.

n = (1.32 * 0.5) &#247 (0.082 * 322) = 0.025 mol

2. molar mass = 2 ÷ 0.025 = 80 g/mol

3. The gas would be Bromine as the molar mass of Bromine is 79.904 g/mol and has the closest molar mass.

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