xaktly | Chemistry

Gas laws

Understanding the behavior of gases in containers

In the early days of the development of chemistry, a few key researchers performed enough experiments on gases to be able to determine some empirical patterns in bulk behavior. They derived a number of "gas laws," including the ideal gas law, which we use to model most gases in most situations.

Much later, researchers in statistical mechanics came at these laws from the microscopic point of view, that is, by focusing on the behavior of small ensembles of atoms and molecules and building up the properties of gases from there.

You can get some sense of that by working through the section on the kinetic molecular theory of gases.

While the gas laws discussed in this section are very useful, it's important to realize that they were developed under a number of assumptions that break down at high pressure, when gas atoms and molecules exert significant attractive forces on one another, and in a few other special circumstances. When that does happen, we have other ways to deal with predicting the behavior of those gases.



An empirical rule or law is one that is based on experiment and observation instead of pure mathematical logic. Empirical observations inform theoretical investigations, and theory invites experiment in order to confirm, disprove or improve the theory.

1. Boyle's law

Boyle's law says that the product of pressure and volume of an ideal gas, all other things (number of moles and temperature) being constant, is always constant

$$PV = \text{constant}$$

Another way to say that is that if a gas system in state 1 is changed to state 2 by adjusting the pressure or volume, then the product (PV) is the same in both:

$$P_1 V_1 = P_2 V_2$$

You can picture it like this. Consider a gas in a cubic container with dimensions L × L × L. With a fixed number of gas particles, there will be a fixed average number of collisions with the walls in a given time.

Now if we shrink that container to half its size, the crowding of the particles increases the rate of collisions with the walls by a factor of two, thus doubling the pressure.

Here's what it looks like graphically. If we plot pressure (P) vs. volume (V), the graph is that of the rational function f(V) = c/V, where c is just a proportionality constant. It's there because we can't say for sure that P = 1/V.

Below are a few examples of how Boyle's law can be used.

Example 1

1 liter of gas at 50,000 Pa of pressure is compressed by moving a piston into the container to decrease the volume to 0.625 liters. Calculate the pressure after the compression.


Pascals (Pa)

The Pascal (Pa) is the SI unit of pressure.

1 Pa = 1 N·m-2 = 1 Kg·m-1s-2

Solution: Here's a picture of the situation before and after (a picture is often a great way to organize your work). This piston arrangement is a nice way to visualize gas-volume changes:

If we call the pressure and volume before the compression P1 and V1, and those after P2 & V2, then we can rearrange Boyle's law to solve for what's missing, P2:

$$P_2 = \frac{P_1 V_1}{V_2}$$

Then it's just a matter of plugging in what we know:

$$P_2 = \frac{50 \, KPa \cdot 1 \, L}{0.625 \, L}$$

... and calculating the result:

$$ \begin{align} P_2 &= 80 \; KPa \\ &= 80,000 \; Pa \end{align}$$

Example 2

The pressure relief valve on a cylinder fitted with a piston activates at a pressure of 7.5 atm. If the initial pressure and volume of the cylinder are P1 = 1.2 atm and V1 = 10 liters, to what volume can the gas be compressed before the relief valve (a safety device to prevent over-pressurization of the cylinder) activates?

Solution: We often engineer gas-handling systems with pressure-relief valves to protect from over-pressurization, which could cause failure of the system and could hurt someone. Here's the sketch of what's going on:

We're looking for V2, so we can rearrange Boyle's law to get it (I like to rearrange formulas before plugging in numbers – then I know I'm on the right track):

$$V_2 = \frac{P_1 V_1}{P_2}$$

Now plugging in what we know gives

$$V_2 = \frac{1.2 \; atm \cdot 10 \, L}{7.5 \, atm}$$

So the volume can be reduced to

$$V_2 = 1.6 \, L$$

before the relief valve opens.

Practice problems


The volume of a sample of CO2 gas, originally at a pressure of 1.2 atm, is tripled. Cacluate the new pressure of the gas in the container at this volume.


If the initial volume is V1, then the final volume is V2 = 3 V1. Now P1V1 = P2V2 (Boyles' law), so we have:

$$ \begin{align} P_1 V_1 &= P_2 V_2 \; \; \leftarrow V_2 = 3 V_1 \\ P_1 V_1 &= 3 \cdot P_2 V_1 \\ P_1 &= 3 P_2 \\ \\ P_2 &= \frac{P_1}{3} \\ \\ &= \frac{1.2 \, atm}{3} = \bf 0.4 \; atm \end{align}$$


A sample of gas in a cylinder with adjustable volume is fitted with a pressure gauge. If the pressure now reads 140 KPa, what will it read when the volume is reduced to 75% of its present volume?


