In a gas, molecules are far from each other and their mutual interactions are negligible except when two molecules collide. Gases at low pressures and high temperatures much above that at which they liquefy (or solidify) approximately satisfy a simple relation between their pressure, temperature and volume given by

Where \(T\) is the temperature in kelvin or (absolute) scale. \(K\) is a constant for the given sample of the gas but varies with the volume of the gas. If we now bring in the idea of atoms or molecules, then \(K\) is proportional to the number of molecules, (say) \(N\) in the sample. We can write \(K=N k\). Observation tells us that this \(k\) is same for all gases. It is called Boltzmann constant and is denoted by \(k_{\mathrm{B}}\).
As \(\frac{P_1 V_1}{N_1 T_1}=\frac{P_2 V_2}{N_2 T_2}=\) constant \(=k_{\mathrm{B}} \dots(13.2)\)

If \(P, V\) and \(T\) are same, then \(N\) is also same for all gases. This is Avogadro’s hypothesis, that the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure.

Avogadro’s Number

The number in 22.4 litres of any gas is \(6.02 \times 10^{23}\). This is known as Avogadro number and is denoted by \(N_{\mathrm{A}}\).

The mass of 22.4 litres of any gas is equal to its molecular weight in grams at S.T.P (standard temperature \(273 \mathrm{~K}\) and pressure \(1 \mathrm{~atm}\) ). This amount of substance is called a mole. Avogadro had guessed the equality of numbers in equal volumes of gas at a fixed temperature and pressure from chemical reactions. Kinetic theory justifies this hypothesis.

The perfect gas equation can be written as \(P V=\mu R T \dots(13.3)\)

where \(\mu\) is the number of moles and \(R=N_{\mathrm{A}}\) \(k_{\mathrm{B}}\) is a universal constant. The temperature \(T\) is the absolute temperature. Choosing kelvin scale for absolute temperature, \(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\). Here
\(
\mu=\frac{M}{M_0}=\frac{N}{N_A} \dots(13.4)
\)
where \(M\) is the mass of the gas containing \(N\) molecules, \(M_0\) is the molar mass and \(N_{\mathrm{A}}\) the Avogadro’s number. Using Eqs. (13.4) and (13.3) can also be written as
\(
P V=k_{\mathrm{B}} N T \quad \text { or } \quad P=k_{\mathrm{B}} n T
\)

where \(n\) is the number density, 1.e. number of molecules per unit volume. \(k_{\mathrm{B}}\) is the Boltzmann constant introduced above. Its value in SI units is \(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\).
Another useful form of Eq. (13.3) is
\(
P=\frac{\rho R T}{M_0} \dots(13.5)
\)
where \(\rho\) is the mass density of the gas.

Ideal Gas

A gas that satisfies Eq. (13.3) exactly at all pressures and temperatures is defined to be an ideal gas. An ideal gas is a simple theoretical model of a gas. No real gas is truly ideal. Fig. 13.1 shows departures from ideal gas behaviour for a real gas at three different temperatures. Notice that all curves approach the ideal gas behaviour for low pressures and high temperatures.

Boyle’s Law

At low pressures or high temperatures, the molecules are far apart and molecular interactions are negligible. Without interactions, the gas behaves like an ideal one.
If we fix \(\mu\) and \(T\) in Eq. (13.3), we get
\(
P V=\text { constant } \dots(13.6)
\)

i.e., keeping the temperature constant, the pressure of a given mass of gas varies inversely with volume. This is the famous Boyle’s law. Fig. 13.2 shows the comparison between experimental \(P\) – \(V\) curves and the theoretical curves predicted by Boyle’s law. Once again you see that the agreement is good at high temperatures and low pressures.

Charles’ Law

If you fix \(P\), Eq. (13.1) shows that \(V \propto T\) i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature \(T\) (Charles’ law). See Fig. 13.3.

Dalton’s Law of Partial Pressures

Consider a mixture of non-interacting ideal gases: \(\mu_1\) moles of gas \(1, \mu_2\) moles of gas 2 , etc. in a vessel of volume \(V\) at temperature \(T\) and pressure \(P\). It is then found that the equation of state of the mixture is:
\(
\begin{aligned}
& P V=\left(\mu_1+\mu_2+\ldots\right) R T \dots(13.7) \\
& \text { 1.e. } P=\mu_1 \frac{R T}{V}+\mu_2 \frac{R T}{V}+\ldots (13.8)\\
& =P_1+P_2+\ldots \dots(13.9)
\end{aligned}
\)
Clearly \(P_1=\mu_1 R T / V\) is the pressure that gas 1 would exert at the same conditions of volume and temperature if no other gases were present. This is called the partial pressure of the gas. Thus, the total pressure of a mixture of ideal gases is the sum of partial pressures. This is Dalton’s law of partial pressures.

