Chapter 7, Problem 7.13P

(a)

Interpretation Introduction

Interpretation:

To prove that $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$ and apply it to Van der Waals equation

Concept Introduction:

Using, $P=\rho ZRT$ to prove $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$

By Joule Thompson inversion curve, ${\left(\frac{\partial Z}{\partial T}\right)}_P=0$

Then applying it to General equation which is given as,

$P=\frac{RT}{V-b}-\frac{a(T)}{\left(V+\in b\right)\times \left(V+\sigma b\right)}$

Answer to Problem 7.13P

  $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$ Hence Proved

$Z=\frac{1}{1-\xi }-q\xi$ (A)

$\xi =1-\frac{1}{\sqrt{2q}}$ (B)

$P_r=\frac{Z\xi T_r}{\Omega }$ (C)

For a given T_r, we can find q and then use equation (B) to find $\xi$

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

${\left(\frac{\partial x}{\partial y}\right)}_z={\left(\frac{\partial x}{\partial y}\right)}_w+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z$

${\left(\frac{\partial Z}{\partial T}\right)}_p={\left(\frac{\partial Z}{\partial T}\right)}_{\rho }+{\left(\frac{\partial Z}{\partial p}\right)}_T{\left(\frac{\partial \rho }{\partial T}\right)}_P$

We get,${\left(\frac{\partial Z}{\partial T}\right)}_{\rho }={\left(\frac{\partial Z}{\partial T}\right)}_P-{\left(\frac{\partial Z}{\partial \rho }\right)}_p{\left(\frac{\partial \rho }{\partial T}\right)}_P$

Now, $P=\rho ZRT$ i.e. $\rho =\frac{P}{ZRT}$

${\left(\frac{\partial \rho }{\partial T}\right)}_P=\frac{P}{R}\left\{\frac{-1}{{\left(ZT\right)}^2}\left[Z+T{\left(\frac{\partial Z}{\partial T}\right)}_P\right]\right\}$

Substituting ${\left(\frac{\partial Z}{\partial T}\right)}_P=0$ in each of the two preceding equations reduces them to:

  ${\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=-{\left(\frac{\partial Z}{\partial \rho }\right)}_T{\left(\frac{\partial \rho }{\partial T}\right)}_P$

  ${\left(\frac{\partial \rho }{\partial T}\right)}_P=-\frac{P}{ZR{T}^2}=-\frac{\rho }{T}$

Combing these two equations yields:

  $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$ Hence Proved

The general equation is as follows,

$P=\frac{RT}{V-b}-\frac{a(T)}{\left(V+\in b\right)\times \left(V+\sigma b\right)}$

With Van der Waals parameters becomes:

$P=\frac{RT}{V-b}-\frac{a}{V^2}$

Multiply the above equation by V/RT,

$\frac{PV}{RT}=\frac{V}{V-b}-\frac{a}{VRT}$

substitute Z = PV/RT and $V=1/\rho$,

$Z=\frac{1}{1-b\rho }-\frac{a\rho }{RT}$

Assume, $q=\frac{a}{bRT}$ and define $\xi =b\rho$, we get,

$Z=\frac{1}{1-\xi }-q\xi$ (A)

Differentiating the above equation, we get

${\left(\frac{\partial Z}{\partial T}\right)}_{\rho }={\left(\frac{\partial Z}{\partial T}\right)}_{\xi }=-\xi \frac{dq}{dT}$

By $q=\frac{\psi a\left(T_r\right)}{\Omega T_r}$ with $\alpha (T_r)=1$ for the van der waals equation, $q=\frac{\psi }{\Omega T_r}$ We get,

$\frac{dq}{dT}=\frac{\psi }{\Omega }\left(\frac{-1}{T_r^2}\right)\frac{d{T}_r}{dT}=-\frac{\psi }{\Omega }\frac{1}{T_r^2{T}_c}=-\frac{\psi }{\Omega }\frac{1}{T{T}_r}=-\frac{q}{T}$

Then,

${\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\left(-\xi \right)\left(-\frac{q}{T}\right)=\frac{q\xi }{T}$

In addition,

${\left(\frac{\partial Z}{\partial \rho }\right)}_T=b{\left(\frac{\partial Z}{\partial \xi }\right)}_T=\frac{b}{{(1-\xi )}^2}-qb$

Substitute for the two partial derivatives in the boxed equation:

$T\frac{q\xi }{T}=\frac{b\rho }{{\left(1-\xi \right)}^2}-qb\rho$ or

$q\xi =\frac{\xi }{{\left(1-\xi \right)}^2}-q\xi$

hence,

$\xi =1-\frac{1}{\sqrt{2q}}$ (B)

Now, $b=\Omega \frac{R{T}_C}{P_C}$

$P_c=\Omega \frac{R{T}_c}{b}$

(I)

$P=Z\rho RT$

(II)

Division of (II) by (I) gives

$P_r=\frac{P}{P_C}=\frac{Z\rho bT}{\Omega T_c}$

$P_r=\frac{Z\xi T_r}{\Omega }$ (C)

