It says that the molecules of gas are in random motion and are continuously colliding with each other and with the walls of the container. All the collisions involved are elastic in nature due to which the total kinetic energy and the total momentum both are conserved. No energy is lost or gained from collisions. The ideal gas equation is as follows

\( P V=\mu R T \)

the ideal gas law relates the pressure, temperature, volume, and number of moles of the ideal gas. Here R is a constant known as the universal gas constant.

Assumptions Of Kinetic Theory Of Gases

• All gases are made of molecules moving randomly in all directions.
• The size of a molecule is much smaller than the average separation between the molecules.
• The molecules exert no force on each other or on the walls of the container except during collision.
• All collisions between two molecules or between a molecule and a wall are perfectly elastic. Also, the time spent during a collision is negligibly small.
• The molecules obey Newton’s laws of motion.
• When a gas is left for sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction, and time. This assumption may be justified if the number of molecules is very large.

Calculation of Pressure of an Ideal Gas

Consider a gas enclosed in a cube of side \(l\). Take the axes to be parallel to the sides of the cube, as shown in Fig. 13.4.

A molecule with velocity \(\left(v_{x^{\prime}}, v_{y^{\prime}}, v_z\right)\) hits the planar wall parallel to \(y z-\) plane of area \(A\left(=l^2\right)\). Since the collision is elastic, the molecule rebounds with the same velocity; its \(y\) and \(z\) components of velocity do not change in the collision but the \(x\)-component reverses sign. That is, the velocity after the collision is \(\left(-v_x, v_y, v_z\right)\).
The change in momentum of the molecule is:
\(\Delta P=-m v_x-\left(m v_x\right)=-2 m v_x\).
By the principle of conservation of momentum, the momentum imparted to the wall in the collision \(\Delta P^{\prime}=2 m v_x \dots(i)\)

The distance travelled parallel to the \(x\)-direction between \(A_1\) and \(A_2=l\). Thus, the time taken by the molecule to go from \(A_1\) to \(A_2=l / v_x\). The molecule rebounds from \(A_2\), travels towards \(A_1\), and collides with it after another time interval \(l / v_x\).
Thus, the time between two consecutive collisions of this molecule with \(A_1\) is \(\Delta t=2 l / v_x\).
The number of collisions of this molecule with \(A_1\) in unit time is
\(
m=\frac{1}{\Delta t}=\frac{v_x}{2 l} \dots(ii)
\)
The momentum imparted per unit time to the wall by this molecule is:
\(\Delta F=m \Delta P^{\prime}\)
This is also the force exerted on the wall \(A_1\) due to this molecule. The total force on the wall \(A_1\) due to all the molecules is
\(
\begin{aligned}
F & =\sum \frac{m}{l} v_x^2 \\
& =\frac{m}{l} \sum v_x^2 \dots(iii)
\end{aligned}
\)

As all directions are equivalent, we have
\(
\begin{aligned}
\sum v_x^2 & =\sum v_y^2=\sum v_z^2 \\
& =\frac{1}{3} \sum\left(v_x^2+v_y^2+v_z^2\right) \\
& =\frac{1}{3} \sum v^2 \dots(13.13)
\end{aligned}
\)
Thus, from (iii), \(F=\frac{1}{3} \frac{m}{l} \sum v^2\).

If \(N\) is the total number of molecules in the sample, we can write
\(
F=\frac{1}{3} \frac{m N}{l} \frac{\Sigma v^2}{N} .
\)
The pressure is force per unit area so that
\(
\begin{aligned}
p & =\frac{F}{l^2} \\
& =\frac{1}{3} \frac{m N}{l^3} \frac{\Sigma v^2}{N} \\
& =\frac{1}{3} \frac{M}{l^3} \frac{\Sigma v^2}{N}=\frac{1}{3} \rho \frac{\Sigma v^2}{N},
\end{aligned}
\)
where \(M\) is the total mass of the gas taken and \(\rho\) is its density. Also, \(\Sigma v^2 / N\) is the average of the speeds squared. It is written as \(\overline{v^2}\) and is called mean square speed. Thus, the pressure is
\(
p=\frac{1}{3} \rho \overline{v^2}
\)

