13.0 moles of gas are in a 5.00 L tank at 20.4 °C. Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a = 2.300 L^{2} ⋅ atm/mol^{2} and b = 0.0430 L/mol.

Express your answer with the appropriate units.

Kinetic molecular theory makes certain assumptions about gases that are, in fact, not true for real gases. Therefore, the measured properties of a gas are often slightly different from the values predicted by the ideal gas law. The van der Waals equation is a more exact way of calculating properties of real gases. The formula includes two constants, *a* and *b*, that are unique for each gas.

The van der Waals equation is

$(P+\frac{{\mathrm{an}}^{2}}{{V}^{2}})\mathbf{(}\mathbf{V}\mathbf{-}\mathbf{nb}\mathbf{)}\mathbf{=}\mathbf{nRT}$

where *P* is the pressure, *n* the number of moles of gas, *V* the volume, *T* the temperature, and *R* the gas constant. All real gases possess intermolecular forces that can slightly decrease the observed pressure.

In the van der Waals equation, the variable *P* is adjusted to

$\mathbf{P}\mathbf{+}\frac{{\mathbf{an}}^{\mathbf{2}}}{{\mathbf{V}}^{\mathbf{2}}}$

where *a* is the attractive force between molecules.

The volume of a gas is defined as the space in which the gas molecules can move. When using the ideal gas law we make the assumption that this space is equal to the volume of the container. However, the gas molecules themselves take up some space in the container. So in the van der Waals equation, the variable *V* is adjusted to be the volume of the container minus the space taken up by the molecules

$\mathbf{V}\mathbf{-}\mathbf{nb}$

where bb is the volume of a mole of molecules.

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