The ideal gas law is useful at very low pressures, but at higher pressures, real gases behave differently from ideal gases.

The gas laws consider ideal gases—gases in which no forces act on the gas and the particles of the gas do not take up space. Real gases are not ideal. Although the ideal gas law and laws derived from it are useful for understanding and predicting the behavior of gases, it is also useful to understand how real gases differ from ideal gases.

The**compressibility factor (**, or compression factor, is a number that indicates how much a gas's behavior differs from an ideal gas.

*Z*)$Z=\frac{PV}{nRT}$

*Z*is equal to 1. The more Z deviates from 1, the less like an ideal gas the given gas is.

#### Nonideal Gases

**van der Waals equation**makes these modifications by accounting for intermolecular (nonideal) interactions between gases. It adjusts the ideal gas law to explain and predict the behavior of real gases.

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

*a*corresponds to the strength of attraction between gas molecules. It is characteristic of each individual gas and has the units pascals meter to the 6th power divided by moles squared (Pa·m

^{6}/mol

^{2}). In the modification for volume, the constant

*b*corresponds to the size of the molecules of a gas. It is also characteristic to each gas and has the units cubic meter divided by mole (m

^{3}/mol). The constants

*a*and

*b*for each gas can be found in chemistry reference guides. At lower pressures the constant

*a*is more important because the attractive or repulsive forces of gas molecules have the greater effect on their behavior. At higher pressures the constant

*b*is more important because the size of the gas molecules has a greater effect on their behavior. Thus, at lower pressures, a gas is more compressible than an ideal gas because its molecules attract one another, but at higher pressures, a gas is less compressible than an ideal gas because its molecules occupy a significant amount of the volume of the gas.

### Description and Unit of each Term in the Van Der Waals Equation

Equation Term | Description | Units |
---|---|---|

P | pressure | atm |

a | constant that corresponds to the strength of attraction between gas molecules | Pa·m^{6}/mol^{2} or atm·L^{2}/mol^{2} |

n | number of molecules | mole |

V | volume | liter |

b | constant that corresponds to the size of the molecules of a gas | m^{3}/mol or L/mol |

R | gas constant | 8.314 m^{3}Pa/K⋅mol or 0.082057 atm⋅L/K⋅mol |

T | temperature | kelvin |

The van der Waals equation is often useful for predicting real systems. For example, consider 1 mole of methane (CH
Then compare this to the calculation of pressure using the van der Waals equation, rearranged to solve for
Because the pressure in these calculations is relatively low, the attractive forces between molecules of the gas result in a pressure lower than an ideal gas would have.
Calculate the pressure exerted by 1 mole of ammonia, NH
Using the van der Waals equation:

_{4}) in a 0.250-L container at standard temperature (273.15 K). First, calculate the pressure using the ideal gas equation.$\begin{aligned} PV&=nRT\\P&=\frac{nRT}{V}\\&=\frac{(1\;\rm{mol})(0.082057\;\rm{atm}{\cdot}{\rm{L/K}}{\cdot}\rm{mol})(273.15\;\rm{K})}{0.250\;\rm L}\\&=89.7\;\rm{atm}\end{aligned}$

*P*.$\begin{aligned} P&=\frac{nRT}{V-nb}-\frac{an^2}{V^2}\\&=\frac{(1\;\rm{mol})(0.082057\;{\rm{atm}{\cdot}\rm{L/K}{\cdot}\rm{mol}})(273.15\;\rm{K})}{0.250\;\rm{L}-(1\;\rm{mol})(0.0431\;\rm{L/mol})}-\frac{(2.303\;\rm{atm}{\cdot}\rm{L}^2\rm{/mol}^{2})(1\;\rm{mol})^2}{(0.250\;\rm{L})^2}\\&=71.5\;\rm{atm}\end{aligned}$

_{3}, in 5.0 L at 480 K using the ideal gas law and the van der Waals equation.$\begin{aligned}PV&=nRT\\P&=\frac{nRT}{V}\\&=\frac{(1\;\rm{ mol)}(0.082057\;\rm{atm}{\cdot}\rm{L/K}{\cdot}\rm{mol})(480\;\rm{K})}{5.0\;\rm{L}}\\&=7.9\;\rm{atm}\end{aligned}$

$\begin{aligned} P&=\frac{nRT}{V-nb}-\frac{an^2}{V^2}\\&=\frac{(1\;\rm{mol})(0.082057\;{\rm{atm}{\cdot}\rm{L/K}{\cdot}\rm{mol}})(480\;\rm{K})}{5.0\;\rm{L}-(1\;\rm{mol})(0.0371\;\rm{L/mol})}-\frac{(4.225\;\rm{atm}{\cdot}\rm{L}^2\rm{/mol}^2)(1\;\rm{mol})^2}{(5.0\;\rm{L})^2}\\&=7.8\;\rm{atm}\end{aligned}$