Question #c71ad

1 Answer
Dec 4, 2016

Answer:

#sf(P=2.84xx10^(7)color(white)(x)"N/m"^2)#

Explanation:

The graphic shows the effective collision cross section, or area:

hyperphysics.phy-astr.gsu.edu

From this we can estimate the diameter of the atoms:

#sf(A=pid^2)#

#:.##sf(d=sqrt(A/pi)=0.36/3.142==0.3385color(white)(x)nm)#

The mean free path #sf(lambda)# is the average distance between collisions.

The expression is:

#sf(lambda=(1)/(sqrt(2)pid^(2)n_v)#

#sf(n_v)# is the number of molecules per unit volume.

We can eliminate this using the Ideal Gas Expression:

#sf(PV=nRT)#

#sf(n_v=(nN_A)/V)#

Where #sf(N_A)# is the Avogadro Constant.

Since #sf(V=(nRT)/P)#

We can write:

#sf(n_v=(cancel(n)N_A)/((cancel(n)RT)/P)=(N_AP)/(RT))#

Substituting this into the expression for #sf(lambdarArr)#

#sf(lambda=(RT)/(sqrt(2)pid^2N_AP))#

#:.##sf(P=(RT)/(lambdasqrt(2)d^2N_A))#

We can now set the condition that #sf(lambda=d)#

#:.##sf(P=(RT)/(sqrt(2)pid^3N_A))#

Putting in the numbers:

#sf(P=(8.31xx298)/(1.414xx3.142xx(0.3385xx10^(-9))^3xx6.02xx10^(23))color(white)(x)"N/m"^2)#

#sf(P=2.837xx10^7color(white)(x)"N/m"^2)#