Question

Arrange the following gases in the increasing order of their van der Waals constant a values: $C{{O}_{2}},{{H}_{2}},{{C}_{6}}{{H}_{6}}$ ​

Answer 1

Van der Waals constant is going to be measured with the symbol ‘a’. Van der Waals constant is going to measure the attractive force which is present in between the gaseous molecules indirectly. Van der Waals constant is inversely proportional to liquification pressure.

Complete answer:
- In the question it is asked to arrange the increasing order of their van der Waals constant a value of the given gases $C{{O}_{2}},{{H}_{2}},{{C}_{6}}{{H}_{6}}$.
- Means we should know the values of the van der Waals constant values means ‘a’ value of the given carbon dioxide, hydrogen and benzene.
- The Van der Waals constant value of the given molecules are as follows.
- ‘a’ value for carbon dioxide is = 3.64, ‘a’ value for hydrogen gas is = 0.246 and ‘a’ value for benzene is 18.24.
- We know that Van der Waals constant ‘a’ is inversely proportional to the liquefaction pressure.
- The order of increasing the van der Waals constant for the given gases is as follows.
${{H}_{2}}< C {{O}_{2}}<{{C}_{6}}{{H}_{6}}$
- Means benzene has higher van der Waals constant value than carbon dioxide and carbon dioxide has higher van der Waals constant value than hydrogen gas.

Note:
As the van der Waals constant value increases, the liquification pressure of the particular gas is going to decrease. As the van der Waals constant value decreases the liquification pressure of the particular gas is going to increase.