Question

If z is a compressibility factor van der waals equation at low pressure can be

• A. $\mathrm{Z}=1+\left(\frac{\mathrm{Pb}}{\mathrm{RT}}\right)$
• B. $\mathrm{Z}=1-\left(\frac{\mathrm{Pb}}{\mathrm{RT}}\right)$
• C. $1+\left(\frac{\mathrm{RT}}{\mathrm{Pb}}\right)$
• D. $1-\left(\frac{\mathrm{a}}{\mathrm{VRT}}\right)$

Answer 1

The correct option is D

$1-\left(\frac{\mathrm{a}}{\mathrm{VRT}}\right)$

Explanation:

Van der Waals gas equation for 1 mole of real gas is:

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

Where ‘a’ represents attractive forces

‘b’ represents repulsive forces

On lowering the pressure, the attractive forces dominate over repulsive forces.

Since, at low pressure $a>>>b$, therefore, we can ignore b term.

Now,

$$\begin{gathered} \left(\mathrm{P}+\frac{\mathrm{a}}{{\mathrm{V}}^2}\right)\mathrm{V}=\mathrm{RT} \\ \mathrm{PV}+\frac{\mathrm{a}}{\mathrm{V}}=\mathrm{RT} \\ \mathrm{or}, \\ \mathrm{PV}=\mathrm{RT}–\frac{\mathrm{a}}{\mathrm{V}} \\ \frac{\mathrm{PV}}{\mathrm{RT}}=1–\frac{\mathrm{a}}{\mathrm{VRT}} \\ \mathrm{Z}=1–\frac{\mathrm{a}}{\mathrm{VRT}} \end{gathered}$$

The value of the compressibility factor is $1-\left(\frac{\mathrm{a}}{\mathrm{VRT}}\right)$

Therefore, Van der Waals equation at low pressure can be $\mathrm{Z}=1-\left(\frac{\mathrm{a}}{\mathrm{VRT}}\right)$.

Hence, the correct answer is option (D).