Question

If $\frac{P}{P_c}=x$ (reduced pressure), $\frac{V}{V_c}=y$ (reduced volume) and $\frac{T}{T_c}=z$ (reduced temperature), then,

• A. $(x-\frac{3}{y^2})(3y-1)=8z$
• B. $(x+\frac{3}{y^2})(y-1)=z$
• C. $(x-\frac{3}{y^2})(3y+1)=8z$
• D. $(x+\frac{3}{y^2})(3y-1)=8z$

Answer 1

The correct option is D $(x+\frac{3}{y^2})(3y-1)=8z$

As we know,
$\text{Critical pressure,}{P}_c=\frac{a}{27b^2}$
$\text{Critical volume,}{V}_c=3b$
$\text{Critical temperature,}{T}_c=\frac{8a}{27Rb}$
For one mole of a real gas
$(P+\frac{a}{V^2})(V-b)=RT...\left(i\right)$
$\frac{P}{P_c}=x\left(\text{given}\right)$
$\therefore P=x{P}_c=\frac{ax}{27b^2}$
$\frac{V}{V_c}=y\left(\text{given}\right)$
$\therefore V=y{V}_c=3by$
$\frac{T}{T_c}=z\left(\text{given}\right)$
$T=z{T}_c=\frac{8az}{27Rb}$
Putting values of $P,V$ and $T$ into Equation $\left(i\right)$, we get
$(\frac{ax}{27b^2}+\frac{a}{{\left(3by\right)}^2})(3by-b)=R\times \frac{8az}{27Rb}$
$(\frac{ax}{27b^2}+\frac{a}{9b^2{y}^2})(3by-b)=R\times \frac{8az}{27Rb}$
$\frac{a}{27b^2}(x+\frac{3}{y^2})\times b(3y-1)=R\times \frac{8az}{27Rb}$
$(x+\frac{3}{y^2})(3y-1)=8z$