Question

What is the compressibility factor (Z) for $0.02$ mole of a Van Der Waals gas at pressure of $0.1$atm. Assume the size of gas molecules is negligible.
Given $RT = 20Latm{\left( {mol} \right)^{ - 1}}$ and $a = 1000atm{L^2}mo{l^{ - 2}}$
A.$2$
B.$1$
C.$0.02$
D.$0.5$

Answer 1

The compressibility factor can be represented by Z and can be defined as ratio of molar volume of a gas to the ideal gas. The molar volume of a gas divided by the ideal gas gives the value of a compressibility factor. The number of moles, pressure, and volume were needed to calculate the compressibility.
Formula used:
$Z = \dfrac{{PV}}{{nRT}}$
Z is compressibility factor
P is pressure
V is volume has to be determined
n is number of moles
RT is given in question

Complete answer:
Given that gas the Van Der Waals gas at pressure of $0.1$atm
Number of moles is $0.02$
The value of a can be given as $a = \dfrac{{P{V^2}}}{{{n^2}}}$
Substitute the values in the above equation to get the value of V
${V^2} = \dfrac{{1000 \times {{\left( {0.02} \right)}^2}}}{{\left( {0.1} \right)}}$
By simplification, we will get
$V = 2$
Substitute the value of V in the compressibility factor equation, we have the value of RT which is the volume of ideal gas and number of moles are given in the question.
Thus, the value of Z will be
$Z = \dfrac{{0.1 \times 2}}{{0.02 \times 20}}$
By simplification we will get $Z = 0.5$
Thus, the compressibility factor is $0.5$

Option D is the correct one.

Note:
The Vander Waal equation is different from the ideal gas equation. The ideal gas equation does not have any constants other than ideal gas constant. But the Vander Waals constant has constants a and b have particular values.