If the initial volume is V1, then the final volume is V2 = 0.75 V1. Rerranging Boyles' law, we get:

$$ \begin{align} P_1 V_1 &= P_2 V_2 \; \; \leftarrow V_2 = 0.75 V_1 \\ P_1 V_1 &= 0.75 \cdot P_2 V_1 \\ P_1 &= 0.75 P_2 \\ \\ P_2 &= \frac{P_1}{0.75} \\ \\ &= \frac{140 \, KPa}{0.75} = \bf 187 \; KPa \end{align}$$


Two identical gas cylinders are coupled by a narrow tube (of negligible volume) with a valve that separates them. One cylinder contains argon at a pressure of 0.7 atm and the other is evacuated (at a vacuum, or P = 0). What will the total volume of the system be if the valve between the cylinders is opened?


Charles' law

Charles' law says that, if we keep the number of particles and the pressure constant, then the ratio of volume (V) to temperature (T) remains constant:

$$\frac{V}{T} = \text{constant}$$

If we hold the number of particles (n moles or N particles) and the pressure (P) constant, then comparing state 1 of a gas system with state 2 gives us Charles' law:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

If we rearrange the first expression to V = cT, where c is some constant of proportionality, and plot a graph of V vs. T, we see that the relationship is linear.

It makes sense, if volume changes, the temperature must change proportionally to keep the ratio constant.

Some examples of how to use Charles' law are given below.

Notice that when we use Charles law, we must use the Kelvin temperature. Kelvin is an absolute scale that can't go all the way to zero, thus we avoid having a zero denominator.

To avoid zero denominators and other problems, use Kelvin temperature when using gas laws.

Example 3

How much (in percent) does the volume of a fixed amount of gas, held at a constant pressure, expand when the temperature is raised from 30˚C to 37˚C

Solution: Here's the situation. It might seem like we don't have enough information, but we do. We don't need the initial volume because we're only trying to find how much any volume of gas would expand (at constant pressure) if the temperature is raised from 30˚ to 37˚C.

We rearrange Charles' law to find the final volume:

$$V_2 = \frac{T_2}{T_1} \cdot V_1$$

Now plug in the temperatures, letting x be the initial volume. The Celsius temperatures are fine because we're calculating a ratio of temperatures, and because the size of the Celsius degree is the same as the Kelvin.

$$V_2 = \frac{310 \, K}{303 \, K} \cdot x$$

The ratio of temperatures gives us our result.

$$V_2 = 1.023 \cdot x$$

So for an initial volume of 1 liter, the final volume would be 1.023 liters, for a 2.3% change in volume.

Example 4

If a gas is expanded to twice its original volume while keeping the pressure constant, by how much, and in which direction (warmer / cooler) must the temperature change?

Solution: Here is the picture, gas expanding in a cylinder with a movable piston:

In this case it seems like we have even less information to go on, but it's still OK. Take a look.

Charles' law can be rearranged to give the final temperature like this:

$$T_2 = \frac{V_2}{V_1} \cdot T_1$$

Then we can plug in the unknown volumes, but their ratio has to be ½ – that's all that's needed.

$$T_2 = \frac{x \, L}{2x \, L} \cdot T_1$$

So the final temperature will be half of the initial temperature, whatever that was.

$$T_2 = 0.5 \cdot T_1$$

The cooling of gases when they expand is the basic idea behind refrigeration and air conditioning.

Practice problems


A gas cylinder is fitted with a piston that maintains constant pressure of the helium gas inside. If the current temperature and volume are 25˚C and 1 liter, respectively, to what volume will the gas expand if the temperature is raised to 35˚C ?



A balloon is filled with air (mostly N2) to a volume of 2.2 liters at room temperature (25˚C). Calculate the new volume of the balloon if it is cooled to 10˚C.



A container holds 100 mL of N2 gas at 25˚C and a pressure of 744 torr. What will the volume be if the pressure is held constant and the temperature is increased to 35˚C ?


Gay-Lussac's law

The Gay-Lussac law of gases says that if we hold volume and number of particles (or moles) of gas constant, then the ratio of pressure to temperature will be constant:

$$\frac{P}{T} = \text{constant}$$

For a gas in state 1, with P1 and T1, we can make this statement, analogous to Charles' law:

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

We can rationalize the Gay-Lussac law by considering what happens to gas atoms or molecules when we raise the temperature.

