What is the ideal gas law? (article) | Gases

Learn how pressure, volume, temperature, and the amount of a gas are related to each other.

What is an ideal gas?

Gases are complicated. They're full of billions and billions of energetic gas molecules that can collide and possibly interact with each other. Since it's hard to exactly describe a real gas, people created the concept of an Ideal gas as an approximation that helps us model and predict the behavior of real gases. The term ideal gas refers to a hypothetical gas composed of molecules which follow a few rules:

Ideal gas molecules do not attract or repel each other. The only interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container.
The phrase elastic collision refers to a collision wherein no kinetic energy is converted to other forms of energy during the collision. In other words, kinetic energy can be exchanged between the colliding objects (e.g. molecules), but the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
A car crash where kinetic energy gets converted into heat energy and sound energy yielding two crumpled bumpers would not be elastic.
Ideal gas molecules themselves take up no volume. The gas takes up volume since the molecules expand into a large region of space, but the Ideal gas molecules are approximated as point particles that have no volume in and of themselves.

If this sounds too ideal to be true, you're right. There are no gases that are exactly ideal, but there are plenty of gases that are close enough that the concept of an ideal gas is an extremely useful approximation for many situations. In fact, for temperatures near room temperature and pressures near atmospheric pressure, many of the gases we care about are very nearly ideal.

If the pressure of the gas is too large (e.g. hundreds of times larger than atmospheric pressure), or the temperature is too low (e.g. $- 200 C$ ‍) there can be significant deviations from the ideal gas law. For more on non-ideal gases read this article.

What is the molar form of the ideal gas law?

The pressure, $P$ ‍, volume $V$ ‍, and temperature $T$ ‍ of an ideal gas are related by a simple formula called the ideal gas law. The simplicity of this relationship is a big reason why we typically treat gases as ideal, unless there is a good reason to do otherwise.

$P V = n R T$ ‍

Where $P$ ‍ is the pressure of the gas, $V$ ‍ is the volume taken up by the gas, $T$ ‍ is the temperature of the gas, $R$ ‍ is the gas constant, and $n$ ‍ is the number of moles of the gas.

$Moles n$ ‍ are a way to describe how many molecules are in a gas.

$1 mole$ ‍ is equal to $6.02 \times 10^{23} molecules$ ‍. This number is called Avogadro's constant $N_{A}$ ‍ and it's a way to convert from $moles$ ‍ to $molecules$ ‍ or vice versa.

$N_{A} = 6.02 \times 10^{23} \frac{molecules}{mole}$ ‍

For a typical room there is likely to be at least $1000 moles$ ‍ of gas molecules. That's an almost unthinkably high number of molecules since,

$1000 \times 6.02 \times 10^{23} molecules = 6.02 \times 10^{27} = six billion billion billion gas molecules$ ‍

**Units to use for $P V = n R T$ ‍**
$R = 8.31 \frac{J}{K \cdot m o l}$ ‍	$R = 0.082 \frac{L \cdot a t m}{K \cdot m o l}$ ‍
Pressure in $pascals P a$ ‍	Pressure in $atmospheres a t m$ ‍
Volume in $m^{3}$ ‍	volume in $liters L$ ‍
Temperature in $kelvin K$ ‍	Temperature in $kelvin K$ ‍

What is the molecular form of the ideal gas law?

If we want to use $N number of molecules$ ‍ instead of $n moles$ ‍, we can write the ideal gas law as,

$P V = N k_{B} T$ ‍

Where $P$ ‍ is the pressure of the gas, $V$ ‍ is the volume taken up by the gas, $T$ ‍ is the temperature of the gas, $N$ ‍ is the number of molecules in the gas, and $k_{B}$ ‍ is Boltzmann's constant,

$k_{B} = 1.38 \times 10^{- 23} \frac{J}{K}$ ‍

When using this form of the ideal gas law with Boltzmann's constant, we have to plug in pressure $P$ ‍ in units of $pascals Pa$ ‍, volume $V$ ‍ in $m^{3}$ ‍, and temperature $T$ ‍ in $kelvin K$ ‍. This information is summarized for convenience in the chart below.

**Units to use for $P V = N k_{B} T$ ‍**
$k_{B} = 1.38 \times 10^{- 23} \frac{J}{K}$ ‍
Pressure in $pascals P a$ ‍
Volume in $m^{3}$ ‍
Temperature in $kelvin K$ ‍

What is the proportional form of the ideal gas law?

There's another really useful way to write the ideal gas law. If the number of moles $n$ ‍ (i.e. molecules $N$ ‍) of the gas doesn't change, then the quantity $n R$ ‍ and $N k_{B}$ ‍ are constant for a gas. This happens frequently since the gas under consideration is often in a sealed container. So, if we move the pressure, volume and temperature onto the same side of the ideal gas law we get,

$n R = N k_{B} = \frac{P V}{T} = constant$ ‍

This shows that, as long as the number of moles (i.e. molecules) of a gas remains the same, the quantity $\frac{P V}{T}$ ‍ is constant for a gas regardless of the process through which the gas is taken. In other words, if a gas starts in state $1$ ‍ (with some value of pressure $P_{1}$ ‍, volume $V_{1}$ ‍, and temperature $T_{1}$ ‍) and is altered to a state $2$ ‍ (with $P_{2}$ ‍, volume $V_{2}$ ‍, and temperature $T_{2}$ ‍), then regardless of the details of the process we know the following relationship holds.

