Clausius-Clapeyron's Equation, Second Latent Heat Equation

CLAUSIUS-CLAPEYRON'S EQUATION

The melting point or boiling point have got a specific value at a specific pressure and vice versa.

It is possible to obtain a relation showing how the melting points and boiling points vary with pressure, by applying the second law of thermodynamics. The relation thus obtained is called Clausius-Clapeyron's equation or the first latent heat equation.

Let ABCD and EFGH represent two isothermals at infinitely close temperature T and (T+dT) respectively. The parts AB and EF represent to the liquid state of the substance. At B and F, substance is purely in liquid state. Along BC and FG, the change of state is in progress and the liquid and vapour state co-exists in equilibrium. At C and G the substance is purely in vapour state. From C to D and G to H the substance is entirely in vapour state.

Let P and P+dP be the saturated vapour pressure of liquid at temperature T and T+dT respectively. Let V₁ and V₂ be the volume of substance at F and G respectively.

The amount of heat Q₁ absorbed along FG is equal to the latent heat of vapourization (L+dL) at tempertaure (T+dT) as liquid changing from liquid state at F to the vapour state of G. From C to B, the amount of heat rejected is Q₂. HereQ₂=L where L is the latent heat at temperature T. By increasing the pressure a little, the original point F is restored. So the cycle FGCBF is completely reversible. Then applying the principle of Carnot's reversible cycle,

\begin{align*}\frac{Q_1}{T_1}=\frac{Q_2}{T_2} \end{align*}\begin{align*}or,\space \frac{Q_1}{Q_2}-1=\frac{T_1}{T_2}-1\end{align*}\begin{align*}or,\space \frac{Q_1-Q_2}{Q_2}=\frac{T_1-T_2}{T_2} \end{align*}we have, Q₁=L+dL; Q₂=L and T₁=T+dT; T₂=T\begin{align*}\therefore \frac{L+dL-L}{L}=\frac{T+dT-T}{T} \end{align*}\begin{align*}or,\space \frac{dL}{L}=\frac{dT}{T}\rightarrow (1) \end{align*}The amount of heat converted into work during cycle FGCBF \(Q_1-Q_2=L\)

But the work done during carnot cycle is given by area of FGCBF,\begin{align*}i.e.\space dL=FG\times perpendicular\space distance\space between\space FG\space and\space CB \end{align*}\begin{align*}=(V_2-V_1)dP \end{align*}Substituting this value in (1), we get\begin{align*}\frac{(V_2-V_1)dP}{L}=\frac{dT}{T} \end{align*}\begin{align*}\therefore \frac{dP}{dT}=\frac{L}{T(V_2-V_1)}\end{align*}

This is called Clausius-Clapeyron's latent heat equation. This is true for both the change of state i.e., from liquid to vapour and solid to liquid. In former case L represents the latent heat of vapourization while in later case it represents the latent heat of fusion.

Application of Clausius-Clapeyron's latent heat equation:

(1) Effect of pressure on boiling point of the liquid:

When a liquid boils, there is increase in volume. So thet \(V_2>V_1\) or \(V_2-V_1\) is the positive. Hence in above relation \(\frac{dP}{dT}\) is a positive quantity, which means the boiling point of liquid rises with increase in pressure and vice versa.

(2) Effect of pressure on melting point of solid:

When solid melts, there may be increase or decrease in volume.

(a)When \(V_2>V_1\)(example:Wax),\(\frac{dP}{dT}\) is a positive quantity, which means the melting point of such substances increases with increase in pressure.

(b) When\(V_2<V_1\)(example:ice, gallium, bismuth, etc.),\(\frac{dP}{dT}\) is a negative quantity, which means the melting point of such substances decreases with increase in pressure.

SECOND LATENT HEAT EQUATION

It gives the variation of latent heat of a substance with temperature and connects it with specific heat of substances in two states. Let C₁ denote the specific heat of a liquid in contact with its vapour and C₂ is the specific heat of saturated vapour it contacts with its liquid.

Let us consider 1 gm of substance is taken round the cycle BFGCB. The quantity of heat absorbed by the substance in passing from B to F, when its temperature rises by dT isC₁dT. In passing along FG, when substances changes from liquid to vapour at constant temperature T+dT it absorbs a quantity of heat L+dL. In passing from G to G, the temperature of substance falls by dT and hence it gives out a quantity of heatC₂dT. While passing along CB, when it condenses from vapour to liqui at constant temperature T, gives out a quantity of heat L.

\(\therefore \)The total amount of heat absorbed during the cycle is\begin{align*}C_1dT+L+dL-C_2dT-L=(C_1-C_2)dT+dL \end{align*}This must be equal to the workdone which is equal to the area of cycle \(=dP(V_2-V_1)=\frac{L}{T}dT\)\begin{align*}\therefore(C_1-C_2)dT+dL= \frac{L}{T}dT\end{align*}\begin{align*}or,\space (C_1-C_2)dT= \frac{L}{T}dT-dL \end{align*}\begin{align*}or,\space C_1-C_2=\frac{L}{T}-\frac{dL}{dT} \end{align*}\begin{align*}or,\space C_2-C_1=\frac{dL}{dT}-\frac{L}{T} \end{align*}

This is the Second Latent Heat Equation.

TRIPLE POINT

The boiling point of water increase with increse in pressure and vice versa. The curve AB represents the relation between pressure and temperature and is called steam line or vapourization line. The liquid and vapour are in stable equilibrium along the line AB. At all points above AB, the substance is all liquid and below it there exists vapour only.

The melting point of ice decreases with increase in pressure i.e., ice melts below \(0^{\circ}\) at a pressure higher than atmospheric pressure. The relationship between pressure and temperature can be represented by a curve CD caled ice line or fusion line. In the figure, the curve C'D' is also shown for wax type of substance whose melting point is increasesd with increase in pressure. The substance is entirely solid on the left of the curve while entirely liquid on the right.

When the pressure on ice is raised, evaporation from its surface slows down. The equilibrium between the solid and vapour states of a substance can be represented by a curve EF called Hoar frost line or sublimation line. Above the curve EF, the substance is all solid and below it all vapour.

These curves when plotted on the same diagram meet in a single point. This point is is called triple point. So at triple point, the pressure and temperature are such that the solid, liquid and vapour states of a substance can co-exist in equilibrium.

In case of water, triple point has been found at \(T=0.0075^{\circ}C\) and P=4.56 mmHg.

To show that there is only one triple point:

Suppose three points do not meet at a point but intersect enclosing an area ACF shown by shaded portion. According to ice line CD, the substance must be entirely solid in the shaded region the substance must be entirely solid as it is left of CD. According to the steam line AB, the substance must be liquid as it is above AB and according to hoar frost line EF the substance in the shaded portion must be vapour as it is below EF. But these three conclusions contradict one another and hence the shaded triangle ACF cannot exist. These three curves should meet in a single point called triple point. The triple point of different substances are different.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Sears, Salinger. Thermodynamics, Kinetic Theory and Statistical Thermodynamics. New Delhi: Narosa Publishing House Pvt. Ltd., 2013.

Singhal, SS, JP Agrawal and Satya Prakash. Heat Thermodynamics and Statistical Physics. Meerut: Pragati Prakashan, 2009.