a) the van der Waals equation is used to correct the ideal gas equation for the

Question

a) the van der Waals equation is used to correct the ideal gas equation for the behavior of real gases. There are two van der Waals constants 'a' and 'b' that are used in the equation. The 'a' constants for two different gases are given below. The 'a' constant measures the deviaion due to forces of interaction existent between gas molecules. Briefly explan the trend based on what you know about these molecules.

NH3=4.17

CH4=2.12

b) Arrange CH3CH2OH, CH3COOH and CH3SO2OH in increasing order of acidity. Explain.

c) Arrange the following in their increasing order of boiling points. explain.

CH3CH2CH2OH, CH3CH2CL, CH3CH2CH2CH3, CH3CH2CH3

Explanation / Answer

The 'a' constant measures the deviation due to forces of interaction existent between gas molecules.

The interaction between NH3 molesucles is hydrogen bonding due to the presence of N which is slightly electronegative and H which is slightly electropositiove so this interaction is fairly large.

In CH4 theere is no charge separtion between C and H so there is almost negligible interaction between molecules except for london forces.

so the a value is larger for NH3 and smaller for CH4

The incresing order of acidity is CH3CH2OH< CH3COOH<CH3SO3H

In SO3H is highly acidic compared to the rest because the negative charge on oxygen generated after H+ is removed is well distributed on the other oxygens due to resonance

In CO2H the acidity is lower than in SO3H as the number of oxygen are less and so numbre of contributing resonce structures are lower so it is less stable hence it is less acidic.

The alcohol does not release H+ easily so it is least acidic.

c) CH3CH2CH3<CH3CH2CH2CH3<CH3CH2Cl<CH3CH2CH2OH

Because the 1st and second are gases and presence of Cl a halogen increases interactions between the CH and CL and hence its boiling point is higher in the case of the alcohol hydrogen bonding is possible so interactions are significantly high

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