Equation of state for an ideal gas :

Let P₁, V₁ and T₁ be the pressure, volume and absolute temperature (thermodynamic temperature) of n moles of a gas (assumed to be ideal).

Suppose the gas is heated at constant pressure (P₁) so that its temperature becomes T and volume becomes V₂.

Then by Charles’ law,  \(\frac{V_2}{V_1}=\frac{T}{T_1}\)    …….. (1)

∴ \(T=\frac{V_2T_1}{V_1}\)

Now, suppose that the gas is heated at constant volume (V₂) so that its temperature becomes T₂ and pressure becomes P₂. Then by Gay—Lussac’s law,

\(\frac{P_2}{P_1}=\frac{T_2}{T}\)

∴ \(T=\frac{P_1T_2}{P_2}\)…….. (2)

From equations (1) and (2), we get,

\(\frac{V_2T_1}{V_1}=\frac{P_1T_2}{P_2}\)

∴ \(\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}\)

When pressure and temperature are constant, V ∝ n. Hence, in general, we can write

\(\frac{PV}{T}=nR\) , where R is a proportionality constant, called the universal gas constant.

Thus, PV = nRT for an ideal gas. This is the equation of state for an ideal gas.

[Note : By Boyle's law, for a fixed mass of gas at constant temperature, V=constant. Writing PV ∝ nT, we get PV = nRT]

(i) Ideal gas equation in terms of the Avogadro number

In the usual notation, PV= nRT;

Now, the number of moles,

n = mass of the gas (M) /molar mass (M₀) = N/N_A 

(Molar mass is the mass of 1 mole of gas)

where N is the number of molecules corresponding to n moles of the gas and N_A is the Avogadro number.

∴ PV = \(\frac{N}{N_A}RT\)

(ii) Ideal gas equation in terms of the Boltzmann constant :

PV = \(\frac{N}{N_A}RT\) = N\(\frac{R}{N_A}T\)

= Nk_BT, where k_B =  \(\frac{R}{N_A}\) is the Boltzmann constant. .

Ideal gas equation in terms of the density of the gas :

In the usual notation, PV= nRT;

The number of moles,

n = mass of the gas (M) /molar mass (M₀)

∴ PV = \(\frac{M}{M_0}RT\)

∴ P =  \(\frac{M/V}{M_0}RT\) = \(\frac{ρ}{M_0}RT\)  , where ρ =M/V is the density of the gas.

The basic assumptions of the kinetic theory of an ideal gas :

• A gas of a pure material consists of an extremely large number of identical molecules.
• A gas molecule behaves as an ideal particle, i.e., it has mass but its structure and size can be ignored as compared with the intermolecular separation in a dilute gas and the dimensions of the container.
• The molecules are in constant random motion with various velocities and obey Newton's laws of motion.
• Intermolecular forces can be ignored on the average so that the only forces between the molecules and the walls of the container are contact forces during collisions. It follows that between successive collisions, a gas molecule travels in a straight line with constant speed.
• The collisions are perfectly elastic conserving total momentum and kinetic energy, and the duration of a collision is very small compared to the time interval between successive collisions.

Distinguish between an ideal gas and a real gas :

Concept of free path of a gas molecule :

Molecules of a real gas are not point like particles. For simplicity, molecules are assumed to be spherical. Because of their finite size, two molecules collide with each other when they come within a molecular diameter of each other, strictly speaking, within the sphere of influence of each molecule.

The molecular collisions, though perfectly elastic, result in significant changes in the speeds and directions of motion of the molecules. Hence, every molecule follows a zigzag path, with abrupt changes in its speed and direction of motion at short and random time intervals. This is called Brownian motion (see Fig.).

Expression for mean free path :

Maxwell, on the basis of the law of distribution of molecular speeds, obtained the formula for the mean free path (λ) as

λ = \(\frac{1}{\sqrt{2}πd^2\frac{N}{V}}\)    where N is the number of molecules in volume V of the gas and d is the diameter of a molecule.

