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# What is the ideal gas law?

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:
1. 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.
2. 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. ) 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.
$PV=nRT$
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.
Perhaps the most confusing thing about using the ideal gas law is making sure we use the right units when plugging in numbers. If you use the gas constant $R=8.31\frac{J}{K\cdot mol}$ then you must plug in the pressure $P$ in units of , volume $V$ in units of ${m}^{3}$, and temperature $T$ in units of .
If you use the gas constant $R=0.082\frac{L\cdot atm}{K\cdot mol}$ then you must plug in the pressure $P$ in units of , volume $V$ in units of , and temperature $T$ in units of .
This information is summarized for convenience in the chart below.
Units to use for $PV=nRT$
$R=8.31\frac{J}{K\cdot mol}$$R=0.082\frac{L\cdot atm}{K\cdot mol}$
Pressure in Pressure in
Volume in ${m}^{3}$volume in
Temperature in Temperature in

## What is the molecular form of the ideal gas law?

If we want to use instead of , we can write the ideal gas law as,
$PV=N{k}_{B}T$
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,
${k}_{B}=1.38×{10}^{-23}\frac{J}{K}$
When using this form of the ideal gas law with Boltzmann's constant, we have to plug in pressure $P$ in units of $\text{pascals Pa}$, volume $V$ in ${\text{m}}^{3}$, and temperature $T$ in $\text{kelvin K}$. This information is summarized for convenience in the chart below.
Units to use for $PV=N{k}_{B}T$
${k}_{B}=1.38×{10}^{-23}\frac{J}{K}$
Pressure in
Volume in ${m}^{3}$
Temperature in

## 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 $nR$ 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,
This shows that, as long as the number of moles (i.e. molecules) of a gas remains the same, the quantity $\frac{PV}{T}$ 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 ${\text{m}}^{3}$ for ${V}_{1}$, we'll have to use ${\text{m}}^{3}$ 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 and the ball has a radius of . Assume the temperature of the air inside the basketball is (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.
$PV=nRT\phantom{\rule{1em}{0ex}}\text{(use the molar form of the ideal gas law)}$
$n=\frac{PV}{RT}\phantom{\rule{1em}{0ex}}\text{(solve for the number of moles)}$
$n=\frac{PV}{\left(8.31\frac{J}{K\cdot mol}\right)T}\phantom{\rule{1em}{0ex}}\text{(decide which gas constant we want to use)}$
Given this choice of gas constant, we need to make sure we use the correct units for pressure ($\text{pascals}$), volume (${\text{m}}^{3}$), and temperature ($\text{kelvin}$).
We can convert the pressure as follows,
.
And we can use the formula for the volume of a sphere $\frac{4}{3}\pi {r}^{3}$ to find the volume of the gas in the basketball.
The temperature can be converted with,
. .
Now we can plug these variables into our solved version of the molar ideal gas law to get,
Now to determine the number of air molecules $N$ in the basketball we can convert $\text{moles}$ into $\text{molecules}$.
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 and atmospheric pressure. The canister is then placed in an ice bath and allowed to cool to a temperature of .
Determine the pressure of the gas after reaching a temperature of
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.
$\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}\phantom{\rule{1em}{0ex}}\text{(start with the proportional version of the ideal gas law)}$
$\frac{{P}_{1}V}{{T}_{1}}=\frac{{P}_{2}V}{{T}_{2}}\phantom{\rule{1em}{0ex}}\text{(volume is the same before and after since the canister is rigid)}$
Notice that we plugged in the pressure in terms of $\text{atmospheres}$ and ended up with our pressure in terms of $\text{atmospheres}$. If we wanted our answer in terms of $\text{pascals}$ we could have plugged in our pressure in terms of $\text{pascals}$, or we can simply convert our answer to $\text{pascals}$ as follows,

## Want to join the conversation?

• Where do R, Na(Avogadro's Number) and k(Boltzmann's constant) come from and why? Is there an explanation for how they have been calculated? Thanks in advance. I wouldn't mind if the answer involved calculus.
• No calculus needed :-) Like most any constants, they are simply needed if there is always that same factor missing in an equation.
For example, in statistical mechanics you have a formula that is: S=k*ln(W). If you know S and W for at least two cases, then you might realize that, for both cases, S = ln(W) only if you multiply the right side by k constant
• When converting, why should we use Kelvin?
• One of the most important formulas in thermodynamics is P1 * V1 / T1= P2 * V2 / T2. However, if we used Celsius or Fahrenheit, what if, for example, the temperature was 0 degrees Celsius? Since you can't divide by 0, the formula would not work.
The Kelvin scale is made with 0 being equal to absolute zero, the coldest possible temperature, where the molecules stop moving completely. Therefore, you will never get a zero or negative temperature in your formula if you use Kelvin.
Kelvin is also the widely accepted temperature scale. If, for example, some people used Celsius and some people used Kelvin, we would all get different answers, so everyone uses Kelvin.
• In the section "What is the molar form of the ideal gas law?" and the first example, shouldn't the atm version of the ideal gas constant be 0.082 L*atm/mol*K instead of 0.082 L*atm/K? Or is there some reason the number of moles isn't included?
• You are right, the R actually does have the "mol" units, and it should read, as you correctly mentioned, L*atm/mol*K.
Nevertheless, the reason why this was probably excluded here is because the units of n are mol, and then if you combine n and R, the mol units will cancel.
• How do I know when a gas behaves like an ideal gas?
• most real gases do as long as the temperature is not too low and the pressure is not too high
• I know that Charles Law need constant moles and constant temperature; Boyles' law needs constant moles and constant temperature; so what does Avogadro's Law and Gay-Lussac's law need?
• Gay-Lussac's law has a constant volume. 'For a given volume of a gas, as the temperature increases, the pressure of the gas is directly proportional'. Volume is not a variable in his formula.

1 mol = 6.02 x 10^23
1 mol = 22.4 L
1 mol = molar mass in grams
Avogadro's law is mostly used for converting from one unit to another so the constant will depend on what you are converting.
What it does require is that you use the correct unit of measurement. For example, you have to use liters, you can't use milliliters.
(1 vote)
• how does the K.E transfer between two molecules (elastic collision) and no loss of energy ?
• In the "Units to use for PV=nRT" section, It says 1 liter=0.001 m​^3​​=1000 cm​^3.
This doesn't make sense to me. Isn't 1000 cm^3 = ​​10m^3 since c is a SI prefix for 10^-2?
10m^3 isn't equal to 0.001m^3. What am I missing? I'm sorry if this is a silly question.
(1 vote)
• Your math is a little bit wrong. Check it:
1 cm = (10^-2) m
(1 cm)^3 = (10^-2 m )^3
1 cm^3 = (10^-6) m^3
(1 cm^3)*1000 = (10^-6) m^3 *1000
1000 cm^3 = (10^-6)*(10^3) m^3
1000 cm^3 = 10^-3 m^3 = 0.001 m^3
• Where do we get the gas constant ,R, from?
Thanks
• Choose any gas, assuming its ideal. For example, 1 mole of Ar = 39.948 = 22.4 L at standard pressure ( 1 atm)

Just solve for R using the same formula, PV=nRT or in this case, R = PV/nT

Subsitue values into the equations :- R = (1atm) (22.4) / (1mole) (273K)

Solve: R = 0.0821 atm L / mol K

There are also alternate values for different units.
If R is needed in units of pressure (kPa) = 8.314 kPa L / mol K

If R is needed in units of pressure (mm Hg) = 62.396 mm Hg L / mol K

I would suggest always using the value of 0.0821 atm L / mol K for R unless stated otherwise.

Hope this helped!