ChemistryStates of matterClass 11

Real gases and liquefaction: interactive simulation

The van der Waals equation corrects the ideal gas law for the volume of the molecules and for their attraction. A gas can be liquefied by pressure alone only below its critical temperature.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards

Try this

  1. 1

    Select CO₂ and set T to 280 K.

    What you should see

    The readout says the isotherm is below Tc (304 K), so liquefaction is possible.

  2. 2

    Switch to H₂ at the same temperature.

    What you should see

    H₂ is far above its critical temperature. Pressure alone will not liquefy it.

Formulas behind it

van der Waals equation

(P+aVm2)(Vm−b)=RT\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT

Critical temperature

Tc=8a27RbT_c = \frac{8a}{27Rb}

Compressibility

Z=PVRTZ = \frac{PV}{RT}

Common mistakes

  • Using the ideal-gas critical formulas with R in the wrong units. Here a is in L² atm mol⁻² and R is 0.0821 L atm K⁻¹ mol⁻¹.
Stuck on a real gases and liquefaction question?Snap it and get a whiteboard video solution.

FAQ

What does Z < 1 mean?

The real pressure is lower than the ideal pressure. Attractions dominate, which is what the constant a describes.

Related simulations

All 65 simulations
Molecular shapes (VSEPR)Chemistry, Class 11Ideal gas in a boxChemistry, Class 11Hydrogen spectrum (Bohr model)Chemistry, Class 11 & Class 12Chemical & ionic equilibrium / pH titrationChemistry, Class 11Periodic table & periodic trendsChemistry, Class 11Qualitative salt analysis & test-tube simulatorChemistry, Class 11 & Class 12

Want a specific topic or simulation added?

Tell us in our feedback box which chapter or concept you want visualised next.

Send feedback