For a given T_r, we can find q and then use equation (B) to find $\xi$

The Equation (A) we can find Z and then P_r with equation (C)

(b)

Interpretation Introduction

Interpretation:

To prove that $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$ and apply it to Redlich Kwong equation

Concept Introduction:

Using, $P=\rho ZRT$ to prove $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$

By Joule Thompson inversion curve, ${\left(\frac{\partial Z}{\partial T}\right)}_P=0$

Then applying it to General equation which is given as,

$P=\frac{RT}{V-b}-\frac{a(T)}{\left(V+\in b\right)\times \left(V+\sigma b\right)}$

Answer to Problem 7.13P

  $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$ Hence Proved

$Z=\frac{1}{1-\xi }-\frac{q\xi }{1+\xi }$ (A)

$q={\left(\frac{1+\xi }{1-\xi }\right)}^2\left(\frac{1}{2.5+1.5\xi }\right)$ (B)

$P_r=\frac{Z\xi T_r}{\Omega }$ (C)

For a given T_r, we can find q and then use equation (B) to find $\xi$

The Equation (A) we can find Z and then P_r with equation (C)

Explanation of Solution

${\left(\frac{\partial x}{\partial y}\right)}_z={\left(\frac{\partial x}{\partial y}\right)}_w+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z$

${\left(\frac{\partial Z}{\partial T}\right)}_p={\left(\frac{\partial Z}{\partial T}\right)}_{\rho }+{\left(\frac{\partial Z}{\partial p}\right)}_T{\left(\frac{\partial \rho }{\partial T}\right)}_P$

We get,${\left(\frac{\partial Z}{\partial T}\right)}_{\rho }={\left(\frac{\partial Z}{\partial T}\right)}_P-{\left(\frac{\partial Z}{\partial \rho }\right)}_p{\left(\frac{\partial \rho }{\partial T}\right)}_P$

Now, $P=\rho ZRT$ i.e. $\rho =\frac{P}{ZRT}$

${\left(\frac{\partial \rho }{\partial T}\right)}_P=\frac{P}{R}\left\{\frac{-1}{{\left(ZT\right)}^2}\left[Z+T{\left(\frac{\partial Z}{\partial T}\right)}_P\right]\right\}$

Substituting ${\left(\frac{\partial Z}{\partial T}\right)}_P=0$ in each of the two preceding equations reduces them to:

  ${\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=-{\left(\frac{\partial Z}{\partial \rho }\right)}_T{\left(\frac{\partial \rho }{\partial T}\right)}_P$

  ${\left(\frac{\partial \rho }{\partial T}\right)}_P=-\frac{P}{ZR{T}^2}=-\frac{\rho }{T}$

Combing these two equations yields:

  $T{\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\rho {\left(\frac{\partial Z}{\partial \rho }\right)}_T$ Hence Proved

The general equation is as follows,

$P=\frac{RT}{V-b}-\frac{a(T)}{\left(V+\in b\right)\times \left(V+b\right)}$

With Redlich Kwong parameters becomes:

$P=\frac{RT}{V-b}-\frac{a}{V(V+b)}$

Multiply the above equation by V/RT,

$\frac{PV}{RT}=\frac{V}{V-b}-\frac{a}{RT(V+b)}$

substitute Z = PV/RT and $V=1/\rho$,

$Z=\frac{1}{1-b\rho }-\frac{ab\rho }{bRT(1+b\rho )}$

Assume, $q=\frac{a}{bRT}$ and define $\xi =b\rho$, we get,

$Z=\frac{1}{1-\xi }-\frac{q\xi }{1+\xi }$ (A)

By $q=\frac{\psi a\left(T_r\right)}{\Omega T_r}$ with $\alpha (T_r)=T_r^{-0.5}$ for the Redlich/Kwong equation, $q=\frac{\psi }{\Omega T_r^{1.5}}$ .

We get,

$\frac{dq}{dT}=-\frac{1.5q}{T}$and ${\left(\frac{\partial Z}{\partial T}\right)}_{\rho }=\frac{1.5q\xi }{T\left(1+\xi \right)}$

${\left(\frac{\partial Z}{\partial \rho }\right)}_T=\frac{b}{{\left(1-\xi \right)}^2}-\frac{bq}{{\left(1+\xi \right)}^2}$

Substitution into the above equation, we get

$q={\left(\frac{1+\xi }{1-\xi }\right)}^2\left(\frac{1}{2.5+1.5\xi }\right)$ (B)

Now, $b=\Omega \frac{R{T}_C}{P_C}$

$P_c=\Omega \frac{R{T}_c}{b}$

(I)

$P=Z\rho RT$

(II)

Division of (II) by (I) gives

$P_r=\frac{P}{P_C}=\frac{Z\rho bT}{\Omega T_c}$

$P_r=\frac{Z\xi T_r}{\Omega }$ (C)

For a given T_r, we can find q and then use equation (B) to find $\xi$

The Equation (A) we can find Z and then P_r with equation (C)