\(
\begin{array}{ll}
\text { or, } & P V=\frac{1}{3} M \overline{v^2} \dots(13.14)\\
\text { or, } & P V=\frac{1}{3} N m \overline{v^2} \dots(13.15)
\end{array}
\)

RMS Speed

The square root of mean square speed is called root-mean-square speed or rms speed. It is denoted by the symbol \(v_{r m s}\). Thus,
we can write the pressure equation as
\(
P=\frac{1}{3} \rho v_{r m s}^2
\)
\(
\text { so that } \quad v_{r m s}=\sqrt{\frac{3 P}{\rho}}=\sqrt{\frac{3 P V}{M}} 
\)

Example 1: Calculate the rms speed of nitrogen at STP (pressure \(=1 \mathrm{~atm}\) and temperature \(=0^{\circ} \mathrm{C}\) ). The density of nitrogen in these conditions is \(1.25 \mathrm{~kg} \mathrm{~m}^{-3}\).

Solution:

At STP, the pressure is \(1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\). The rms speed is
\(
\begin{aligned}
v_{r m s} & =\sqrt{\frac{3 p}{\rho}} \\
& =\sqrt{\frac{3 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}}{1 \cdot 25 \mathrm{~kg} \mathrm{~m}^{-3}}} \\
& =490 \mathrm{~m} \mathrm{~s}^{-1} .
\end{aligned}
\)

Translational Kinetic Energy of a Gas

GasThe total translational kinetic energy of all the molecules of the gas is 
\(E=\sum \frac{1}{2} m v^2=\frac{1}{2} m N \frac{\Sigma v^2}{N}=\frac{1}{2} M v_{r m s}^2 \dots(13.16)\)
The average kinetic energy of a molecule is
\(
E / N=\frac{1}{2} \frac{M}{N} v_{r m s}^2=\frac{1}{2} m v_{r m s}^2
\)
From equation (13.14)
\(
\begin{aligned}
P V & =\frac{2}{3} \cdot \frac{1}{2} M v_{r m s}^2 \\
P V & =\frac{2}{3} E \\
E & =\frac{3}{2} P V
\end{aligned}
\)

We know that \(P V=\mu R T\)
where \(\mu\) is the number of moles and \(R=N_{\mathrm{A}}\) \(k_{\mathrm{B}}\) is a untversal constant. The temperature \(T\) is the absolute temperature. 

Combining Eq. \(P V =(\frac{2}{3}) E\) with the ideal gas \(P V=\mu R T\), we get
\(
\begin{aligned}
& E=(3 / 2) k_B N T \\
& \text { or } E / N=1 / 2 m \overline{v^2}=(3 / 2) k_B T
\end{aligned}
\)

Kinetic Interpretation of Temperature

We know that a hotter body has larger internal energy than an otherwise similar colder body. Thus, higher temperature means higher internal energy and lower temperature means lower internal energy. According to the kinetic theory of gases, the internal energy of an ideal gas is the same as the total translational kinetic energy of its molecules which is, from the equation

\(E=\frac{1}{2} M v_{r m s}^2\)

Thus, for a given sample of a gas, higher temperature means higher value of \(v_{r m s}\) and lower temperature means lower value of \(v_{r m s}\). We can write,
\(
T=f\left(v_{r m s}\right)
\)
for a given sample of a gas.
Let \(P\) and \(v\) be the pressure of the gas and the rms speed of the molecules at a temperature \(T\) respectively. Let \(P_{t r}\) and \(v_{t r}\) be the values of these quantities at a temperature \(273.16 \mathrm{~K}\), keeping the volume \(V\) the same as that at \(T\).
From equation (13.14),
and
Thus,
\(
\begin{aligned}
P V & =\frac{1}{3} M v^2 \\
P_{t r} V & =\frac{1}{3} M v_{t r}^2 . \\
\frac{P}{P_{t r}} & =\frac{v^2}{v_{t r}^2} \dots(iv)
\end{aligned}
\)
From the definition of the absolute temperature scale,
\(
\frac{P}{P_{t r}}=\frac{T}{273 \cdot 16 \mathrm{~K}} \dots(v)
\)
From (iv) and (v),
\(
T=\left(\frac{273 \cdot 16 \mathrm{~K}}{v_{t r}^2}\right) v^2 \dots(13.17)
\)

Now, \(v_{t r}\) is the rms speed of the molecules at \(273.16 \mathrm{~K}\) and hence is a constant for a given gas. Equation (13.17) shows that the absolute temperature of \(a\) given gas is proportional to the square of the rms speed of its molecules. As the total translational kinetic energy of the molecules is \(E=\frac{1}{2} M v_{r m s}^2\), we see that \(T \propto E\) for a given sample of a gas.

