Question

If a real gas is following equation $P(V-nb)=nRT$, at low pressure then find the intercept and slope of graph between $\frac{d}{p}v/sP$.

• A. $\frac{MR}{T},\frac{M(RT{)}^2}{b}$
• B. $\frac{M}{RT},-\frac{Mb}{(RT{)}^2}$
• C. $\frac{Mb}{RT},-\frac{M}{(RT{)}^2}$
• D. $\frac{RT}{M},-\frac{b}{M(RT{)}^2}$

Answer 1

The correct option is B $\frac{M}{RT},-\frac{Mb}{(RT{)}^2}$

$P(V-\frac{w}{M}b)=\frac{w}{M}RT$
$P[1-\frac{db}{M}]=\frac{dRT}{M}$
$P[M-db]=dRT$
$\frac{d}{P}=\frac{M}{RT}\times \frac{-db}{RT}\cdots \left(1\right)$
$\frac{d}{P}=y,P=x,\therefore d=yx$
$\therefore y=\frac{M}{RT}-\frac{b}{RT}\left(yx\right)$
$y[1+\frac{bx}{RT}]=\frac{M}{RT}\cdots \left(2\right)$
For y-intercept, $x=0$
$\therefore y=\frac{M}{RT}\cdots \left(3\right)$
For slope, $\frac{dy}{dx}=\frac{M}{RT}\left(\frac{1}{{(1+\frac{bx}{RT})}^2}\right)\left(\frac{b}{RT}\right)$
At low pressure, $P=0\Rightarrow x=0$
$\therefore$ Slope$=\frac{dy}{dx}=\frac{-Mb}{(RT{)}^2}$