Thermodynamic Potentials and their Relations with Thermodynamic Variables

Thermodynamic Potentials:

The thermodynamic state of a homogeneous system may be represented by means of certain selected variables, such as pressure P, volume V, temperature T and entropy S. Out of these four variables, any two may vary independently and when known enable the others to be determined. Thus there are only two independent variables and the others may be considered as their functions.

There exists certain relations between these thermodynamical variables.

From first law of thermodynamics, \(dQ=dU+dW\rightarrow 1\)

From second law of thermodynamics, \(dQ=TdS\rightarrow 2\)

Combining 1 and 2, we get

\begin{align*}TdS=dU+dW\end{align*}\begin{align*}\therefore dU=TdS-PdV(\because dW=PdV)\rightarrow 3 \end{align*}

This expresses the change in internal energy of the system in terms of thermodynamical variables. However for the complete knowledge of the system, certain other relations are required and for this purpose we introduce some functions of variables P,V,T and S, known as thermodynamical potentials or the thermodynamic functions.

There are four principal thermodynamic potentials:

(1)Internal Energy or Intrinsic Energy:

From equation (3), we have\begin{align*}dU=TdS-PdV \end{align*}Taking partial differentials of the internal energy w.r.to the variables S and V, we get\begin{align*}\left ( \frac{\partial U}{\partial S}\right )_V=T and\space\left (\frac{\partial U}{\partial V} \right )_S=-P \end{align*}These are the relations connecting the internal energy U with the thermodynamic variables S, V, T and P.

Since dU is a perfect differential, we must have\begin{align*}\frac{\partial}{\partial V} \left (\frac{\partial U}{\partial S} \right )_V=\frac{\partial}{\partial S} \left(\frac{\partial U}{\partial V} \right )_S \end{align*}\begin{align*}\left (\frac{\partial T}{\partial V} \right )_S=-\left (\frac{\partial T}{\partial S} \right )_V \end{align*} This result is the first thermodynamical relation of Maxwell.

(2)Helmholtz's Function:

From equation (3), we have\begin{align*}dU=TdS-PdV \end{align*}\begin{align*} dU-TdS=-dW[\because dW=PdV]\end{align*}\begin{align*}or,dU-d(TS)=-dW \end{align*}\begin{align*}or,d(U-TS)=-dW \end{align*}\begin{align*}or,dF=-dW \end{align*}Where F=U-TS is known as Helmholtz free energy or work function. It shows that for reversible changes the work done by system is equal to the decrease in this function F. \begin{align*}Now,\space dF=dU-d(TS) \end{align*}\begin{align*}=dU-TdS-SdT \end{align*}But dU=TdS-PdV\begin{align*}\therefore dF=TdS-PdV-TdS-SdT \end{align*}\begin{align*} dF=-PdV-SdT\end{align*}Here T and V are independent variables. Taking partial differential of F, \begin{align*}\left ( \frac{\partial F }{\partial V}\right )_T=-P\space and\space \left (\frac{\partial F }{\partial T } \right )_V=-S \end{align*}Since dF is a perfect differential\begin{align*} \frac{\partial }{\partial V}\left ( \frac{\partial F }{\partial V}\right )_T=\frac{\partial }{\partial T}\left (\frac{\partial F }{\partial T } \right )_V\end{align*}\begin{align*}or,\space -\left (\frac{\partial P}{\partial T } \right )_V =-\left (\frac{\partial S }{\partial V } \right )_T \end{align*}\begin{align*}\left (\frac{\partial P}{\partial T } \right )_V =\left (\frac{\partial S }{\partial V } \right )_T \end{align*}This is the second thermodynamic relation of Maxwell.

(3) Enthalpy or total heat (H):

It is a certain quantity of heat of a system which is expressed by the following equation,

H=U+PV

The differentiation of which yeilds \begin{align*}dH=dU+d(PV) \end{align*}\begin{align*} or,\space dH=dU+PdV+VdP\end{align*}\begin{align*}or,\space dH=TdS-PdV+PdV+VdP [from 3]\end{align*}\begin{align*}dH=TdS+VdP\rightarrow (4) \end{align*}Now taking partial differential of H treating S and P as independent variable.\begin{align*}\left (\frac{\partial H }{\partial S } \right )_P=\left (\frac{\partial H }{\partial P } \right )_S=V \end{align*}And since dH is a perfect differential\begin{align*}\frac{\partial}{\partial P}\left (\frac{\partial H }{\partial S } \right )_P=\frac{\partial}{\partial S}\left (\frac{\partial H }{\partial P } \right )_S \end{align*}\begin{align*}\left (\frac{\partial T }{\partial P} \right )_S=\left (\frac{\partial V }{\partial S} \right )_P \end{align*}This is the third thermodynamical relation of Maxwell.