Thus, the absolute temperature of a given sample of a gas is proportional to the total translational kinetic energy of its molecules.

Now consider a mixture of two gases \(A\) and \(B\). Let \(m_1\) be the mass of a molecule of the first gas and \(m_2\) be that of the second. As the molecules collide with each other, they exchange energy. On an average, the molecules with higher kinetic energy lose energy to those with lower kinetic energy. In thermal equilibrium, the average kinetic energy of all molecules are equal. If \(v_1\) and \(v_2\) be the rms speeds of the molecules of \(A\) and \(B\) respectively,
\(
\frac{1}{2} m_1 v_1^2=\frac{1}{2} m_2 v_2^2=(3 / 2) k_B T \dots(13.18)
\)
We find that for different kinds of gases, it is not the rms speed but the average kinetic energy of individual molecules that has a fixed value at a given temperature. The heavier molecules move with smaller rms speed and the lighter molecules move with larger rms speed.

We know that We can also write \(P=(1 / 3) n m \overline{v^2}\), where \(n = N/V\) is known as where \(n\) is the number of molecules per unit volume.

For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas in the mixture. Thus the pressure equation becomes:
\(
P=(1 / 3)\left[n_1 m_1 \overline{v_1^2}+n_2 m_2 \overline{v_2^2}+\ldots\right] \dots(13.19)
\)
In equilibrium, the average kinetic energy of the molecules of different gases will be equal. That is,
\(
\begin{aligned}
& 1 / 2 m_1 \overline{v_1^2}=1 / 2 m_2 \overline{v_2^2}=(3 / 2) k_B T \\
& \text { so that } \\
& P=\left(n_1+n_2+\ldots\right) k_B T \dots(13.20)
\end{aligned}
\)
which is Dalton’s law of partial pressures.

Example 2: A flask contains argon and chlorine in the ratio of \(2: 1\) by mass. The temperature of the mixture is \(27^{\circ} \mathrm{C}\). Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed \(v_{\mathrm{rms}}\) of the molecules of the two gases. Atomic mass of argon \(=39.9 \mathrm{u}\); Molecular mass of chlorine \(=70.9 \mathrm{u}\).

Solution:

The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to \((3 / 2) k_B T\). It depends only on temperature and is independent of the nature of the gas.
(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is \(1: 1\).
(ii) Now \(1 / 2 m v_{\text {rms }}^2=\) average kinetic energy per molecule \(=(3 / 2) k_{\mathrm{B}} T\) where \(m\) is the mass of a molecule of the gas. Therefore,
\(
\frac{\left(\mathbf{v}_{\mathrm{rms}}^2\right)_{\mathrm{Ar}}}{\left(\mathbf{v}_{\mathrm{rms}}^2\right)_{\mathrm{Cl}}}=\frac{(m)_{\mathrm{Cl}}}{(m)_{\mathrm{Ar}}}=\frac{(M)_{\mathrm{Cl}}}{(M)_{\mathrm{Ar}}}=\frac{70.9}{39.9}=1.77
\)
You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers
to (i) and (ii), provided the temperature remains unaltered.

Example 3: Uranium has two isotopes of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed? If the atomic mass of fluorine is 19 units. estimate the percentage difference in speeds at any temperature.

Solution:

At a fixed temperature the average energy \(=1 / 2 m <v^2>\) is constant. So smaller the mass of the molecule the faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So \(v_{349} / v_{352}=(352 / 349)^{1 / 2}=1.0044\).
Hence difference \(\frac{\Delta V}{V}=0.44 \%\).

Example 4: If the rms speed of nitrogen molecules is \(490 \mathrm{~m} \mathrm{~s}^{-1}\) at \(273 \mathrm{~K}\), find the rms speed of hydrogen molecules at the same temperature.

