Question

A gas described by van der Waals equation is:
A. it behaves similar to an ideal gas in the limit of large molar volume.
B. it behaves similar to an ideal gas in the limit of large pressures.
C. it is characterized by van der Waals' constant that is dependent on the identity of the gas but is independent of the temperature.
D. it has the pressure that is lower than the pressure exerted by the same behaving ideally.

Answer 1

Hint: The van der Waals equation is an equation of state that tells us about the ideal gas law based on plausible reasons that real gases do not act ideally.
Van der wall's equation is \[PV = nRT\] , where P=pressure, V= volume, n= no. of moles, R=gas constant (\[8.31432 \times \mathop {10}\nolimits^3 {\text{ }}N \cdot m \cdot \mathop {kmol}\nolimits^{ - 1} \cdot \mathop K\nolimits^{ - 1} \] ), T= temperature.

Complete step by step answer:
In this question we have asked the Van Der Waal equation to describe which type of gas. So, first of all what is Van Der Waals equation. Van der wall's equation is \[PV = nRT\], where P=pressure, V= volume, n= no. of moles, R=gas constant (\[8.31432 \times \mathop {10}\nolimits^3 {\text{ }}N \cdot m \cdot \mathop {kmol}\nolimits^{ - 1} \cdot \mathop K\nolimits^{ - 1} \] ), T= temperature and the real gas equation is (P + an2)(V-nb) = nRT. Where a is a constant whose value depends on the attraction between the gas molecules and b is the volume that is occupied by one mole of the molecules.
Our 1st option is (A) it behaves similar to an ideal gas in the limit of large molar volumes. When the volume is very large or let's say infinite then the attraction between the particles becomes nil, so a≈0 and the volume of one mole molecule compared to such a large volume will also be nil. So, b~0. Now, we can neglect a and b and the gas behave ideally, as \[\left( {P{\text{ }} + {\text{ }}a\mathop n\nolimits^2 } \right)\left( {V - nb} \right){\text{ }} = {\text{ }}nRT\] now becomes \[PV = nRT\]. Thus, the option A is correct.
(B) It behaves similar to an ideal gas in the limit of low pressures and high temperatures. Now, the pressure is very low. So, the attraction between the particles decreases and we can neglect a, but we can’t neglect ‘b’. So, this gas will not behave like real gas as the equation becomes\[P\left( {V - nb} \right){\text{ }} = {\text{ }}nRT\]. So, this option is wrong.
(C) It is characterized by van der Waals coefficients that are dependent on the identity of the gas but are independent of the temperature. Yes this is true we have two Van der Waals coefficients- ‘a’ and ‘b’. Where ‘a’ is a constant whose value depends on the attraction between the gas molecules and ‘b’ is the volume that is occupied by one mole of the molecules and they are independent of the temperature. So, this option is correct.
(D) It has the pressure that is lower than the pressure exerted by the same gas behaving ideally. It is also true because in ideal gas we do not have factor a, that means attraction between the particles is nil. Now we have a gas, whose particles exerting less pressure on each other as compared to ideal gas. So, the coefficient a, for this gas will also be nil and the gas will behave as ideal gas. So, this is also true.
So, the correct option will be A, C and D.

Note:
When the value of ‘a’ and ‘b’ becomes nil for a real gas, then the real gas becomes ideal gas and will behave ideally according to Van der waals equation.