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

  • A
    $V_m > 22.4 \, L$
  • B
    $V_m < 22.4 \, L$
  • C
    $V_m = 22.4 \, L$
  • D
    $V_m = 44.82 \, L$

Explore More

Similar Questions

Under which one of the following conditions do real gases approach the ideal gas behaviour?

The compressibility factor $(Z)$ of a gas is less than unity at $STP$. Therefore:

What is compressibility factor $(Z)$? Explain $(i)$ deviation factor $(Z = 1)$,$(ii)$ $Z > 1$,$(iii)$ $Z < 1$,$(iv)$ deviation graph,and $(v)$ the relation between molar volume and $Z$.

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:

Identify the conditions at which the van der Waals equation of state changes to the ideal gas equation.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo