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Section 21.1 Ideal 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.

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

where \(k_B\) is called Boltzmann's constant that has the following value.

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

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 \(N_A\text{.}\)

\begin{equation} n = \dfrac{N}{N_A},\tag{21.1.3} \end{equation}

where \(N_A\) is the number of moecules in one mole.

\begin{equation} N_A = 6.022\times 10^{23},\tag{21.1.4} \end{equation}

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

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

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,

\begin{equation} pV = n R T.\tag{21.1.6} \end{equation}
Remark 21.1.1. 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*}

Subsection 21.1.1 Specifying 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?


The temperature in ideal gas law is in kelvin scale.




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 \(6.4 \times 10^{-29}\text{ m}^3\text{.}\) \(R = 0.082\text{ L.atm.K}^{-1}\text{.}\)


Find the number of moles first.




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.