## Section21.1Ideal Gas Law

Gaseous phase is a rarefied state of matter. In this state the interaction between molecules is often weak. If a gas is dilute, that is, if the density is low, we can completely ignore inter-molecular interactions without losing much physics. The model gas with no interaction between molecules is called the ideal gas.

The thermodynamic state of a given amount of “ideal” gas, say containing $N$ molecules, is completely determined by its temperature $T\text{,}$ pressure $p\text{,}$ and volume $V\text{,}$ where temperature is in kelvin scale. These thermodynamic variables are not all independent, but are related by an equation of state.

$$p V = N k_B T,\label{eq-ideal-gas-law-physics}\tag{21.1.1}$$

where $k_B$ is called Boltzmann's constant that has the following value.

$$k_B = 1.38 \times 10^{-23} \text{ J}\,\text{K}^{-1}.\tag{21.1.2}$$

In Chemistry, one writes the equation of state in another form. In place of number of molecules $N \text{,}$ we can defines number of moles $n$ by diving $N$ by Avogadro number $N_A\text{.}$

$$n = \dfrac{N}{N_A},\tag{21.1.3}$$

where $N_A$ is the number of moecules in one mole.

$$N_A = 6.022\times 10^{23},\tag{21.1.4}$$

and in place of $k_B\text{,}$ we define another constant, called the universal gas constant, $R\text{,}$ by

$$R = N_A k_B = 8.31\text{ J}\,\text{K}^{-1}.\tag{21.1.5}$$

Often the unit of $R$ is given as $\text{ J}\,\text{K}^{-1}\, \text{mol}^{-1}$ to reflect the fact that $R$ is often multiplied by $n \text{,}$ the number of moles. It is not necessary to do that as long as we rememeber the relation of $R$ and $n$ to the total number of molecules $N$ and the Boltzmann constant $k_B\text{.}$

\begin{equation*} n R = Nk_B. \end{equation*}

In terms of $n$ and $R\text{,}$ the equation of state (21.1.1) takes the following form.

\begin{equation*} p V = N k_B T = \left(\dfrac{N}{N_A}\right) \left( N_Ak_B\right) T = n R T. \end{equation*}

That is,

$$pV = n R T.\tag{21.1.6}$$
###### Remark21.1.1.Units.

If you express $p$ in $\text{Pa}$ and $V$ in $\text{m}^3\text{,}$ the product $pV$ will be $\text{Joules}\text{.}$

\begin{equation*} 1\ \text{Pa} \times \text{m}^3 =1\ \dfrac{\text{N}}{\text{m}^2} \times \text{m}^3 =1\ \text{N}\,\text{m}= 1\ \text{J}. \end{equation*}

Sometimes, it is more convenient to express $p$ in $\text{atm}$ and $V$ in $\text{liter (L)}\text{.}$ In that case, we need to convert units with

\begin{equation*} 1\ \text{L}\times \text{atm} = 10^{-3}\text{m}^3\times 1.013\times 10^{5}\,\text{Pa} = 101.3\text{ J}. \end{equation*}

### Subsection21.1.1Specifying Thermodynamic State of Ideal Gas

Suppose we have a fixed quantity of a gas, say $n$ moles, and we wish to specify its thermodynamic state. The three thermodynamic variables that are well-defined in the equilibrium are volume $V\text{,}$ pressure $p\text{,}$ and temperature $T\text{.}$ A state of this system will have particular values of these three variables.

You need to independently measure only two of the variables, $(p,\ V,\ T)$ since you can use equation of state to find the third. For instance, if $p$ and $T$ are known for an ideal gas, then, we can immediately say that its volume must be

\begin{equation*} V = \dfrac{nRT}{p}. \end{equation*}

That means, we have three choices for stating a thermodynamic state: $(p,\ T)\text{,}$ $(p,\ V)\text{,}$ or $(T,\ V)\text{,}$ whichever is convenient.

For instance, suppose we place $n = 1\text{ mole}$ of helium into a $50\text{ L}$ container at $300\text{K}\text{.}$ Then, this specifies the state fully with pressure in the gas tank given by

\begin{equation*} p = \dfrac{nRT}{V} = \dfrac{1\times8.31 \times 300}{50\times 10^{-3}} = 49,860\text{ Pa}. \end{equation*}

An ideal gas occupies a container of volume $0.03\text{ m}^3$ at $30^{\circ}\text{C}$ and $1.0\times 10^5\text{ Pa}\text{,}$ how many molecules of the gas are there in the container?

Hint

The temperature in ideal gas law is in kelvin scale.

$0.175$

Solution

We use the ideal gas law. But, here we need to be careful in converting the temperature to the kelvin scale first.

\begin{equation*} T = 30 + 273 = 303\text{K}. \end{equation*}

Now, using the variables in the ideal gas law, again making sure of having correct units, gives

\begin{align*} n \amp= \dfrac{pV}{RT} = \dfrac{1.0\times 10^5\text{ Pa} \times 0.03\text{ J}}{8.31 \text{ J/K} \times 303\text{K}} = 0.175. \end{align*}

What volume fraction of a 1 L of the Nitrogen gas is occupied by the molecules if the pressure is 1 atm and the temperature $27^{\circ}\text{C}\text{?}$

Data: Assume volume of each $\text{N}_2$ molecule to be $6.4 \times 10^{-29}\text{ m}^3\text{.}$ $R = 0.082\text{ L.atm.K}^{-1}\text{.}$

Hint

Find the number of moles first.

$0.2\%$

Solution

From the ideal gas equation, we get number of moles in the container. We will do the moles first since $R$ in unit $\text{L-atm/K}$ is usually listed.

\begin{align*} n \amp = \dfrac{pV}{RT} \\ \amp = \frac{1\ \text{atm}\times 1\ \text{L}}{0.082\ \text{L.atm.K}^{-1}.\text{mol}^{-1} 303.15\ \text{K}} = 0.04\text{ mol}. \end{align*}

Therefore, the volume of the molecules is

\begin{align*} v \amp = 0.04\times 6.022\times 10^{23} \times 6.4 \times 10^{-29}\text{ m}^3 \\ \amp = 2\times 10^{-6}\text{ m}^3 = 2\times 10^{-3}\text{ L}. \end{align*}

This says that only 0.2% of the space occupied by the gas is occupied by the molecules themselves, the rest of the space is empty.