(4) Gibb's potential G:

From equation (4),dH=TdS+VdP

When the process is isothermal, TdS=d(TS)

and if the process is isobaric, dP=0

Hence for the process having both isothermal as well as isobaric,\begin{align*} dH=d(TS)+0\end{align*}\begin{align*}or,\space dH-d(TS)=0 \end{align*}\begin{align*}or,\space d(H-TS)=0 \end{align*}\begin{align*}or,\space dG=0 \end{align*}where G=H-TS=constant, known as Gibb's Potential.\begin{align*}or,\space G=U+PV-TS \end{align*}\begin{align*} G=U-TS+PV\rightarrow (5)\end{align*}\begin{align*}\therefore G=F+PV[\because U-TS=F] \end{align*}This is known as thermodynamic potential at constant pressure or Gibb's function.

Differentiation of equation (5) gives\begin{align*}dG=dU-d(TS)+d(PV) \end{align*}\begin{align*}=dU-TdS-SdT+PdV+VdP \end{align*}\begin{align*}But\space dU=-PdV+TdS \end{align*}\begin{align*}\therefore dG=TdS-PdV-TdS-SdT+PdV+VdP \end{align*}\begin{align*}or,\space dG=VdP-SdT \end{align*}NOw, taking partial derivative of G treating P and T as independent variables,\begin{align*}\left (\frac{\partial G }{\partial P} \right )_T=V\space and\space\left (\frac{\partial G }{\partial T} \right )_P =-S\end{align*}Since dG is perfect differential,\begin{align*}\frac{\partial}{\partial T} \left (\frac{\partial G }{\partial P} \right )_T=\frac{\partial}{\partial P}\left (\frac{\partial G }{\partial T} \right )_P\end{align*}\begin{align*}\left (\frac{\partial V}{\partial T} \right )_P=-\left (\frac{\partial S }{\partial P } \right )_T \end{align*}This is the fourth thermodynamical relation of Maxwell.

Thus, we see that the four thermodynamical potential U(S,V); F(T,V); H(S,P) and G(T,P) lead us to four thermodynamical relation known as Maxwell relation.

Clausius Clapeyron's Equation from Maxwell's Thermodynamical Relation:

onsider Maxwell's second thermodynamical relation\begin{align*}\left (\frac{\partial P}{\partial T } \right )_V =\left (\frac{\partial S }{\partial V } \right )_T \end{align*}Multiplying both sides by T, we have, \begin{align*}T\left (\frac{\partial P}{\partial T } \right )_V =T\left (\frac{\partial S }{\partial V } \right )_T \end{align*}But TdS=dQ(from the second law of thermodynamics)\begin{align*}\therefore \left (\frac{\partial Q }{\partial V } \right )_T=T\left (\frac{\partial P}{\partial T } \right )_V\end{align*}

Here \(\frac{\partial Q}{\partial V}\) represents the quantity of heat absorbed or liberated per unit change in volume at constant tempertaure. The quantity of heat absorbed or liberated at constant temperature must be the latent heat and the change in volume due to change of state.

Consider a unit mass of substance. Let L be the latent heat when the substance changes in volume from V₁ to V₂ at constant temperature, then\begin{align*}\partial Q=L, \space dV=V_2-V_1 \end{align*}\begin{align*}\therefore \left (\frac{L }{V_2-V_1 } \right )_T= T\left (\frac{\partial P}{\partial T } \right )_V\end{align*}\begin{align*}\frac{L }{V_2-V_1}=T\frac{\partial P}{\partial T } \end{align*}\begin{align*}or,\space \frac{\partial P}{\partial T }=\frac{L }{T(V_2-V_1) } \end{align*} This is known as Clausius-Clapeyron latent heat equation.

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.