When can I not use the ideal gas law?

The ideal gas law breaks down under extreme conditions, such as very high pressures or extremely low temperatures. Under these conditions, gas particles get forced close together, making their actual volume and intermolecular forces significant. Real gas laws, like the Van der Waals equation, are used instead.

Why must temperature be in Kelvin?

Kelvin is an absolute temperature scale, meaning it starts at absolute zero (where all atomic motion stops). If you used Celsius, you could plug a zero or a negative number into the equation, which would mathematically suggest a gas could have zero or negative volume. Kelvin ensures all values remain positive and proportional.

How do I know which R value to use?

Look strictly at the units of pressure given in the problem. If it is in atm, use 0.08206. If it is in kPa, use 8.314. If it is in mmHg or torr, use 62.36. As long as your volume is in liters and temperature is in Kelvin, the pressure unit dictates your R value.

How Do You Use the Ideal Gas Law?

You use the ideal gas law by plugging in three known properties of a gas to solve for the fourth unknown property. The equation is mathematically written as $PV = nRT$ . It allows you to calculate the pressure, volume, temperature, or amount (in moles) of a gas at a specific moment in time.

Think of the ideal gas law like a recipe for a perfect balloon. If you know how much air you pumped in, the room's temperature, and how hard the balloon's rubber is squeezing, this formula tells you exactly how big the balloon will get. It ties all the physical properties of a gas together into one neat equation.

What is the Ideal Gas Law?

The ideal gas law describes the behavior of a hypothetical 'ideal' gas where particles don't attract or repel each other and take up no space. The formula is $PV = nRT$ .

In this equation, $P$ stands for pressure, $V$ stands for volume, $n$ is the number of moles of gas, $T$ is the absolute temperature, and $R$ is the ideal gas constant. If you are given any three of these variables, you can rearrange the equation using basic algebra to solve for the missing piece.

Choosing the Right Gas Constant (R)

The gas constant, $R$ , is the glue that holds the equation together. Its value changes depending on the units of pressure and volume you are using. This is a crucial step that you must not skip when setting up your problem.

If your pressure is in atmospheres (atm) and volume is in liters (L), you must use $R = 0.08206$ L atm / (mol K). If your pressure is in kilopascals (kPa), you use $R = 8.314$ J / (mol K) or L kPa / (mol K). Always match your $R$ value to the units provided in your homework problem.

Where Students Slip Up

The most common mistake when using the ideal gas law is forgetting to convert the temperature to Kelvin. Celsius will not work. Because Celsius can be negative, using it in the formula could result in a negative volume or pressure, which is physically impossible. Always add 273.15 to your Celsius temperature to get Kelvin.

Another frequent error is unit mismatch. If your volume is given in milliliters (mL) but your chosen $R$ value uses liters (L), you must convert your volume to liters before plugging it into the equation. Taking a moment to write down all your given variables and their units will save you from these easy-to-make mistakes.

Worked through

Calculate the volume occupied by 2.50 moles of nitrogen gas at a temperature of 25.0 degrees Celsius and a pressure of 1.20 atm.

First, convert the temperature from Celsius to Kelvin: $T = 25.0 + 273.15 = 298.15$ K.

Second, identify your known variables: $n = 2.50$ moles, $P = 1.20$ atm, and $T = 298.15$ K. Because pressure is in atm, we use the gas constant $R = 0.08206$ L atm / (mol K).

Third, rearrange the ideal gas law to solve for volume: $V = \frac{nRT}{P}$ .

Finally, plug in the values: $V = \frac{2.50 imes 0.08206 imes 298.15}{1.20}$ .

Calculating the result gives $V = 50.97$ L. The volume occupied by the nitrogen gas is 51.0 L (rounded to three significant figures).

Questions students ask

Ask about this topic

Explain at

Where this comes from: OpenStax Chemistry 2e, Chapter 9: Gases · Khan Academy, Unit: Gases and kinetic molecular theory