The compressibility factor $(Z)$ of a gas is less than unity at $STP$. Therefore:
- A$V_m > 22.4 \ L$
- B$V_m < 22.4 \ L$
- C$V_m = 22.4 \ L$
- D$V_m = 44.8 \ L$
Explore More
Similar Questions
Van der Waals equation for a gas is stated as,
$p = \frac{nRT}{V - nb} - a\left( \frac{n}{V} \right)^2$
This equation reduces to the perfect gas equation,$p = \frac{nRT}{V}$,when:
$p = \frac{nRT}{V - nb} - a\left( \frac{n}{V} \right)^2$
This equation reduces to the perfect gas equation,$p = \frac{nRT}{V}$,when:
DifficultJEE MAIN 2014
View SolutionThe units of constant $a$ in the Van der Waals equation are:
Easy
View SolutionWhen is deviation more in the behaviour of a gas from the ideal gas equation $PV = nRT$?
If the compressibility factor $(Z)$ of a real gas is $1.05$ at $STP$,what is the molar volume of the real gas (in $dm^3$)?
Which of the given sets of temperature and pressure will cause a gas to exhibit the greatest deviation from ideal gas behavior?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo