When helium is allowed to expand into vacuum,a heating effect is observed. The reason for this is that

  • A
    Helium is an ideal gas
  • B
    Helium is an inert gas
  • C
    The inversion temperature of helium is very low
  • D
    The boiling point of helium is the lowest among the elements

Explore More

Similar Questions

Maximum deviation from ideal gas behavior is expected from which of the following gases?

Assertion : Greater the value of van der Waal's constant $a$,greater is the liquefaction of gas.
Reason : $a$ indirectly measures the magnitude of attractive forces between the molecules.

For real gases,the van der Waals equation is written as $\left( p + \frac{a n^2}{V^2} \right) (V - nb) = nRT$,where $a$ and $b$ are van der Waals constants. Two sets of gases are:
$(I)$ $O_2, CO_2, H_2, He$
$(II)$ $CH_4, O_2, H_2$
The gases given in set-$I$ are arranged in increasing order of $b$,and the gases given in set-$II$ are arranged in decreasing order of $a$. Select the correct order from the following:

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?

The correct equation for one mole of a real gas is ($a, b$ are constants)

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