Question

Given that the reduced temperature, $\theta =\frac{T}{T_C}$ the reduced pressure, $\pi =\frac{F}{F_C}$ the reduced volume, $\varphi =\frac{V}{V_C}$. Thus, it can be said that the reduced equation of state may be given as :

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

Answer 1

Answer: (D)

$(\frac{\pi }{3}+\frac{1}{{\phi }^2})(3\varphi -1)=\frac{8}{3}\theta$

The reduced temperature $\theta =\frac{T}{T_C}$

The reduced pressure $\pi =\frac{P}{P_C}$

The reduced volume $\varphi =\frac{V}{V_C}$
The van der waal's equation is:

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

Substitute the values of $P,V$ and $T$ in the van der Waal's equation and replace the terms $P_c,V_c$ and $T_c$ by van der Waal's constant
$(\pi \frac{a}{27b^2}+\frac{a}{{\varphi }^{2}9b^2})(\varphi 3b-b)=R\theta \frac{8a}{27Rb}$
$(\pi a+\frac{3a}{{\varphi }^2})(3b\varphi -b)=8ab\theta$
$(\pi +\frac{3}{{\varphi }^2})(3\varphi -1)=8\theta$
Thus, it can be said that the reduced equation of state may be given as:

$(\frac{\pi }{3}+\frac{1}{{\phi }^2})(3\varphi -1)=\frac{8}{3}\theta$

Hence, the correct answer is option $\text{D}$.