Gases

##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. [[☃ explanation 1]]

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. $-200 \text { C}$) 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.

$\Large PV=nRT$

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.
[[☃ explanation 1]]

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 $R=8.31 \dfrac{J}{K\cdot mol}$ then you must plug in the pressure $P$ in units of $\text{pascals } Pa$, volume $V$ in units of $m^3$, and temperature $T$ in units of $\text{kelvin } K$.

If you use the gas constant $R=0.082 \dfrac{L\cdot atm}{K \cdot mol}$ then you must plug in the pressure $P$ in units of $\text{atmospheres } atm$, volume $V$ in units of $\text{liters } L$, and temperature $T$ in units of $\text{kelvin } K$.

This information is summarized for convenience in the chart below.

|| Units to use for $PV=nRT$ ||
$R=8.31 \dfrac{J}{K\cdot mol}$| $R=0.082 \dfrac{L\cdot atm}{K \cdot mol}$

• | -
  Pressure in $\text{pascals } Pa$ | Pressure in $\text{atmospheres } atm$ |
  Volume in $m^3$ | volume in $\text{liters } L$ |
  Temperature in $\text{kelvin } K$| Temperature in $\text{kelvin } K$ |

[[☃ explanation 2]]

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

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

$\Large PV=Nk_BT$

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,

$k_B=1.38\times 10^{-23} \dfrac{J}{K}$

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 $PV=Nk_BT$ ||
$k_B=1.38\times 10^{-23} \dfrac{J}{K}$|

• | -
  Pressure in $\text{pascals } Pa$ |
  Volume in $m^3$ |
  Temperature in $\text{kelvin } K$|

##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 $nR$ and $Nk_B$ 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,

$nR=Nk_B=\dfrac{PV}{T}=\text{ constant}$

This shows that, as long as the number of moles (i.e. molecules) of a gas remains the same, the quantity $\dfrac{PV}{T}$ 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.

$\Large \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

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 $1.54 \text{ atm}$ and the ball has a radius of $0.119\text{ m}$. Assume the temperature of the air inside the basketball is $25^o \text{ C}$ (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.

$PV=nRT \quad \text{(use the molar form of the ideal gas law)}$

$n=\dfrac{PV}{RT} \quad \text{(solve for the number of moles)}$

$n=\dfrac{PV}{(8.31 \dfrac{J}{K\cdot mol})T} \quad \text{(decide which gas constant we want to use)}$

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 ($\text{kelvin}$). [[☃ explanation 1]]

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

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

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

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

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

$n=0.445 \text{ moles}$

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

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

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 $T=293 \text{ K}$ and atmospheric pressure. The canister is then placed in an ice bath and allowed to cool to a temperature of $T=255\text { K}$.

Determine the pressure of the gas after reaching a temperature of $255 \text{ K}.$

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.

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2} \quad\text{(start with the proportional version of the ideal gas law)}$

$\dfrac{P_1V}{T_1}=\dfrac{P_2V}{T_2} \quad\text{(volume is the same before and after since the canister is rigid)}$

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2} \quad\text{(divide both sides by }V)$

$P_2=T_2\dfrac{P_1}{T_1} \quad\text{(solve for the pressure }P_2)$

$P_2=(255\text{ K})\dfrac{1 \text{ atm}}{293 \text{ K}} \quad\text{(plug in values for pressure and temperature})$

$P_2=0.87 \text{ atm}\quad\text{(calculate and celebrate})$

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,

$P_2=0.87 \text{ atm} \times (\dfrac{1.013\times 10^5 \text{ Pa}}{1 \text{ atm}})=88,200 \text{ Pa}\quad\text{(convert from atmospheres to pascals})$

##What is the Maxwell-Boltzmann distribution?

The air molecules surrounding us are not all traveling at the same speed, even if the air is all at a single temperature. Some of the air molecules will be moving extremely fast, some will be moving with moderate speeds, and some of the air molecules will hardly be moving at all. Because of this, we can't ask questions like "What is the speed of an air molecule in a gas?" since a molecule in a gas could have any one of a huge number of possible speeds.

So instead of asking about any one particular gas molecule, we ask questions like, "What is the distribution of speeds in a gas at a certain temperature?" In the mid to late 1800s, James Clerk Maxwell and Ludwig Boltzmann figured out the answer to this question. Their result is referred to as the **Maxwell-Boltzmann distribution** , because it shows how the speeds of molecules are distributed for an ideal gas. The Maxwell-Boltzmann distribution is often represented with the following graph. [[☃ image 1]]

The y-axis of the Maxwell-Boltzmann graph can be thought of as giving the **number of molecules** per unit speed. So, if the graph is higher in a given region, it means that there are more gas molecules moving with those speeds. [[☃ explanation 1]]

Notice that the graph is not symmetrical. There is a longer "tail" on the high speed right end of the graph. The graph continues to the right to extremely large speeds, but to the left the graph must end at zero (since a molecule can't have a speed less than zero).

