Given van der Waals constant a for NH_{3},H_{ | vedclass

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

Question

(A) The ease of liquefaction of a gas is directly proportional to the magnitude of the van der Waals constant $a$,which represents the intermolecular

Vedclass · Accepted Answer

$NH_{3}$

Vedclass · Answer

(A) The ease of liquefaction of a gas is directly proportional to the magnitude of the van der Waals constant $a$,which represents the intermolecular forces of attraction.Greater value of $a$ implies stronger intermolecular forces,leading to a higher critical temperature $(T_{C})$.Comparing the given values: $a(NH_{3}) = 4.17$,$a(H_{2}) = 0.244$,$a(O_{2}) = 1.36$,and $a(CO_{2}) = 3.59$.Since $NH_{3}$ has the highest value of $a$ $(4.17)$,it has the strongest intermolecular forces and is the most easily liquefied gas among the given options.

Given van der Waals constant $a$ for $NH_{3}$,$H_{2}$,$O_{2}$ and $CO_{2}$ are respectively $4.17$,$0.244$,$1.36$ and $3.59$,which one of the following gases is most easily liquefied?

Similar Questions

$A$ real gas would be most likely to obey the ideal gas laws at

Which of the following can be most easily liquefied? (Value of $a$ for $NH_3 = 4.17, CO_2 = 3.59, SO_2 = 6.71, Cl_2 = 6.49$)

The compressibility factor $(Z)$ of a gas is greater than unity at $1 \, atm$ and $273 \, K$. Therefore:

Which of the given sets of temperature and pressure will cause a gas to exhibit the greatest deviation from ideal gas behavior?

At low pressure,Vander Waals equation is written as $(P + \frac{a}{V^2})V = RT$. The compressibility factor is then equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module