Question

A general from of equation of state gases is $PV=RT[A+B/V+C/V^2+]$, where $V$ is the molar volume of the gas and $A,B,C$ are constants for the gas. The value of $A$ and $B$, if the gas obeys van der Waals equation, are respectively:

• A. $1,\frac{a}{27b}$
• B. $1,b-\frac{a}{RT}$
• C. $\frac{a}{RT},b$
• D. $\frac{2a}{RT},b$

Answer 1

Answer: (B)

$1,b-\frac{a}{RT}$

The correct option is $B$.

$\left(P+\frac{a}{v^2}\right)(V-b)=RT$ Van der Waals equation after correction.

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

$\frac{PV}{RT}=\frac{1}{1-\frac{b}{v}}-\frac{a}{v}$

$\frac{PV}{RT}=(1-\frac{b}{v}{)}^{-1}-\frac{a}{v}$

$\frac{PV}{RT}=1+\frac{b}{v}+(\frac{b}{v}{)}^2+(\frac{b}{v}{)}^3+...............-\frac{a}{v}$

$\because (1-x{)}^{-1}=1+x+x^2+x^3+...........$

$=1+\frac{1}{v}(b-\frac{a}{RT})+(\frac{b}{v}{)}^2+(\frac{b}{v}{)}^3+.............$

$\therefore$ Comparing with given equation we get

$PV=RT[A+\frac{B}{V}+\frac{C}{V^2}+.............]$

$A=1,B=(b-\frac{a}{RT})$