Solution:

The molecular weight of nitrogen is \(28 \mathrm{~g} \mathrm{~mol}^{-1}\) and that of hydrogen is \(2 \mathrm{~g} \mathrm{~mol}^{-1}\). Let \(m_1, m_2\) be the masses and \(v_1, v_2\) be the rms speeds of a nitrogen molecule and a hydrogen molecule respectively. Then \(m_1=14 m_2\). Using equation
\(
\frac{1}{2} m_1 v_1^2=\frac{1}{2} m_2 v_2^2
\)
or, \(\quad v_2=v_1 \sqrt{m_1 / m_2}=490 \mathrm{~m} \mathrm{~s}^{-1} \times \sqrt{14} \approx 1830 \mathrm{~m} \mathrm{~s}^{-1}\).

Deductions From Kinetic Theory

Boyle’s Law: At a given temperature, the pressure of a given mass of a gas is inversely proportional to its volume. This is known as Boyle’s law.
We know
\(
P V=\frac{1}{3} m N v_{r m s}^2 \dots(vi)
\)
As for a given gas \(v_{r m s}^2 \propto T\), the value of \(v_{r m s}^2\) is constant at a given temperature. Also, for a given mass of the gas, \(m\) and \(N\) are constants. Thus, from (vi), \(P V=\text { constant }\)

or, \(\quad P \propto \frac{1}{V}\) which is Boyle’s law.

Charles’s Law: Charles’s LawAt a given pressure, the volume of a given mass of a gas is proportional to its absolute temperature. This is known as Charles’s law.

From (vi), if \(P\) is constant,
\(
V \propto v_{r m s}^2 .
\)
As \(v_{r m s}^2 \propto T\), we get \(V \propto T\) which is Charles’s law.

Charles’s Law of Pressure: At a given volume, the pressure of a given mass of a gas is proportional to its absolute temperature. This is known as Charles’s law for pressure.

In fact, this is the definition of the absolute temperature \(T\). If one starts from the fact that \(v_{r m s}^2 \propto T\) and uses the fact that \(V\) is constant, one gets from (vi),
\(\begin{array}{ll} & P \propto v_{r m s}^2 \\ \text { or, } & P \propto T .\end{array}\)

Avogadro’s Law: At the same temperature and pressure, equal volumes of all gases contain equal number of molecules. This is known as Avogadro’s law.

Consider equal volumes of two gases kept at the same pressure and temperature. Let,

\(m_1=\) mass of a molecule of the first gas
\(m_2=\) mass of a molecule of the second gas
\(N_1=\) number of molecules of the first gas
\(N_2=\) number of molecules of the second gas
\(P=\) common pressure of the two gases
\(V=\) common volume of the two gases.
From the PV equation,
\(
\begin{aligned}
& P V=\frac{1}{3} N_1 m_1 v_1^2 \\
& \text { and } \quad P V=\frac{1}{3} N_2 m_2 v_2^2 \text {, } \\
&
\end{aligned}
\)
where \(v_1\) and \(v_2\) are rms speeds of the molecules of the first and the second gas respectively. Thus,
\(N_1 m_1 v_1^2=N_2 m_2 v_2^2 \dots(vii)\)
As the temperatures of the gases are the same, the average kinetic energy of the molecules is same for the two gases, i.e.,
\(
\frac{1}{2} m_1 v_1^2=\frac{1}{2} m_2 v_2^2 \dots(viii)
\)
From (vii) and (viii),
\(
N_1=N_2
\)
which proves Avogadro’s law.

Graham’s Law of Diffusion: When two gases at the same pressure and temperature are allowed to diffuse into each other, the rate of diffusion of each gas is inversely proportional to the square root of the density of the gas. This is known as Graham’s law of diffusion.

It is reasonable to assume that the rate of diffusion is proportional to the rms speed of the molecules of the gas. Then if \(r_1\) and \(r_2\) be the rates of diffusion of the two gases,
\(
\frac{r_1}{r_2}=\frac{v_{1, r m s}}{v_{2, r m s}} \dots(ix)
\)
We know,
\(
v_{r m s}=\sqrt{\frac{3 P}{\rho}} .
\)
If the pressure of the two gases are the same,
\(
\frac{v_{1, r m s}}{v_{2, r m s}}=\sqrt{\frac{\rho_2}{\rho_1}}
\)
so that from (ix)
\(
\frac{r_1}{r_2}=\sqrt{\frac{\rho_2}{\rho_1}}
\)
which is Graham’s law.

