For real gases the relation between p, V and T is given by van der Waals equation:
\left ( p+\frac{an^{2}}{v^{2}} \right )(v-nb)=nRT
where ‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas. ‘a’ is the measure of magnitude of intermolecular attraction.
(i) Arrange the following gases in the increasing order of ‘b’. Give reason.O_{2}, CO_{2}, H_{2}, He
(ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason.CH_{4}, O_{2}, H_{2}
 

Answers (1)

(i) We know that volume α size of the molecules. Hence, the increasing order of the value of 'b' will be:

H_{2}<He<O_{2}<CO_{2}

(ii) 'a' is the Van der Waal's constant which represents the magnitude of intermolecular attraction. An \alpha size of the electron cloud. Hence, the greater the size of the electron cloud, the greater will be the dispersion forces and polarizability of the molecule. Hence, the gases in decreasing order of the magnitude of 'a' will be: CH_{4} > O_{2} > H_{2}

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions