# What causes gas molecules to spread through air?

##### 1 Answer

Mostly entropy. The process of mixing gases, particularly ideal gases, is favored by entropy, and for ideal gases, it is *always* spontaneous.

Below is a thermodynamics proof of why entropy and Gibbs' free energy of mixing are positive and negative, respectively, for ideal gases.

The **entropy of mixing** is given by the difference in entropy between the unmixed (

#DeltabarS_(mix) = barS - barS^"*"#

#= sum_i n_ibarS_i - sum_i n_ibarS_i^"*"#

#= sum_i n_i(barS_i - barS_i^"*")# #" "" "bb((1))# where:

#n_i# is the mols of gas#i# .#barS_i# is the molar entropy,#S_i/n_i# , of gas#i# .#"*"# indicates the unmixed substance.

The entropy is related to the **Gibbs' free energy** **enthalpy**

#DeltaG = DeltaH - TDeltaS# #" "" "bb((2))#

These have their own thermodynamic mixing expressions:

#DeltabarG_(mix) = sum_i n_i(barG_i - barG_i^"*")# #" "" "bb((3))#

#DeltabarH_(mix) = sum_i n_i(barH_i - barH_i^"*")# #" "" "bb((4))#

For ideal gases, there is zero heat released or absorbed after mixing (no intermolecular forces exchanged), so

From

#barG_i = barG_i^"*" + RTlnchi_i# #" "" "bb((5))# where

#chi_i# is the mol fraction of gas#i# .

As a result, plugging

#color(blue)(DeltabarG_(mix) = RTsum_i n_ilnchi_i) ~~ -TDeltabarS_(mix)#

or, comparing

#color(blue)(DeltabarS_(mix) ~~ -Rsum_i n_ilnchi_i)#

But since mol fractions are always less than or equal to

#DeltabarG_(mix) < 0# . Therefore, mixing ideal gases is**spontaneous**.#DeltabarS_(mix) > 0# . Therefore, mixing ideal gases**increases**the entropy of the system.