Dalton’s Law of Partial Pressure: Dalton’s law of partial pressure says that the pressure exerted by a mixture of several gases equals the sum of the pressures exerted by each gas occupying the same volume as that of the mixture.

In kinetic theory, we assume that the pressure exerted by a gas on the walls of a container is due to the collisions of the molecules with the walls. The total force on the wall is the sum of the forces exerted by the individual molecules. Suppose there are \(N_1\) molecules of gas \(1, N_2\) molecules of gas 2, etc., in the mixture.
Thus, the force on a wall of the surface area \(A\) is
\(
\begin{aligned}
F= & \text { force by } N_1 \text { molecules of gas } 1 \\
& + \text { force by } N_2 \text { molecules of gas } 2+\ldots \\
= & F_1+F_2+\ldots
\end{aligned}
\)
Thus, the pressure is
\(
P=\frac{F_1}{A}+\frac{F_2}{A}+\ldots
\)
If the first gas alone is kept in the container, its \(N_1\) molecules will exert a force \(F_1\) on the wall. If the pressure, in this case, is \(P_1\),
\(
P_1=F_1 / A
\)
Similar is the case for other gases.
Thus,
\(
P=P_1+P_2+P_3+\ldots
\)

Ideal Gas Equation

Consider a sample of an ideal gas at pressure \(P\), volume \(V\) and temperature \(T\). Let \(m\) be the mass of each molecule and \(v\) be the rms speed of the molecules. Also, let \(v_{t r}\) be the rms speed of the gas at the triple point \(273 \cdot 16 \mathrm{~K}\). From equation (13.15),
\(
P V=\frac{1}{3} N m v^2 \dots(1)
\)
and we know
\(
\begin{gathered}
T=\left(\frac{273 \cdot 16 \mathrm{~K}}{v_{t r}^2}\right) v^2 \\
v^2=\left(\frac{v_{t r}^2}{273 \cdot 16 \mathrm{~K}}\right) T
\end{gathered}
\)
Putting this expression for \(v^2\) in (1),
\(
P V=N\left(\frac{1}{3} \frac{m v_{t r}^2}{273 \cdot 16 \mathrm{~K}}\right) T \dots(2)
\)
Now \(\frac{1}{2} m v_{t r}^2\) is the average kinetic energy of a molecule at the triple point \(273 \cdot 16 \mathrm{~K}\). As the average kinetic energy of a molecule is the same for all gases at a fixed temperature, \(\frac{1}{2} m v_{t r}^2\) is a universal constant. Accordingly, the quantity in the bracket in equation (2) above is also a universal constant. Writing this constant as \(k\), equation (2) becomes,
\(
P V=N k T \dots(3)
\)
The universal constant \(k\) is known as the Boltzmann constant and its value is
\(
k=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}
\)
up to three significant digits. If the gas contains \(\mu\) moles, the number of molecules is
\(
N=\mu N_A
\)
where \(N_A=6.02 \times 10^{23} \mathrm{~mol}^{-1}\) is the Avogadro constant.
Using this, equation (3) becomes
\(
\begin{aligned}
& P V=n N_A k T \\
& \text { or, } \quad P V=\mu R T \dots(4)\\
&
\end{aligned}
\)
where \(R=N_A k\) is another universal constant known as the universal gas constant. Its value is
\(
R=8 \cdot 314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\)
Equation (4) is known as the equation of the state of an ideal gas.

Example 5: Calculate the number of molecules in each cubic metre of a gas at \(1 \mathrm{~atm}\) and \(27^{\circ} \mathrm{C}\).

Solution: We have \(P V=N k T\)
or,
\(
\begin{aligned}
N & =\frac{P V}{k T} \\
& =\frac{\left(1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\right)\left(1 \mathrm{~m}^3\right)}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(300 \mathrm{~K})} \\
& \approx 2.4 \times 10^{25}
\end{aligned}
\)

Rms Speed in Terms of Temperature

We are now in a position to write the rms speed of the molecules in terms of the absolute temperature. From equation (13.15)

\(
P V=\frac{1}{3} N m v_{r m s}^2
\)
and from equation (3),
\(
P V=N k T \text {. }
\)
From these two,
\(
\begin{aligned}
\frac{1}{3} m v_{r m s}^2 & =k T \\
\text { or, } \quad v_{r m s} & =\sqrt{\frac{3 k T}{m}} \dots(5)
\end{aligned}
\)
This may also be written as,
\(
v_{r m s}=\sqrt{\frac{3 k N_A T}{m N_A}}
\)
\(
=\sqrt{\frac{3 R T}{M_o}} \dots(6)
\)
where \(M_o=m N_A\) is the molecular weight.

