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

Answer 1

The correct option is D $(\frac{\pi }{3}+\frac{1}{{\varphi }^2})(3\varphi -1)=\frac{8}{3}\theta$

the reduced temperature, $\theta =\frac{T}{T_C}$,the reduced pressure $\pi =\frac{F}{F_C}$, 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 $P_c,V_c,T_c$ by van der Waal's constant.
$(\pi \frac{a}{27b^2}+\frac{a}{{\varphi }^{2}9b^2})(\varphi 3b-b)=R\theta \frac{8a}{27Rb}$
solving:
$(\pi +\frac{3}{{\varphi }^2})(3\varphi -1)=8\theta$
Thus, it can be said that the reduced equation of state may be given as:
$(\frac{\pi }{3}+\frac{1}{{\varphi }^2})(3\varphi -1)=\frac{8}{3}\theta$
Hence, the correct answer is option D.