The actual mathematical equation for the Maxwell-Boltzmann distribution is a little intimidating and not typically needed for many introductory algebra classes. [[☃ explanation 4]]

##What does root-mean-square speed mean?

You might think that the speed located directly under the peak of the Maxwell-Boltzmann graph is the average speed of a molecule in the gas, but that's not true. The speed located directly under the peak is the $\redD{ \text{most probable speed } v_{p}}$, since it is the speed that is most likely to be found for a molecule in a gas.

[[☃ image 1]]

The $\blueD{ \text{average speed } v_{avg}}$ of a molecule in the gas is actually located a bit to the right of the peak. The reason the average speed is located to the right of the peak is due to the longer "tail" on the right side of the Maxwell-Boltzmann distribution graph. This longer tail pulls the average speed slightly to the right of the peak of the graph.

Another useful quantity is known as the $\greenD{ \text{root-mean-square speed } v_{rms}}$. This quantity is interesting because the definition is hidden in the name itself. The *root-mean-square* speed is the *square root* of the *mean* of the *squares* of the velocities. *Mean* is just another word for *average* here. We can write the root-mean-square speed mathematically as,

$v_{rms}=\sqrt {\dfrac{1}{N}(v_1^2+v_2^2+v_3^2+...)}$

It might seem like this technique of finding an average value is unnecessarily complicated since we squared all the velocities, only to later take a square root. You might wonder, ""Why not just take an average of the velocities?" But remember that velocity is a vector and has a direction. The average gas molecule velocity is zero, since there are just as many gas molecules going right (+ velocity) as there are going left (- velocity). This is why we square the velocities first, making them all positive. This ensures that taking the mean (i.e. average value) will not give us zero. Physicists use this trick often to find average values over quantities that can take positive and negative values (e.g. voltage and current in an alternating current circuit). [[☃ explanation 1]]

It should be noted that all three of these quantities ($\redD{v_{p}}$, $\blueD{v_{avg}}$, and $\greenD{v_{rms}}$) are quite large, even for a gas at room temperature. For example, Neon gas at room temperature ($293\text{ K}$) has a most probable speed, average speed, and root-mean-square speed of about,

$\redD{v_p}=491\dfrac{\text m}{\text s}\qquad $ (or $1100 \dfrac{\text{mi}}{\text{hr}})$

$\blueD{v_{avg}}=554\dfrac{\text m}{\text s}\quad $ (or $1240 \dfrac{\text{mi}}{\text{hr}})$

$\greenD{v_{rms}}=602\dfrac{\text m}{\text s}\qquad $ (or $1350\dfrac{\text{mi}}{\text{hr}})$

##What does the area under a Maxwell-Boltzmann distribution represent?

The y-axis of the Maxwell-Boltzmann distribution graph gives the **number of molecules** per unit speed. The total area under the entire curve is equal to the total number of molecules in the gas.

[[☃ image 2]]

If we heat the gas to a higher temperature, the peak of the graph will shift to the right (since the average molecular speed will increase). As the graph shifts to the right, the height of the graph has to decrease in order to maintain the same total area under the curve. Similarly, as a gas cools to a lower temperature, the peak of the graph shifts to the left. As the graph shifts to the left, the height of the graph has to increase in order to maintain the same area under the curve. This can be seen in the curves below which represent a sample of gas (with a constant amount of molecules) at different temperatures.

[[☃ image 1]]

As the gas gets colder, the graph becomes taller and more narrow. Similarly, as the gas gets hotter the graph becomes shorter and wider. This is required for the area under the curve (i.e. total number of molecules) to stay constant.

If molecules enter the sample, the total area under the curve would increase. Similarly, if molecules were to leave the sample, the total area under the curve would decrease.

##What do solved examples involving the Maxwell-Boltzmann distribution look like?

###Example 1: Cooling a gas

A gas of diatomic nitrogen is in a sealed container. The sealed container is then placed in an ice bath and reaches a lower equilibrium temperature with the ice bath.

**What happens to the following quantities as the gas cools? (select two correct statements)**

[[☃ graded-group 1]]

[[☃ explanation 1]]

###Example 2: Change in the gas

A gas has a distribution of speeds that looks like the following.

[[☃ image 2]]

**Which one of the following series of actions could cause the distribution graph to change from curve 1 to curve 2, as seen below?**

[[☃ image 3]]

[[☃ graded-group 2]]

[[☃ explanation 2]]

[[☃ explanation 1]]