# Show that #b = 4V# for a hard sphere potential?

##### 1 Answer

I assume you mean the second virial coefficient

#Z = (PbarV)/(RT) = 1 + (B_(2V)(T))/barV + (B_(3V)(T))/barV^2 + . . . # where

#barV = V/n# is the molar volume,#Z# is the compressibility factor,#B_(nV)(T)# is the#n# th virial coefficient, and the rest of the variables are known from the ideal gas law.For example,

#B_(2V)(T) = 0.00064# at#"1 bar"# of pressure, and#0.00648# at#"10 bar"# of pressure.

The **molar second virial coefficient** is given by (*Physical Chemistry*, McQuarrie & Simon, Ch. 16-5):

#B_(2V)(T) = -2piN_A int_(0)^(oo) (e^(-u(r)//k_BT) - 1)r^2dr# where:

#u(r)# is theintermolecular pair potential, or the potential energy of two molecules separated by a distance#r# . This has more than one functional form.#k_B# is the Boltzmann constant.#T# is the temperature in#"K"# .#N_A = 6.0221413 xx 10^(23)# #"mol"^(-1)# is Avogadro's number. Thus,#B_(2V)(T)# is in units of#"[ . . . ]/mol"# .

In general, this integral is very hard to evaluate, but for the **hard sphere potential** (i.e. for molecules that are assumed to have spherical shapes and collide elastically), we can use:

#u(r) = {(oo, " "r < sigma),(0," "r > sigma):}# where

#sigma# is thedistancefrom thecenterof one molecule to thecenterof the other (sometimes called the "collision diameter").

This applies well at *high temperatures*, because intermolecular forces are minimized as collisions get stronger and faster. We can easily obtain

Under a transformation of integral bounds,

#B_(2V)(T) = -2piN_A int_(0)^(sigma) (0 - 1)r^2dr#

#= -2piN_Acdot-[(sigma^3)/3 - (0^3)/3]#

#= 2/3 pisigma^3 N_A#

For one molecule, we then have that

So, we can rewrite this as:

#(B_(2V)(T))/(N_A) = 2/3 pi (2r)^3#

#= 2/3 pi cdot 8 cdot r^3 = 16/3 pi r^3#

The volume of a single sphere is

Thus, the hard sphere potential shows that for a single molecule,

#color(blue)(barul(|stackrel(" ")(" "(B_(2V)(T))/(N_A) = 4V" ")|))# ,where

#V# is the volume of one molecular sphere.