Average Kinetic Energy of a Molecule

The average kinetic energy of a molecule is
\(
\begin{aligned}
\frac{1}{2} m v_{r m s}^2 & =\frac{1}{2} m \cdot \frac{3 k T}{m} \\
& =\frac{3}{2} k T \dots(7)
\end{aligned}
\)
The total kinetic energy of all the molecules is
\(
U=N\left(\frac{3}{2} k T\right)=\frac{3}{2} n R T \dots(8)
\)
The average speed \(\bar{v}=\Sigma v / N\) is somewhat smaller than the rms speed. It can be shown that
\(
\bar{v}=\sqrt{\frac{8 k T}{\pi m}}=\sqrt{\frac{8 R T}{\pi M_o}}
\)

Example 6: Find the rms speed of oxygen molecules in a gas at 300 K.

Solution:

\(
\begin{aligned}
v_{r m s} & =\sqrt{\frac{3 R T}{M_o}} \\
& =\sqrt{\frac{3 \times\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(300 \mathrm{~K})}{32 \mathrm{~g} \mathrm{~mol}^{-1}}} \\
& =\sqrt{\frac{3 \times 8.3 \times 300}{0.032}} \mathrm{~m} \mathrm{~s}^{-1}=483 \mathrm{~m} \mathrm{~s}^{-1} .
\end{aligned}
\)

Maxwell’s Speed Distribution Law

The rms speed of an oxygen molecule in a sample at \(300 \mathrm{~K}\) is about \(480 \mathrm{~m} \mathrm{~s}^{-1}\). This does not mean that the speed of each molecule is \(480 \mathrm{~m} \mathrm{~s}^{-1}\). Many of the molecules have speed less than \(480 \mathrm{~m} \mathrm{~s}^{-1}\) and many have speed more than \(480 \mathrm{~m} \mathrm{~s}^{-1}\). Maxwell derived an equation giving the distribution of molecules in different speeds. If \(d N\) represents the number of molecules with speeds between \(v\) and \(v+d v\) then
\(
d N=4 \pi N\left(\frac{m}{2 \pi k T}\right)^{3 / 2} v^2 e^{-m v^2 / 2 k T} d v
\)
The figure above shows plots of \(d N / d v\) against \(v\) at two different temperatures. We see that there are some molecules which have speeds many times greater than the mean speed. This fact helps in making nuclear fusion reactions in a laboratory. The speed \(v_p\) at which \(d N / d v\) is maximum is called the most probable speed. Its value is given by
\(
v_p=\sqrt{\frac{2 k T}{m}} \dots(9)
\)

Thermodynamic State

A given sample of a substance has a number of parameters which can be physically measured. When these parameters are uniquely specified, we say that the thermodynamic state of the system is specified. However, not all of these parameters are independent of each other. For example, we can measure pressure, volume, temperature, internal energy, and many other parameters of an ideal gas. But if pressure and volume are specified, the rest of the parameters may be calculated. Thus, a thermodynamic state of a given sample of an ideal gas is completely described if its pressure and its volume are given. When a process is performed on a system, it passes from one thermodynamic state to another.

Equation Of State

The pressure, volume and temperature of a given sample of a substance are related to each other. An equation describing this relation is called the equation of state for that substance. For an ideal gas it is
\(
P V=n R T (n = \mu)
\)
where the symbols have their usual meanings. For a real gas the equation of state is different. The size of a molecule is not negligible in comparison to the average separation between them. Also, the molecular attraction is not negligible. Taking these two facts into account, van der Waals derived the following equation of state for a real gas:
\(
\left(P+\frac{a}{V^2}\right)(V-b)=n R T \dots(10)
\)
where \(a\) and \(b\) are small positive constants. The constant \(a\) is related to the average force of attraction between the molecules and \(b\) is related to the total volume of the molecules.

