Question 24.1: Show that if μ = G(T, P) = λ(T) + RTlnP for a one component ...
Show that if \mu= G ( T , P )=\lambda( T )+ RT \ln P for a one component gas, then, the gas is ideal because it satisfies the two requirements below:
(i) PV = RT
(ii) U = U(T) and H = H(T)
Choosing T and P as the two independent intensive variables for a one component gas, it follows that G = G(T,P). In addition, it follows that dG = -SdT + VdP, and therefore, that:
\left(\frac{\partial G }{\partial P }\right)_{ T }= V (\text { General result for } n =1) (24.1)
Using \mu= G ( T , P )=\lambda( T )+ RT \ln P in Eq. (24.1), it follows that:
\left(\frac{\partial G }{\partial P }\right)_{ T }=0+\frac{ RT }{ P }=\frac{ RT }{ P } (24.2)
Equating Eqs. (24.1) and (24.2), including rearranging, yields:
PV = RT \text { (Requirement (i) is satisfied) } (24.3)
Next, we show that H = H(T). Recall that:
G = H – TS \Rightarrow \frac{ G }{ T }=\frac{ H }{ T }- S (24.4)
Taking the temperature partial derivative, at constant pressure, of the expression on the right-hand side of the arrow in Eq. (24.4) yields:
\frac{\partial}{\partial T }\left(\frac{ G }{ T }\right)_{ P }=\frac{\partial}{\partial T }\left(\frac{ H }{ T }\right)_{ P }-\left(\frac{\partial S }{\partial T }\right)_{ P } (24.5)
The first partial derivative on the right-hand side of Eq. (24.5) is given by:
\frac{\partial}{\partial T }\left(\frac{ H }{ T }\right)_{ P }=\frac{1}{ T }\left(\frac{\partial H }{\partial T }\right)_{ P }-\left(\frac{ H }{ T ^{2}}\right) (24.6)
The second partial derivative on the right-hand side of Eq. (24.5) is given by:
\left(\frac{\partial S }{\partial T }\right)_{ P }=\frac{ C _{ P }}{ T }=\frac{1}{ T }\left(\frac{\partial H }{\partial T }\right)_{ P } (24.7)
Using Eqs. (24.6) and (24.7) in Eq. (24.5), including cancelling the equal terms, yields:
\frac{\partial}{\partial T}\left(\frac{ G }{ T }\right)_{ P }=-\frac{ H }{ T ^{2}} (24.8)
Equation (24.8) is known as the Gibbs-Helmholtz equation, and is a general relation between G and H for n = 1. If we know G(T,P), Eq.(24.8) allows us to compute H(T,P) by differentiation. Alternatively, if we know H(T,P), Eq. (24.8) allows us to compute G(T,P) by integration.
In extensive form, the Gibbs-Helmholtz equation is expressed as follows:
\frac{\partial}{\partial T }\left(\frac{\underline{ G }}{ T }\right)_{ P , N }=-\frac{\underline{ H }}{ T ^{2}} (24.9)
According to the Problem Statement, for a one component (n = 1) Ideal Gas, \mu= G ( T , P )=\lambda( T )+ RT \ln P, and therefore:
\frac{ G }{ T }=\frac{\lambda( T )}{ T }+ R \ln P (24.10)
Taking the partial derivative of Eq. (24.10) with respect to temperature, at constant pressure, yields:
\frac{\partial}{\partial T }\left(\frac{ G }{ T }\right)_{ P }=\frac{\partial}{\partial T }\left(\frac{\lambda( T )}{ T }\right)+0 (24.11)
Using Eq. (24.11) in Eq. (24.8), including rearranging, yields:
H =- T ^{2} \frac{\partial}{\partial T }\left(\frac{ G }{ T }\right)_{ P }=- T ^{2} \frac{\partial}{\partial T }\left(\frac{\lambda( T )}{ T }\right) (24.12)
Equation (24.12) shows that H is only a function of temperature. In addition, because U = H – PV = H(T) – RT, it follows that U = U(T). To derive the last result, we used the fact that, for an ideal gas, PV = RT.