For the following gases,the values of 'a' (va | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The van der Waals constant $'a'$ is a measure of the magnitude of intermolecular attractive forces between gas molecules.Greater the value of $'a'

Vedclass · Accepted Answer

$N_2$

Vedclass · Answer

(B) The van der Waals constant $'a'$ is a measure of the magnitude of intermolecular attractive forces between gas molecules.Greater the value of $'a'$,stronger are the intermolecular forces,and easier it is to liquefy the gas.Conversely,a smaller value of $'a'$ indicates weaker intermolecular forces,making the gas more difficult to liquefy.Comparing the given values: $CH_4 (2.25) > CO (1.46) > O_2 (1.36) > N_2 (1.35)$.Since $N_2$ has the smallest value of $'a'$ $(1.35)$,it has the weakest intermolecular forces and will be the most difficult to liquefy.

For the following gases,the values of $'a'$ (van der Waals constant) are given in parentheses: $CH_4 (2.25)$,$N_2 (1.35)$,$O_2 (1.36)$,and $CO (1.46)$. Which gas will be the most difficult to liquefy?

Similar Questions

$A$ real gas most closely approaches the behaviour of an ideal gas at

Gases deviate from ideal gas behavior because the molecules $...$

$A$ gas at $350 \; K$ and $15 \; bar$ has a molar volume $20$ percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is

The compressibility factor $(Z)$ of one mole of a van der Waals' gas of negligible '$a$' value is

The Van der Waals constant '$a$' for the gases $O_2$,$N_2$,$NH_3$ and $CH_4$ are $1.3$,$1.390$,$4.170$ and $2.253 \ L^2 \ atm \ mol^{-2}$ respectively. The gas which can be most easily liquefied is

Vedclass Test Series

Exam Paper Generator

Online Exam Module