Question

The reduced temperature $=\theta =\frac{T}{T_{C}}$

The reduced pressure $=\pi =\frac{P}{P_{C}}$

The reduced volume $=\phi=\frac{V}{V_{C}}$

Hence, it can be said that the reduced equation of state may be given as

• A. $\left(\frac{\pi }{3} + \frac{1}{\left(\phi\right)^{2}}\right)\left(3 \phi - 1\right)=\frac{8}{3}\theta $
• B. $\left(\frac{\pi }{3} + \frac{1}{\phi}\right)\left(\phi - 1\right)=\frac{8}{3}\theta $
• C. $\left(\frac{\pi }{4} + \frac{1}{\phi}\right)\left(3 \theta - 1\right)=\frac{8}{3}\phi$
• D. $\left(\frac{\pi }{3} + \frac{1}{\phi}\right)\left(3 \phi - 1\right)=\frac{8}{3}\theta $

Answer 1

Answer: (A)

$P=P_{c}\pi =\left(\frac{a}{27 b^{2}}\right)\pi $
$V=V_{c}\phi=\left(3 b\right)\phi$
$T=T_{c}\theta =\left(\frac{8 a}{27 R b}\right)\theta $
Hence substituting in the van der Waal's equation
$ \Rightarrow\left(\pi+\frac{3}{\phi^{2}}\right)(3 \phi-1)=8 \theta $
This is reduced state equation
$ \begin{array}{l} \left(\pi+\frac{3}{\phi^{2}}\right)\left(\frac{3 \phi}{8 \theta}-\frac{1}{8 \theta}\right)=1 \\ \left(\frac{\pi}{3}+\frac{1}{\phi^{2}}\right)(3 \phi-1)=\frac{8}{3} \theta \end{array} $