Brownian Motion

We have assumed in the kinetic theory of gases that the molecules of a gas are in constant random motion, colliding with each other and with the walls of the container. This is also valid for a liquid. Robert Brown, a botanist, accidentally came across evidence of this type of molecular motion in 1827. He was observing all pollen grains suspended in water, under a powerful microscope. He observed that although the water appeared to be at complete rest, the grains were moving randomly in the water, occasionally changing their directions of motion. A typical path of a grain looks as shown in figure (24.3). Such a phenomenon is called Brownian motion. The molecules strike the particles of the pollen grains and kick them to move in a direction. Another collision with some other molecules changes the direction of the grain.

Brownian motion increases if we increase the temperature. Comparing between different liquids, one with smaller viscosity and smaller density will show more intense Brownian motion. Einstein developed a theoretical model for Brownian motion in 1905 and deduced the average size of the molecules from it.

Saturated and Unsaturated Air

The air is said to be saturated when the maximum possible amount of water vapours are present in it. The pressure of the water vapours in the saturated air is called saturation vapour pressure. If the air contains vapours less than the maximum possible amount possible in the air, then it (air) is said to be unsaturated.

Vapour Pressure

A vapour differs from a gas in that the former can be liquified by pressure alone, whereas the latter cannot be liquified unless it is first cooled. If a liquid is kept in a closed vessel, the space above the liquid becomes saturated with vapour. The pressure exerted by this vapour is called saturated vapour pressure. Its value depends only on temperature. It is independent of any external pressure.

Humidity

The humidity shows the presence of water vapours in the atmosphere. The amount of water vapours present per unit volume of the air is called absolute humidity.
Relative humidity \((R H)\) is defined as the ratio of the mass of water vapour \((\mathrm{m})\) actually present in the given volume of air to the mass of water vapour \((M)\) required to saturate the same volume at the same temperature. Normally \(\mathrm{RH}\) is expressed in percentage
ie., \(\% R H=\frac{m}{M} \times 100 \%\)
\(\mathrm{RH}\) is also defined as the ratio of actual pressure (p) of water vapour to the saturated vapour pressure \((\mathrm{P})\) of the water at the same temperature
i.e., \(\% R H=\frac{p}{P} \times 100 \%\)
Also,
\(
R H=\frac{\text { Saturated vapour pressure of water at dew point }}{\text { Saturated vapour pressure of water at room temperature }}
\)

As the pressure exerted by the vapour is directly proportional to the amount of vapour present in a given volume, the relative humidity may also be defined as
\(
R H=\frac{\text { Vapour pressure of air }}{\text { SVP } \text { at the same temperature }} .
\)
The vapour pressure of air at the actual temperature is equal to the saturation vapour pressure at the dew point. Thus, the relative humidity may be redefined as
\(
R H=\frac{\text { SVP at the dew point }}{\text { SVP at the air-temperature }}
\)

Dew Point

It is the temperature at which the amount of water vapours actually present in a certain volume of the air which is sufficient to saturate that volume of air.
At the dew point the actual vapour pressure becomes the saturated vapour pressure.
It means that vapour pressure at room temperature \(=\) saturated vapour pressure(S.V.P) at dew point
If room temperature \(=\) dew point
then Relative humidity (R.H.) \(=100 \%\)

If \(f\) and \(F\) be the saturation vapour pressures at the dew point and at the air-temperature respectively, the relative humidity is
\(
\frac{f}{F} \times 100 \%
\)

Evaporation

The molecules of ether move with random speeds and in random directions. A molecule collides frequently with other molecules and the container walls to change its direction and speed. Occasionally, a molecule starting in an upward direction near the surface of the liquid may escape collisions and move out of the liquid. This process is called evaporation. Thus, evaporation is a process in which molecules escape slowly from the surface of a liquid.

Boiling

If we heat the liquid, the average kinetic energy of the entire liquid increases and at a certain stage, the energy becomes sufficient to break the molecular attraction. The molecules anywhere in the liquid can form vapour bubbles. These bubbles float to the surface of the liquid and finally come out of the liquid. This phenomenon is called boiling and the temperature at which boiling occurs is called the boiling point. Thus, in evaporation, only the molecules near the surface which have kinetic energy greater than the average escape from the liquid, whereas, in boiling, the molecules all over the liquid gain enough energy to become vapour.