Example 1: The density of water is 1000 \(\mathrm{kg} \mathrm{m}^3\). The density of water vapour at \(100^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure is \(0.6 \mathrm{~kg} \mathrm{~m}^{-3}\). The volume of a molecule multiplied by the total number gives, what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.

Solution:

For a given mass of water molecules, the density is less if the volume is large. So the volume of the vapour is \(1000 / 0.6=1 /\left(6 \times 10^{-4}\right)\) times larger. If densities of bulk water and water molecules are the same, then the fraction of molecular volume to the total volume in the liquid state is 1. As the volume in vapour state has increased, the fractional volume is less by the same amount, 1.e. \(6 \times 10^{-4}\).

Example 2: The density of water is 1000 \(\mathrm{kg} \mathrm{m}^3\). The density of water vapour at \(100^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure is \(0.6 \mathrm{~kg} \mathrm{~m}^{-3}\). The volume of a molecule multiplied by the total number gives, what is called, molecular volume. Estimate the volume of a water molecule.

Solution:

In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\). To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to \((2+16) g=18 g=0.018 \mathrm{~kg}\).
Since 1 mole contains about \(6 \times 10^{23}\) molecules (Avogadro’s number), the mass of a molecule of water is \((0.018) /\left(6 \times 10^{23}\right) \mathrm{kg}=\) \(3 \times 10^{-26} \mathrm{~kg}\). Therefore, a rough estimate of the volume of a water molecule is as follows :
The volume of a water molecule
\(
\begin{aligned}
& =\left(3 \times 10^{-26} \mathrm{~kg}\right) /\left(1000 \mathrm{~kg} \mathrm{~m}^{-3}\right) \\
& =3 \times 10^{-29} \mathrm{~m}^3 \\
& =(4 / 3) \pi \text { (Radius }^3 \\
&
\end{aligned}
\)
Hence, Radius \(\approx 2 \times 10^{-10} \mathrm{~m}=2 Å\)

Example 3: What is the average distance between atoms (interatomic distance) in water? Use the data given in Examples 1 and 2.

Solution:

A given mass of water in vapour state has \(1.67 \times 10^3\) times the volume of the same mass of water in a liquid state (Example 1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by \(10^3\) times the radius increases by \(V^{1 / 3}\) or 10 times, 1.e., \(10 \times 2 Å=20 Å\). So the average distance is \(2 \times 20=40 Å\).

Example 4: A vessel contains two nonreactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is \(3: 2\). Estimate the ratio of (i) the number of molecules and (ii) the mass density of neon and oxygen in the vessel. Atomic mass of \(\mathrm{Ne}=20.2 \mathrm{u}\), molecular mass of \(\mathrm{O}_2\) \(=32.0 \mathrm{u}\)

Solution:

Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since \(V\) and \(T\) are common to the two gases, we have \(P_1 V=\mu_1 R T\) and \(P_2 V=\) \(\mu_2 R T\), 1.e. \(\left(P_1 / P_2\right)=\left(\mu_1 / \mu_2\right)\). Here 1 and 2 refer to neon and oxygen respectively. Since \(\left(P_1 / P_2\right)=\) (3/2) (given), \(\left(\mu_1 / \mu_2\right)=3 / 2\).

(i) By definition \(\mu_1=\left(N_1 / N_{\mathrm{A}}\right)\) and \(\mu_2=\left(N_2 / N_{\mathrm{A}}\right)\) where \(N_1\) and \(N_2\) are the number of molecules of 1 and 2 , and \(N_{\mathrm{A}}\) is the Avogadro’s number. Therefore, \(\left(N_1 / N_2\right)=\left(\mu_1 / \mu_2\right)=3 / 2\).

(ii) We can also write \(\mu_1=\left(m_1 / M_1\right)\) and \(\mu_2=\) \(\left(m_2 / M_2\right.\) ) where \(m_1\) and \(m_2\) are the masses of 1 and 2; and \(M_1\) and \(M_2\) are their molecular masses. (Both \(m_1\) and \(M_1\); as well as \(m_2\) and \(M_2\) should be expressed in the same units). If \(\rho_1\) and \(\rho_2\) are the mass densities of 1 and 2 respectively, we have
\(
\begin{aligned}
& \frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left(\frac{M_1}{M_2}\right) \\
& =\frac{3}{2} \times \frac{20.2}{32.0}=0.947
\end{aligned}
\)