Expression for the mean free path of a gas molecule in terms of pressure and temperature :

In the usual notation, the mean free path of a gas molecule, λ = \(\frac{1}{\sqrt{2}πd^2\frac{N}{V}}\)

Now PV=nRT

PV = \(\frac{N}{N_A}\)RT = N\(\frac{R}{N_A}\)T = Nk_BT

∴ \(\frac{N}{V}\) = \(\frac{P}{k_BT}\)

∴ λ = \(\frac{1}{\sqrt{2}πd^2P}\)

Expression for the mean free path of a gas molecule in terms of the density of the gas :

 In the usual notation, the mean free path of a gas molecule, λ = \(\frac{1}{\sqrt{2}πd^2\frac{N}{V}}\)

Density, ρ of the gas = mass/volume = mN/V , where m is the molecular mass.

∴ N/V = ρ/m

∴ λ = \(\frac{m}{\sqrt{2}πd^2ρ}\)

Expressions :

If there are N molecules in an enclosed pure gas and v₁, v₂, v₃, ..., v_N are the speeds of different molecules,

 (i) the mean speed, v = \(\frac{v_1+v_2+.....v_N}{N}\)

(ii) the mean square speed, \(\vec{v^2}=\frac{v_1^2+v_2^2+.....v_N^2}{N}\)

(iii) the rms speed, v_rms = \(\sqrt{v^2}\)

The mean square velocity is numerically equal to the mean square speed. Similarly, the rms velocity is numerically equal to the rms speed. But in random motion, the mean velocity would be statistically zero, but the mean speed cannot be zero.

On the basis of the kinetic theory of gases, derive an expression for the pressure exerted by the gas. The pressure exerted by a gas on the walls of a container results from the momentum transfer during the collisions of the gas molecules with the walls. The normal force on the wall by all the molecules, per unit wall area, is the gas pressure on the wall.

Consider a very dilute gas at a steady state temperature enclosed in a cubical container of Side L

The coordinate axes are aligned with the edges of this cube, as shown in above Fig. Suppose that there are molecules in the container, each of mass m.

Consider a molecule moving with velocity  \(\vec{v_1}\)

  \(\vec{v_1}=v_{1x}i+v_{1y}j+v_{1z}k\)  ………. (1)

Where  v_1x, v_1y, and v_1z are the velocity components along the x—, y- and z—axes, respectively, so that

\(v_1^2=v_{1x}^2+v_{1y}^2+v_{1z}^2\) 

The change in momentum of the molecule on collision with the right wall

= final momentum — initial momentum

= (-m| v_1x |) - m|v_1x| = - 2m|v_1x|

Assuming the collision to be elastic, the change in momentum of the right wall clue to this collision is 2m |v_1x|.

Since the distance between the right and left walls is L, the time between successive collisions with the right wall is,

Δt = \(\frac{2L}{|v_1x|}\)

Therefore, the frequency with which the molecule collides with the right wall is \(\frac{|v_1x|}{2L}\)

∴ Rate of change of momentum of the right wall

= (change in momentum in one collision)-(frequency of collision)

= (2m |v_1x|) \(-\frac{|v_1x|}{2L}\) = \(\frac{mv_1^2x}{L}\)……… (2)

By Newton's second law of motion, this is the force fix exerted by the molecule on the right wall.

Thus, the net force on the right wall by all the N molecules is

F_x = f_1x +f_2x+….+f_Nx

= \(\frac{mv_{1x}^2}{L}+\frac{mv_{2x}^2}{L}+......\frac{mv_{Nx}^2}{L}\)

= \(\frac{m}{L}(v_{1x}^2x+v_{2x}^2+......v_{Nx}^2)\)  ……. (3)

The pressure P_x exerted by all the molecules on a wall perpendicular to the x-axis is

P_x = \(\frac{F_x}{L^2}=\frac{m}{L^3}(v_{1x}^2+v_{2x}^2+......v_{Nx}^2)\)

=   \(\frac{m}{V}(v_{1x}^2+v_{2x}^2+......v_{Nx}^2)\)………….(4)

Where V = L³ is the volume of the container.

