Question

At high temperature and low pressure, the Vander Waals equation is reduced to:
(A) $ \left( {{\text{p + }}\dfrac{{\text{a}}}{{{{\text{V}}_{\text{m}}}^{\text{2}}}}} \right)\left( {{{\text{V}}_{\text{m}}}} \right){\text{ = RT}} $
(B) $ {\text{p}}{{\text{V}}_{\text{m}}}{\text{ = RT}} $
(C) $ {\text{p}}\left( {{{\text{V}}_{\text{m}}} - {\text{b}}} \right){\text{ = RT}} $
(D) $ \left( {{\text{p + }}\dfrac{{\text{a}}}{{{{\text{V}}_{\text{m}}}^{\text{2}}}}} \right)\left( {{{\text{V}}_{\text{m}}} - {\text{b}}} \right){\text{ = RT}} $

Answer 1

Hint: In the above question, we are asked how Van Der Waals equation gets reduced at high temperature and low pressure. So, first we have to see why the constants are added in the Vander Waals equation, then only we can predict its nature at high temperature and low pressure. The constant a is used to account for the intermolecular forces and b is used to account for the volume of the gas molecules. At high temperature and low pressure, these both are neglected.

Complete step by step solution
We know that the Vander Waals equation is applicable for real gas which is not ideal.
According to ideal gas law, $ {\text{PV = nRT}} $
To account for volume of real gas V is replaced by $ {{\text{V}}_{\text{m}}}{\text{ - b}} $ where $ {{\text{V}}_{\text{m}}} $ denotes molar volume of the gas and b is the volume that is occupied by one mole of the molecules.
 $ {\text{P}}\left( {{{\text{V}}_{\text{m}}}{\text{ - b}}} \right){\text{ = RT}} $
Second modification is made to account for the fact that gas molecules don’t interact with each other. But in reality attraction between molecules is observed at low pressure and repulsion at high pressure which results in addition of $ \dfrac{{\text{a}}}{{{{\text{V}}_{\text{m}}}^{\text{2}}}} $ , where a is constant and $ {{\text{V}}_{\text{m}}} $ is the molar volume of the gas.
So we have, $ \left( {{\text{P + }}\dfrac{{\text{a}}}{{{{\text{V}}_{\text{m}}}^{\text{2}}}}} \right)\left( {{{\text{V}}_{\text{m}}} - {\text{b}}} \right){\text{ = RT}} $ .
Molar volume means $ \dfrac{{\text{V}}}{{\text{n}}} $ , where n =number of moles. So, we can rewrite the equation as: $ \left( {{\text{P + }}\dfrac{{{\text{a}}{{\text{n}}^{\text{2}}}}}{{{{\text{V}}_{\text{m}}}^{\text{2}}}}} \right)\left( {\dfrac{{\text{V}}}{{\text{n}}} - {\text{b}}} \right){\text{ = RT}} $
  $ \left( {{\text{P + }}\dfrac{{{\text{a}}{{\text{n}}^{\text{2}}}}}{{{{\text{V}}_{\text{m}}}^{\text{2}}}}} \right)\left( {\dfrac{{{\text{V}} - {\text{nb}}}}{{\text{n}}}} \right){\text{ = RT}} $
Rearranging the equation, we get:
 $ \left( {{\text{P + }}\dfrac{{{\text{a}}{{\text{n}}^{\text{2}}}}}{{{{\text{V}}_{\text{m}}}^{\text{2}}}}} \right)\left( {{\text{V}} - {\text{nb}}} \right){\text{ = nRT}} $
At high temperature, molar pressure is very less and at low pressure, repulsion and attraction between molecules are negligible. And hence, Vander Waals equation reduced to-
  $ {\text{PV = nRT}} $
$ \therefore $ The correct option is option B.

Note
-Ideal gas equation is valid for ideal gas only.
-Vander Waals equation can be applied for all real gases.
-The higher value of a, more is the attraction between molecules and gas can be compressed easily.