Question

The compressibility factor for 1 mole of a Vander Waal's gas Boyle temperature is

• A. 1+$\frac{b^2}{v(v-b)}$
• B. 1-$\frac{b}{v}$
• C. 1+$\frac{b}{v}$
• D. 1-$\frac{b^2}{v^2}$

Answer 1

Answer: (A)

1+$\frac{b^2}{v(v-b)}$

Answer 2

The van der Waals equation for 1 mole of a real gas is given by:

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

The compressibility factor, $Z$, is defined as:

$Z = \frac{PV}{RT}$

From the van der Waals equation, we can express pressure $P$:

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

Now, substitute this expression for $P$ into the definition of $Z$:

$Z = \frac{\left(\frac{RT}{V-b} - \frac{a}{V^2}\right)V}{RT}$

Distribute $V$ and divide by $RT$:

$Z = \frac{RT \cdot V}{(V-b)RT} - \frac{aV}{V^2 RT}$

$Z = \frac{V}{V-b} - \frac{a}{VRT}$

The Boyle temperature ($T_B$) is the temperature at which the second virial coefficient of a real gas is zero. For a van der Waals gas, the virial expansion of $Z$ is:

$Z = 1 + \left(b - \frac{a}{RT}\right)\frac{1}{V} + \frac{b^2}{V^2} + \dots$

The second virial coefficient is $B = b - \frac{a}{RT}$. At Boyle temperature, $B = 0$:

$b - \frac{a}{RT_B} = 0$

$\frac{a}{RT_B} = b$

So, $RT_B = \frac{a}{b}$

Now, substitute $RT = \frac{a}{b}$ (since $T=T_B$) into the expression for $Z$:

$Z = \frac{V}{V-b} - \frac{a}{V \left(\frac{a}{b}\right)}$

$Z = \frac{V}{V-b} - \frac{b}{V}$

To simplify, combine the terms on the right-hand side:

$Z = \frac{V \cdot V - b(V-b)}{V(V-b)}$

$Z = \frac{V^2 - bV + b^2}{V(V-b)}$

This can be rewritten by separating the first term:

$Z = \frac{V}{V-b} - \frac{b}{V}$

We can write $\frac{V}{V-b}$ as $\frac{V-b+b}{V-b} = 1 + \frac{b}{V-b}$.

So,

$Z = 1 + \frac{b}{V-b} - \frac{b}{V}$

Now combine the last two terms:

$Z = 1 + \left(\frac{b}{V-b} - \frac{b}{V}\right)$

$Z = 1 + \left(\frac{bV - b(V-b)}{V(V-b)}\right)$

$Z = 1 + \left(\frac{bV - bV + b^2}{V(V-b)}\right)$

$Z = 1 + \frac{b^2}{V(V-b)}$

Comparing this with the given options, option (A) matches our result.