The boiling point of a liquid depends on the external pressure over its surface. In fact, boiling occurs at a temperature where the SVP equals the external pressure. Thus, from the figure below, the boiling point of water at \(1 \mathrm{~atm}\) (760 \(\mathrm{mm}\) of mercury) is \(100^{\circ} \mathrm{C}\) but at \(0.5 \mathrm{~atm}\) it is \(82^{\circ} \mathrm{C}\).

Example 7: In an experiment with Regnault’s hygrometer, dew appears at \(10^{\circ} \mathrm{C}\) when the atmospheric temperature is \(40^{\circ} \mathrm{C}\). Find the relative humidity.

Solution:

The dew point is \(10^{\circ} \mathrm{C}\). The saturation vapour pressure at this temperature is \(8.94 \mathrm{~mm}\) of \(\mathrm{Hg}\). Also, the saturation vapour pressure of air at \(40^{\circ} \mathrm{C}\) is \(55 \cdot 1 \mathrm{~mm}\) of \(\mathrm{Hg}\). The relative humidity expressed in percentage
\(
\begin{aligned}
& =\frac{\text { vapour pressure at the dew point }}{\text { SVP at the air-temperature }} \times 100 \% \\
& =\frac{8.94}{55.1} \times 100 \%=16.2 \% .
\end{aligned}
\)

Example 8: The vapour pressure of air at \(20^{\circ} \mathrm{C}\) is found to be \(12 \mathrm{~mm}\) of \(\mathrm{Hg}\) on a particular day. Find the relative humidity.

Solution:

The saturation vapour pressure of water at \(20^{\circ} \mathrm{C}\) is \(17.5 \mathrm{~mm}\) of \(\mathrm{Hg}\). Thus, the relative humidity is
\(
\begin{aligned}
& \frac{\text { vapour pressure of air }}{\text { SVP at the same temperature }} \\
= & \frac{12 \mathrm{~mm} \text { of } \mathrm{Hg}}{17 \cdot 5 \mathrm{~mm} \text { of } \mathrm{Hg}}=0.69,
\end{aligned}
\)
that is, \(69 \%\).

Example 9: At what external pressure will water boil at \(140^{\circ} \mathrm{C}\)? Express the answer in \(\mathrm{atm}\).

Solution:

The saturation vapour pressure of water at \(140^{\circ} \mathrm{C}\) is \(2710 \mathrm{~mm}\) of \(\mathrm{Hg}\). Thus, water will boil at \(140^{\circ} \mathrm{C}\) at this pressure. Now \(760 \mathrm{~mm}\) of \(\mathrm{Hg}=1 \mathrm{~atm}\). Thus, \(2710 \mathrm{~mm}\) of \(\mathrm{Hg}=\frac{2710}{760} \mathrm{~atm}=3.56 \mathrm{~atm}\).

The pressure inside a pressure cooker is of this order when it whistles. So, the temperature inside is of the order of \(140^{\circ} \mathrm{C}\) which helps in cooking the food much faster.

Phase Diagrams: Triple Point

The combination of the temperature and the pressure at which the three phases (gas, liquid, and solid) of a substance coexist in thermodynamic equilibrium.

The three curves meet at one point labelled as Triple point. At the pressure and temperature corresponding to this point, all the three phases may remain together in equilibrium. This point is known as the triple point. For water, the triple point occurs at the pressure 4.58 mm of mercury and temperature \(273 \cdot 16 \mathrm{~K}\).

In one’s everyday life, one experiences three phases of materials: solids, liquids, and gases. Water provides some of the most familiar examples of phase transitions. One often observes the freezing or melting of ice, the boiling of water, and the condensation of steam. These are often attributed to changes in temperature, however, pressure also plays an important role.

Phases are only stable in certain ranges of pressure and temperature. Transition phases normally occur under conditions of phase equilibrium between two phases, which are indicated by the curves on the phase diagram above (figure above). For instance, near the bottom of the phase diagram, there is a curve indicating the phase transition between a solid and a gas. At this transition, gas sublimates from the material’s solid-state, skipping the liquid transition entirely, as seen in dry ice at normal atmospheric pressure.
The triple point occurs where the solid, liquid, and gas transition curves meet. The triple point is the only condition in which all three phases can coexist, and is unique for every material.