Phase Transition & Low Temperature Physics | Kinetic Theory & Thermodynamics PDF Download

Third Law of Thermodynamics & Attainable of Low Temperature

The third law of thermodynamics is some time stated as follows:

It is impossible for any process no matter how idealized the process is to reduce the entropy of a system to its zero point value in a finite number of operations.

Properties of material at low temperature

At T → 0

C_P = C_V = 0

At T → 0

S→0

At T → 0

Thermal expansion coefficient α =

Production of Low Temperature by the Joule – Kelvin Expansion

The Joule – Kelvin Expansion is essentially a continuous steady state flow process in which a compressed gas is made to expand adiabatically irreversibly through a porous

plug and do work.

Fig: A schematic diagram of the porous-plug experiment for Joule-Kelvin

Let us assume that we start with a gas of internal energy U₁ and volume V₁ . After passing through the porous plug let final internal energy and volume of the gas be U₂ and V₂ respectively. No heat enters the system so this work has to performed at the expense of internal energy

Hence U₁ + P₁V₁ = U₂ + P₂V₂

H, = H₂

Joule – Kelvin expansion is isenthalpic process

H = H(T,P)

dH =

dH = TdS + VdP

So dH =

Since H does not change i.e. dH = 0 as the process is isoenthalpic the change in temperature w.r.t. pressure is given by

 = μ =  =

or  μ =

where μ is known as the Joule – Kelvin Coefficient.

This equation defines a curve in the (TP) plane and is known as the inversion curve when μ = 0 .

Figure: The inversion curve. Inside it the gas cools on undergoing Joule-Thomson expansion.

The temperature change ΔT = T₁ - T₂  produced in a series of isoenthalphs i.e. curve with H(T, P) = constant.

At T = T_i μ = 0 known as inversion temperature.

T < T_i μ = Positive . There is cooling effect of gases i.e. temperature and pressure move in same direction.

T > T_i μ = Negative . There is heating effect of gases i.e., temperature and pressure move in opposite direction.

Example 1: (a) For van der Waals gas prove that inversion temperature T_i = , where a and b are parameter used in van der Waals gas.

(b) Why Hydrogen and Helium shows heating effect as pressure increased at constant enthalphy.

(a) μ =

For van der Waals gas

 ⇒

Since, b << V

 =  =

As,

μ =

For inversion temperature, μ = 0 ⇒  = b ⇒ T_i =

(b) Since inversion temperature of Hydrogen and Helium is very small. For cooling effect initial temperature must be smaller than inversion temperature but for Helium and Hydrogen inversion temperature is very small, so it is not possible to achieve initial temperature lower than inversion temperature, so Helium and Hydrogen give heated effect.

 ⇒

 ⇒

Phase Transition

A phase of a thermodynamic system and the state of matter has uniform physical properties.

A phase transition is the transformation of a thermodynamic system from one phase or state of matter to another.

During a phase transition of a given medium certain properties of the medium changes often discontinuously as a result of some external condition such as  temperature pressure and others.

Gibbs phase rule: It is proposed by Josiah Willard Gibbs which is given by

F = C - P + 2

Where C is number of components P is the number of phase in thermodynamic equilibrium with each other and F is number of degree of freedom.

Phase: A phase is form of matter that is homogeneous in chemical composition and physical state.

Typical phases are solid liquid and gas. Two immiscible (example-liquid mixture with different compositions) separated by distinct boundary are considered as two different phase.

Components: The number of components is the number of chemically independent constituents of the system i.e. minimum number of independent species necessary to define the composition of all phase of the system.

The number of degree of freedom (F) in the context is the number of intensive variable which are independent to each other.

First Order Phase Transition

Let us consider one component system in which system having only one kind of constituent particles. For first order phase transition, P - T diagram are shown in figure1.

Depending on the system at some values for temperature and pressure the three phases of the system may be found in equilibrium. In the P - T diagram the line OA represents equilibrium between solid and liquid phases the line OB represents equilibrium between solid and gas phases and the line OC represents equilibrium between liquid and gas phases. The point O where all the three phases are in equilibrium is known as a triple point. The line OC terminates at the point called the critical point. Beyond this point the gas phase cannot be converted into the liquid phase. In figure 1 the point C is at the apex of the P - T curve at the critical temperature T_c . For the temperature T > T_c the gas phase of the matter cannot be converted into the liquid phase but for T < T_c the gas phase in general can be converted into the liquid phase.

Figure 1: P-T phase of one component

Equilibrium between Two Phases

Let us consider an isolated system having a matter which is existing in two phases denoted by 1 and 2 simultaneously in equilibrium with each other (Figure 2). Suppose V₁ and V₂ are volumes N₁ and N₂ the number of particles E₁ and E₂ the internal energies, S₁ and S₂ the entropies of the two phases respectively. For each phase entropy is a function of its volume number of particles (mass) and internal energy.

Figure 2: Equilibrium of two phases of an isolated one component system From these relations it follows that

From these relations it follows that

P₁ = P₂ (mechanical equilibrium)

μ₁ = μ₂ (chemical equilibrium)

Hence when two different phases of the matter are in equilibrium their temperatures pressures and chemical potentials must be equal. If the chemical potentials are expressed as function of pressure and temperature we have

where P = P₁ = P₂ and T = T₁ = T₂ are the common pressure and temperature respectively of the two phases in equilibrium. Thus from above equation we have 

where G₁(P, T) and G₂(P, T) are the Gibbs free energies and N₁ and N₂ the number of particles in the two phases respectively. Since during the phase transition the number of

particles is not changing (i.e. N₁ = N₂) we have

G₁(P, T) = G₂(P, T)

Hence during the phase transition the Gibbs free energy does not change. Gibbs energies G₁ and G₂ of the two phases 1 and 2, respectively can be exhibited as shown in figure 3.

