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I'm trying to fill in the steps from my textbook on the derivation of enthalpy departure from ideal gas behavior. My textbook gives the variation of mass-specific enthalpy with temperature pressure as $$\left ( \frac{\partial h}{\partial p} \right )_T=v-T\left ( \frac{\partial v}{\partial T} \right )_p$$

Integrating from pressure $p'$ to $p$ at fixed temperature $T$, $$h(T,p)-h(T,p')=\int_{p'}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$

If we use the superscript $^*$ to denote ideal gas property values, adding and subtracting $h^*(T)$ from the left hand side of the equation gives $$\left [h(T,p)-h^*(T) \right ]-\left [h(T,p')-h^*(T) \right ]=\int_{p'}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$

By the assumptions of the ideal gas model, we have

$$\lim_{p'\rightarrow 0}\left [h(T,p')-h^*(T) \right ]=0$$

and in this limit, the following expression is obtained $$h(T,p)-h^*(T)=\int_{0}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$ This can be thought of as the change in specific enthalpy as the pressure is increased from zero to the given pressure isothermally. This equation can be evaluated with $pvT$ data.

The integral in the last equation can be expressed in terms of the compressibility factor $Z$ and the reduced temperature $T_R$ and reduced pressure $p_R$. Solving $Z=pv/RT$ gives

$$v=\frac{ZRT}{p}$$

On differentiation,

$$\left ( \frac{\partial v}{\partial T} \right )_p=\frac{RZ}{p}+\frac{RT}{p}\left ( \frac{\partial Z}{\partial T} \right )_p$$

$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=\frac{ZRT}{p}-T\left [ \frac{RZ}{p}+\frac{RT}{p}\left ( \frac{\partial Z}{\partial T} \right )_p \right ]=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p$$

This equation can be written in terms of the reduced properties $T_R$ ad $p_R$ as

$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=-\frac{RT_c}{p_c}\cdot \frac{T_R^2}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}$$

Introducing this last equation into the equation

$$h(T,p)-h^*(T)=\int_{0}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$

gives, on rearrangement,

$$\frac{h^*(T)-h(T,p)}{RT_c}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}\frac{dp_R}{p_R}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d\ln p_R$$

What I'm failing to understand is how we get from the equation

$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=\frac{ZRT}{p}-T\left [ \frac{RZ}{p}+\frac{RT}{p}\left ( \frac{\partial Z}{\partial T} \right )_p \right ]=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p$$

to the equation

$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=-\frac{RT_c}{p_c}\cdot \frac{T_R^2}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}$$

and then to the equation

$$\frac{h^*(T)-h(T,p)}{RT_c}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}\frac{dp_R}{p_R}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d\ln p_R$$

In a couple books I've looked at, the algebraic and calculus manipulations are just glossed over and ignored. I'm hoping someone can help me fill in the steps.

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    $\begingroup$ Why is enthalpy and volume denoted by small letters? h is usually considered as height, enthalpy is H. Similarly, v is usually velocity or speed, volume is V. $\endgroup$
    – Poutnik
    Feb 14, 2022 at 5:52
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    $\begingroup$ I see, you are right, it is used like that. But it is strange seeing that, I would expect rather H_s and V_s. $\endgroup$
    – Poutnik
    Feb 14, 2022 at 6:33
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    $\begingroup$ The first step you are missing seems trivial, just substitute $T=T_RT_c$ and same for p and remember that $T_c$ can be treated as a constant. $\endgroup$
    – Buck Thorn
    Feb 14, 2022 at 6:47
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    $\begingroup$ It seems very straightforward. Is it the algebra that you are having trouble with? $\endgroup$ Feb 14, 2022 at 10:24
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    $\begingroup$ It looks right. Use the following general identity for k=constant: $\left ( \frac{\partial f}{\partial( kx)}\right )_y = \frac1k \left ( \frac{\partial f}{\partial( x)}\right )_y $ $\endgroup$
    – Buck Thorn
    Feb 14, 2022 at 15:31

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Conclusion:

$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p=-\frac{RT_R^2T_c^2}{p_Rp_c}\left ( \frac{\partial Z}{\partial( T_RT_c)}\right )_p$$

$$\left( \frac{\partial Z}{\partial( T_RT_c)}\right )_p=\frac{1}{T_c}\left( \frac{\partial Z}{\partial( T_R)}\right )_p=\frac{1}{T_c}\left( \frac{\partial Z}{\partial T_R}\right )_{p_R}$$ $$-\frac{RT_R^2T_c^2}{p_Rp_c}\left ( \frac{\partial Z}{\partial( T_RT_c)}\right )_p=-\frac{RT_c}{p_c}\cdot \frac{T^2_R}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}$$

$$h(T,p)-h^*(T)=\int_{0}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp=-\int_{0}^{p_R}\frac{RT_c}{p_c}\cdot \frac{T_R^2}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d(p_Rp_c)$$

$$\Rightarrow \frac{p_c\left [h^*(T)-h(T,p) \right ]}{RT_c}=\frac{T_R^2}{p_R}\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}p_cdp_R$$

$$\Rightarrow\frac{h^*(T)-h(T,p)}{RT_c}= T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}\frac{dp_R}{p_R}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d\ln p_R$$

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