Similarly, P_y = \(\frac{m}{V}(v_{1y}^2+v_{2y}^2+......v_{Ny}^2)\)  ………(5)

and P_z = \(\frac{m}{V}(v_{1z}^2+v_{2z}^2+......v_{Nz}^2)\) ………..(6)

As pressure is the same in all directions,

P_x = P_y = P_z = P (say)

∴ P_x + P_y + P_z = 3P  ……….(7)

P = \(\frac{1}{3}\)( P_x + P_y + P_z )

= \(\frac{m}{3V}[(v_{1x}^2+v_{2x}^2+......v_{Nx}^2)+(v_{1y}^2+v_{2y}^2+......v_{Ny}^2)+.....(v_{1z}^2+v_{2z}^2+......v_{Nz}^2)]\)

= \(\frac{m}{3V}[(v_{1x}^2+v_{1y}^2+......v_{1zx}^2)+(v_{2x}^2+v_{2y}^2+......v_{2z}^2)+.....(v_{Nx}^2+v_{Ny}^2+......v_{Nz}^2)]\)

=   \(\frac{m}{3V}(v_1^2+v_2^2+....v_N^2)\) ……. (8)

By definition, the mean square speed of the gas molecules is

\(\bar{v^2}=\frac{v_1^2+v_2^2+....v_N^2}{N}\)

∴ \(v_1^2+v_2^2+....v_N^2\) = \(N\bar{v^2}\)   ….……. (9)

∴ P = \(\frac{1Nm}{3V}\vec{v^2}\)

Here, Nm = (number of molecules)x(molecular mass) =  mass of the gas, M

∴ \(\frac{Nm}{V}\) = \(\frac{M}{V}\)= density ρ

∴ P = \(\frac{1}{3}ρ\bar{v^2}\)  ….. (11)

Writing \(\bar{v^2}\) = \(v_{rms}^2\)

∴ P = \(\frac{1}{3}ρv_{rms}^2\)   ….. (12) where v_rms is the root-mean-square speed of molecules of the at a given temperature.

Equation, P = \(\frac{1}{3}ρv_{rms}^2\) gives the pressure exerted by an ideal gas in terms of its density and the root—mean-square speed of its molecules.

Deduce Boyle's law using the expression for pressure exerted by an ideal gas :

According to the kinetic theory of gases, the pressure exerted by the gas is

P =  \(\frac{1}{3}\frac{Nm}{V}v_{rms}^2\)

where N is the number of molecules of the gas, m is the mass of a single molecule, v_rms is the rms speed of the molecules and V is the volume occupied by the gas.

 PV =  \(\frac{2}{3}N(\frac{1}{2}mv_{rms}^2)\)

= (KE of a gas molecule)x\(\frac{2}{3}N\)       ……..(1)

For a fixed mass of gas, N is constant. Further, inter-molecular forces are ignored so that the corresponding potential energy of the gas molecules may be assumed to be zero. Therefore, \(\frac{1}{2}mv_{rms}^2\) is the total energy of a gas molecule and N\(\frac{1}{2}mv_{rms}^2\)  is the total energy of the gas molecules, which is proportional to the absolute temperature of the gas. Then, the right—hand side of Eq. (1) will be constant if its temperature is constant. Hence, it follows that PV = constant for a fixed mass of gas at constant temperature, which is Boyle's law.

Distribution of speeds of molecules of gas :

Figure shows a typical distribution of speeds for a gas at a temperature T. This is known as Maxwell's distribution of molecular speeds. Here, the shaded area n_v dv is the number of molecules having speed between v and v + dv. Average values of physical quantities like \(\bar{v^2}\) can be calculated once the distribution is known.

• Maxwell-Boltzmann distribution of molecular speeds is a relation that describes the distribution of speeds among the molecules of a gas at a given temperature.
• The root-mean-square speed v_rms gives us a general idea of molecular speeds in a gas at a given temperature. However, not all molecules have the same speed.
• At any instant, some molecules move slowly and some very rapidly. In classical physics, molecular speeds may be considered to cover the range from 0 to ∞
• The molecules constantly collide with each other and with the walls of the container and their speeds change on collisions.
• Also the number of molecules under consideration is very large statistically. Hence, there is an equilibrium distribution of speeds.

Degrees of freedom for a diatomic molecule : A diatomic molecule, in addition to translation, can rotate about axes perpendicular to the line connecting the atoms, as shown in Fig. but not about that line itself. Therefore, it has only two degrees of rotational freedom.