Clausius-Clapeyron equation

When the two phases denoted by 1 and 2 of the given matter are in equilibrium we have

G₁(P, T) = G₂(P, T)

where G₁ and G₂ are Gibbs free energies of the two phases respectively and P = P1 = P2 and T = T1 = T2 are the common pressure and temperature respectively of the two phases.

In the P - T diagram along the phase-transition line let us consider a point where the

pressure is P + dP and the temperature is T + dT, so that we have

Using Taylor series expansion and neglecting the higher order terms we have

 =

Using above two equation we get,

ΔS =

where ΔH(= H₂ - H₁) is the change in heat (enthalpy) which is the molar latent heat L.

Thus from above equation we have,

for V₂ > V₁, we have

further for S₂ > S₁ we have

Liquid-Vapour Phase Transition

Let us consider a phase transition from a liquid state to a vapour state. If  V_l and V_g respectively denote the volume in the liquid and gases phases and L_v is the heat of vaporization (latent heat for the transition from liquid to vapour) the Clausius-Clapeyro equation is

Since, in the phase transition V_g is always greater than V_l and the heat of vaporization L_v is positive, hence we have

It shows that the boiling point of a liquid increases with the increase in pressure.

Now if the vapour pressure is low i.e. V_g >> V_l in comparison to V_g , then we have 

Using the ideal gas equation g PV_g = RT we have

∴

where C is a constant of integration. At the critical point we have P = P_c , T = T_c and equation is

Here we have assumed that the heat of vapourisation L_v is independent of the temperature. However it depends on the temperature. Suppose it varies as L_v = a - bT then for an ideal gas at low pressure we have

where  C is a constant of integration. At the critical point we have P = P_c , T = T_c and equation is

On putting the value of C , we have

Properties of First Order Phase Transition

(1) Gibbs free energy is continuous

(2) First order derivative with respect to temperature and pressure have finite discontinuity i.e. entropy (S) and pressures (P) have finite discontinuity.

(3) Second and more higher order differential is infinite

Example 2: For a two phase system in equilibrium, P is a function of T only so that

, show that

Let us take T and V as independent variables and write

S = S(T, V)

so that

For an adiabatic process it yields

Using first Maxwell relation we obtain

Since,  , so we can write

 =

=  =

where, β_s is adiabatic compressibility.

Example 3: Calculate under what Pressure water would boil at 120^oC . One gram of steam occupies a volume of 1677 cm³ . Latent heat of steam = 540 cal / g j = 4.2 x10⁷erg / cal, atmospheric pressure = 1.0x10⁶ dyne / cm³

V₂ = 1677cm³ / g V₁ = 1cm³ / g

L = 4.2 x10⁷ x 540erg / g, dT = 20^o K

dP = 0.688 x 10⁶ ⇒ P₂ - P₁ = 0.688 x 10⁶ dyne / cm³

P2 = 0.688 x 10⁶ + P₁ = 1.688 x 10⁶ dyne / cm³

Example 4: Liquid helium - 4 has normal boiling point of 4.2 K . However at pressure of 1 mm of mercury, it boils at 1.2 K. Estimate the average latent heat of vaporization of helium in this temperature range.

∴

⇒  ⇒

Now, and , P₀ = 760 mm  and T₀ = 4.2K, P = 1mm and T₀ = 1.2K

⇒ L = 93.49 J /mol

Example 5: Liquid helium boils at temperature T0 when its vapour pressure is equal to P0, we now pump on the vapour and reduce the pressure to much smaller value P. Assume that the Latent heat L is approximately independent of temperature and helium vapour density is much smaller than that of liquid calculate the approximate temperature T_m of the liquid in equilibrium with its vapour at pressure P. 

Express your answer in terms of L,T₀,P₀,P_m and any other required constants.

 , where

∴ T_m =

Example 6: In the phase transition from a liquid state to a vapour state. The heat of vapourisation L_v varies with temperature T as  . Considering the gas as an ideal one at low pressure show that the pressure P(T) at temperature T in terms of the critical pressure P_c(T_c) at critical temperature T_c is given by

Clausius-Clapeyron equation for the phase transition from liquid to vapour is

where L_v is the heat of vapourisation and V_l and V_g, respectively denote the volume of liquid and gas phase. For low pressure V_g >> V_l

Using  , we have

For an ideal gas equation PV_g = RT, and thus

At critical point, pressure and temperature are P_c and T_c, respectively.

From (i) and (ii), we have

Derive Clausius-Clapeyron equation from Maxwell relation,

 ⇒ ⇒

∂Q = Ldm

Second Order Phase Transition

In some cases the state of matter does not change but the arrangement of its constituent particle changes. This kind of phase transition is known as second order phase transition.

In the case of second order phase transition no heat is evolved or absorbed. In second order phase transition-

1. Gibbs free energy is continuous

2. First order differential of Gibbs energy with respect to temperature ie entropy are changes smoothly

3. Second order differential of Gibbs energy with respect to temperature i.e., specific heat and second order differential of Gibbs energy with respect to pressure i.e. isothermal and isobaric expansivity have finite discontinuity at critical temperature.

4. The thermodynamic properties which are determined by more than second order derivative will be infinite at critical temperature