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ ‍

This formula is particularly useful when describing an ideal gas that changes from one state to another. Since this formula does not use any gas constants, we can use whichever units we want, but we must be consistent between the two sides (e.g. if we use $m^{3}$ ‍ for $V_{1}$ ‍, we'll have to use $m^{3}$ ‍ for $V_{2}$ ‍). [Temperature, however, must be in Kelvins]

What do solved examples involving the ideal gas law look like?

Example 1: How many moles in an NBA basketball?

The air in a regulation NBA basketball has a pressure of $1.54 atm$ ‍ and the ball has a radius of $0.119 m$ ‍. Assume the temperature of the air inside the basketball is $25^{o} C$ ‍ (i.e. near room temperature).

a. Determine the number of moles of air inside an NBA basketball.

b. Determine the number of molecules of air inside an NBA basketball.

We'll solve by using the ideal gas law. To solve for the number of moles we'll use the molar form of the ideal gas law.

$P V = n R T (use the molar form of the ideal gas law)$ ‍

$n = \frac{P V}{R T} (solve for the number of moles)$ ‍

$n = \frac{P V}{(8.31 \frac{J}{K \cdot m o l}) T} (decide which gas constant we want to use)$ ‍

Given this choice of gas constant, we need to make sure we use the correct units for pressure ( $pascals$ ‍), volume ( $m^{3}$ ‍), and temperature ( $kelvin$ ‍).

Yes, we could have used the gas constant $R = 0.082 \frac{L \cdot a t m}{K \cdot m o l}$ ‍. We would have just had to be careful to plug in pressure in terms of $atmospheres$ ‍ and volume in terms of $liters$ ‍.

We can convert the pressure as follows,
$1.54 atm \times (\frac{1.013 \times 10^{5} Pa}{1 atm}) = 156, 000 Pa$ ‍.

And we can use the formula for the volume of a sphere $\frac{4}{3} π r^{3}$ ‍ to find the volume of the gas in the basketball.
$V = \frac{4}{3} π r^{3} = \frac{4}{3} π (0.119 m)^{3} = 0.00706 m^{3}$ ‍

The temperature $25^{o} C$ ‍ can be converted with,
$T_{K} = T_{C} + 273 K$ ‍. $T = 25^{o} C + 273 K = 298 K$ ‍.

Now we can plug these variables into our solved version of the molar ideal gas law to get,

$n = \frac{(156, 000 Pa) (0.00706 m^{3})}{(8.31 \frac{J}{K \cdot m o l}) (298 K)} (plug in correct units for this gas constant)$ ‍

$n = 0.445 moles$ ‍

Now to determine the number of air molecules $N$ ‍ in the basketball we can convert $moles$ ‍ into $molecules$ ‍.

$N = 0.445 moles \times (\frac{6.02 \times 10^{23} molecules}{1 mole}) = 2.68 \times 10^{23} molecules$ ‍

Alternatively, we could have solved this problems by using the molecular version of the ideal gas law with Boltzmann's constant to find the number of molecules first, and then converted to find the number of moles.

Example 2: Gas takes an ice bath

A gas in a sealed rigid canister starts at room temperature $T = 293 K$ ‍ and atmospheric pressure. The canister is then placed in an ice bath and allowed to cool to a temperature of $T = 255 K$ ‍.

Determine the pressure of the gas after reaching a temperature of $255 K .$ ‍

Since we know the temperature and pressure at one point, and are trying to relate it to the pressure at another point we'll use the proportional version of the ideal gas law. We can do this since the number of molecules in the sealed container is constant.

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} (start with the proportional version of the ideal gas law)$ ‍

$\frac{P_{1} V}{T_{1}} = \frac{P_{2} V}{T_{2}} (volume is the same before and after since the canister is rigid)$ ‍

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} (divide both sides by V)$ ‍

$P_{2} = T_{2} \frac{P_{1}}{T_{1}} (solve for the pressure P_{2})$ ‍

$P_{2} = (255 K) \frac{1 atm}{293 K} (plug in values for pressure and temperature)$ ‍

$P_{2} = 0.87 atm (calculate and celebrate)$ ‍

Notice that we plugged in the pressure in terms of $atmospheres$ ‍ and ended up with our pressure in terms of $atmospheres$ ‍. If we wanted our answer in terms of $pascals$ ‍ we could have plugged in our pressure in terms of $pascals$ ‍, or we can simply convert our answer to $pascals$ ‍ as follows,

$P_{2} = 0.87 atm \times (\frac{1.013 \times 10^{5} Pa}{1 atm}) = 88, 200 Pa (convert from atmospheres to pascals)$ ‍

What is the ideal gas law? (article) | Gases | Khan Academy (2024)