The compressibility factor (Z) of one mole of a van der Waals gas of negligible ‘a’ value is

Question

The compressibility factor (Z) of one mole of a van der Waals gas of negligible ‘a’ value is (A) 1 (B) (bp/RT) (C) 1+(bp/RT) (D) 1-(bp/RT)

Tardigrade Admin · Accepted Answer

van der Waals' equation is (p+(a/V2))(V-b)=R T (for 1 mole) If a is negligible, p+(a/V2) ≈ p p(V-b)=R T ⇒ p V-p b=R T or (p V/R T)-(p b/R T)=1 ⇒ (p V/R T)=1+(p b/R T)

Q. The compressibility factor $(Z)$ of one mole of a van der Waals gas of negligible $‘ a ’$ value is

$1$

$\frac{b p}{RT}$

$1 + \frac{b p}{RT}$

$1 - \frac{b p}{RT}$

van der Waals' equation is
$(p + \frac{a}{V ^{2}}) (V - b) = RT$ (for 1 mole)
If $a$ is negligible, $p + \frac{a}{V ^{2}} \approx p$
$p (V - b) = RT$
$\Rightarrow p V - p b = RT$
or $\frac{p V}{RT} - \frac{p b}{RT} = 1$
$\Rightarrow \frac{p V}{RT} = 1 + \frac{p b}{RT}$

Q. The compressibility factor (Z) of one mole of a van der Waals gas of negligible ‘a’ value is

1

RTbp​

1+RTbp​

1−RTbp​

van der Waals' equation is (p+V2a​)(V−b)=RT(for 1 mole) If a is negligible, p+V2a​≈p p(V−b)=RT ⇒pV−pb=RT or RTpV​−RTpb​=1 ⇒RTpV​=1+RTpb​

Q. The compressibility factor $(Z)$ of one mole of a van der Waals gas of negligible $‘ a ’$ value is

$1$

$\frac{b p}{RT}$

$1 + \frac{b p}{RT}$

$1 - \frac{b p}{RT}$

van der Waals' equation is
$(p + \frac{a}{V ^{2}}) (V - b) = RT$ (for 1 mole)
If $a$ is negligible, $p + \frac{a}{V ^{2}} \approx p$
$p (V - b) = RT$
$\Rightarrow p V - p b = RT$
or $\frac{p V}{RT} - \frac{p b}{RT} = 1$
$\Rightarrow \frac{p V}{RT} = 1 + \frac{p b}{RT}$