As T increases, the average velocity of particles increases, therefore collisions with the container walls exert more force on the walls, and the collisions are more frequent. That all adds up to higher pressure.

As for Charles' law, this relationship is linear. We would expect a graph of P vs. T to be linear, with a slope equal to the proportionality constant.

Note: As with all gas laws, it is important to use the Kelvin temperature scale, an absolute scale with no zero, so we avoid expressions like this "blowing up" when the denominator goes to zero.

Example 5

Calculate the pressure drop we would expect to see if a quantity of a gas is is cooled from 42˚C to 20˚C at constant volume.

Solution: Here's the picture for this problem. Remember that we're forcing volume to remain constant, so we're not letting the piston move in this case.

The Gay-Lussac law can be rearranged to solve for the final pressure, P2, like this:

$$P_2 = \frac{T_2}{T_1} \cdot P_1$$

We don't need to know the initial pressure, just that the equation shows that it will be reduced (in this case) by the ratio of the final and initial temperatures:

$$P_2 = \frac{293 \, K}{315 \, K} \cdot x$$

The result is:

$$P_2 = 0.930 \cdot x$$

... so this reduction of temperature would result in the pressure dropping to about 93% of its original value. Notice that moving to the absolute temperature scale, which we must do, makes the 20-42˚C jump not as dramatic.

Practice problems


Calculate the pressure change when a gas in a container of constant volume, originally at a pressure of 1.2 atm, is heated from 25˚C to 30˚C



The pressure of N2 gas in a cylinder sitting in the sun is observed to rise from 1200 psi (pounds per square inch) to 1290 psi. In the morning the temperature of the air and the cylinder was 18˚C. What must the temperature of the cylinder be now, while it's in the sun ?



A gas in a metal cylinder is currently at a pressure of 50 torr (760 torr = 1 atm.) and at the temperature of liquid nitrogen (77 K). What will the pressure measure in the cylinder after it warms up to room temperature, 25˚C ?


Avogadro's law

Finally, Avogadro showed that if we hold pressure and temperature constant, the ratio of volume to number of moles of gas (or number of particles, N) is constant:

$$\frac{V}{n} = \text{constant}$$

Avogadro actually used this observation to develop the concept of the mole in chemistry.

Just like we did for the previous gas laws, we can rewrite Avogadro's law as an equation involving initial and final states of a gas – in this case as gas is added or removed from the container

$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$

Examples of using Avogadro's law are very similar to the previous examples involving ratios.

Molar gas volume

Avogadro showed that the volume of one mole of any ideal gas is 22.4 L at standard temperature and pressure (STP),



In chemistry and physics, STP stands for Standard Temperature & Pressure.

T = 0˚C = 273.15 K

P = 1 atm = 101.325 KPa = 760 torr

The ideal gas law

Finally, we can put all of these observations together to develop the ideal gas law. It says that the pressure-volume product, PV, is proportional to the product of the number of particles (or moles) and the temperature. The constant of proportionality really just helps us get the units right.

The ideal gas law can be expressed in two main ways, in terms of particles

$$PV = Nk_B T$$

where N is the number of particles, and in terms of moles

$$PV = nRT$$

where n is the number of moles, k = 1.381 x 10-23 J/K is the Boltzmann constant and R = 8.314 J/(mol·K) is the molar gas constant in SI units.

Often it's more convenient not to use the SI units of pressure and volume (Pascals and m3), and to use atmospheres (atm) and liters (L) instead. The gas constant R = 0.08314 L·atm/(mol·K) makes that easy.

Ideal gas law

The ideal gas law applies to a gas of particles for which we make some key assumlptions:

  1. Gas particles are small spheres, obey Newton's laws and undergo elastic collisions,
  2. Gas particles are very small compared to the volume of the gas,
  3. Intermolecular forces between particles are of negligible strength.

Ideal gas law:     PV = nRT,

where P = pressure, V = volume, n = number of moles of gas, R = gas constant & T = Kelvin temperature.

R = 8.314 m3·Pa·mol-1K-1   or   R = 0.0821 L·atm·mol-1K-1

For more (plus practice problems) on the ideal gas law, go here.

To see how the ideal gas law is derived from first principles, see the Kinetic-molecular theory of gases.

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