Courses for Kids
Free study material
Offline Centres
Store Icon

Arrange the vander Waals constant for the gases.

${C_6}{H_6}\left( g \right)$ $0.217$
${C_6}{H_5}C{H_3}\left( g \right)$$5.464$
$Ne\left( g \right)$ $18.000$
${H_2}O\left( g \right)$ $24.060$

A.$I - A,II - D,III - C,IV - B$
B.$I - D,II - A,III - B,IV - C$
C.$I - C,II - D,III - A,IV - B$
D.$I - B,II - C,III - A,IV - D$

Last updated date: 23rd Jul 2024
Total views: 348k
Views today: 5.48k
348k+ views
Hint: We have to know that an equation which relates the relationship between the pressure, volume, temperature, and amount of real gases is known to van der Waals equation of gases. In the alterations that are made interconnected to the equation of ideal gas for real gases, the drop in pressure is because of forces of attractions among the molecules is straightly proportionate to $\dfrac{{{n^2}}}{{{V^2}}}$.

Complete answer:
We know that equation for ideal gas could be given as,
$PV = nRT$
We can give the altered expression as,
$\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)\left( {v - nb} \right) = nRT$
We can represent the pressure as P.
We can represent the molar volume as V.
We can represent the temperature of the gaseous sample as T.
We can indicate the gas constant as R.
We can indicate a and b are van der Waals constant.
We can indicate n as the moles of the gas.
We have to know when there is an increase in van der Waals constant “b”, there is an increase in size of gas which is nothing but volume. Volume of one mole of atoms (or) molecules is volume of B.
${C_6}{H_5}C{H_3}\left( g \right)$ $24.060$
${C_6}{H_6}\left( g \right)$ $18.000$
$Ne\left( g \right)$ $0.217$
${H_2}O\left( g \right)$ $5.464$

Option (C) is correct.

We have to know that there are certain of the benefits of van der Waals equation are,
Obtains the nature of gases in a better way when compared to the ideal gas equation.
Valid for gases as well as fluids.
Certain of the difficulties of van der Waals expression are,
We could obtain better conclusions of all real gases only at higher critical temperatures with the help of van der Waals equation.
Vander Waals equation totally fails in the alteration phase of gas to the liquid less than a critical temperature.