Further, the two atoms may oscillate alternately toward and away from one another along the line joining them, as if connected by a spring. As a harmonic oscillator can have potential energy as well as kinetic energy, a diatomic molecule is regarded to have two degrees of vibrational freedom. Thus, at high enough temperatures, a diatomic molecule has seven degrees of freedom: three of translation, and two each of rotation and vibration.

Mayer's relation between the molar specific heat of a gas at constant pressure and that at constant volume :                                 

(Q. Using the first law of thermodynamics, show that for an ideal gas, the difference between the molar specific heat capacities at constant pressure and at constant volume is equal to the molar gas constant R. )

Consider a cylinder of volume V containing H moles of an ideal gas at pressure P, fitted with a piston of area A.

Suppose, the gas is heated at constant pressure which raises its temperature by dT.

The gas exerts a total force F =PA on the piston which moves outward a small distance dx, see Fig.

The work done by the force in moving the piston is

dW = Fdx = PAdx = PdV  ……….(1)

where Adx = dV is the increase in volume of the gas during the expansion. dW is the work done by the gas on the surroundings as a result of the expansion. If the heat supplied to the gas is dQ_P and the increase in its internal energy is dE then, by the first law of thermodynamics,

dQ_P = dE+dW = dE+PdV

If C_P is the molar specific heat capacity of the gas at constant pressure,

dQ_P = nC_P dT.

∴ nC_PdT = dE+PdV     …….(2)

On the other hand, if the gas was heated at constant volume (instead of at constant pressure) from the initial state such that its temperature increases by the same amount dT, then dW = 0.

Since the internal energy of an ideal gas depends only on the temperature, the increase in internal energy would again be dE.

If dQ_V was the heat supplied to the gas in this case, by the first law of thermodynamics and the definition of molar specific heat capacity at constant volume (C_V),

dQ_V = dE = nC_VdT ……. (3)

From Eqs  (2) and (3),

∴ nC_PdT = nC_VdT + PdV

∴ C_P - C_V =  \(\frac{P}{n}\frac{dV}{dT}\)  ……. (4)

The equation of state of an ideal gas is PV= nRT

Therefore, at a constant pressure,

PdV = nRdT

∴ \(\frac{dV}{dT}\) = \(\frac{nR}{P}\)   ……. (5)

From Eqs. (4) and (5),

C_P - C_V = \(\frac{P}{n}\frac{nR}{P}\) = R   ……. (6)

This is Mayer’ s relation between C_P and C_V

Here, heat and work are expressed in the same units. If heat is expressed in calorie or kilocalorie and work is expressed in erg or joule, the above relation becomes

C_P - C_V = \(\frac RJ\)  where J is the mechanical equivalent of heat.

Mayer’s relation in terms of the principal specific heats, S_P and S_V :

Mayer's relation : C_P — C_V = R. Let M₀ = molar mass of the gas, S_P=specific heat of the gas at constant pressure and S_V= specific heat of the gas at constant volume.

Now:  C_P = M₀S_P and C_V = M₀S_V

 M₀S_P — M₀S_V = R

S_P —S_V = \(\frac{R}{M_0}\)  when heat and work are expressed in the same units.

If heat is expressed in calorie or kilocalorie and work is expressed in erg or joule, we get, S_P —S_V = \(\frac{R}{M_0J}\)  where J is the mechanical equivalent of heat.

[Notes : C_P= \(\frac{1}{n}\frac{dQP}{dT}\)  and  S_P = \(\frac{1}{M}\frac{dQP}{dT}\)

∴ \(\frac{C_P}{S_P}=\frac{M}{n}=\frac{nM_0}{n}=M_0\)

C_P = M₀S_P Similarly C_V = M₀S_V

Hence expression for,

(i) The energy per molecule

= \(3(\frac{1}{2}k_bT)\)+\(3(\frac{1}{2}k_bT)\)+f(k_BT) = (3+f)k_BT

(ii) The energy per mole,

E = (3+f)k_BTxN_A = (3+f)RT

 (iii) C_V = \(\frac{dE}{dT}\) = (3+f)R

(iv) C_P= C_V +R = (3+f)R+R = (4+f)R

(v) The adiabatic constant, λ = \(\frac{C_P}{C_V}=\frac{4+f}{3+f}\)

[As f increases, λ decreases]