# I do not understand this " rate of diffusion of gases is inversely proportional to the square root of their molar masses provided that pressure and temperature remain constant. ?

##### 1 Answer

**DEPENDENCE OF RATE OF EFFUSION ON TEMPERATURE**

Consider the **root-mean-square speed** of a gas following Boltzmann statistics (we could have chosen the most probable speed or average speed; it doesn't matter).

#v_(RMS) = sqrt((3RT)/M)# where

#R# and#T# are known from the ideal gas law, and#M# is the molar mass in#"kg/mol"# .

Two gases with speeds

#z_A/z_B = v_A/v_B# ,

Thus, two gases at the *same temperature* will have effusion rates given by:

#color(blue)(z_A/z_B) = (sqrt((3RT)/M_A))/(sqrt((3RT)/M_B))#

#= color(blue)(sqrt(M_B/M_A))# Or,

#z prop 1/sqrtM# .

And this is known as **Graham's Law of effusion**. This holds true as long as the gases are held under the ** same** pressure and temperature.

*At constant temperature and pressure, the rates of effusion themselves also remain the same.*

The temperature dependence is clear from the derivation. However, the **pressure dependence** is implicit at the moment. Let's see how this comes in.

**DEPENDENCE OF RATE OF EFFUSION ON PRESSURE**

One way to look at this is to use the ideal gas law

#ul(v_(RMS) = sqrt((3PV)/(nM)))#

So, in this manner, **gas speed (and therefore effusion rate) is a function of pressure**, and pressure therefore matters.

Alternatively, consider that

#P = F/A# where

#F# is the force the gas imparts onto a surface and#A# is the surface area that it collides with.

The effusion can be modeled by a "collision" with a pinhole:

And one would have used the Maxwell-Boltzmann speed distribution to set up the following integral:

#P = int_(0)^(2pi) int_(0)^(pi) int_(0)^(oo) 2mvcostheta rho (m/(2pik_BT))^(3//2) v^3 e^(-mv^2//2k_BT)costhetasinthetadvd theta d phi#

#= [ . . . ]#

#= 4/3pi m rho (m/(2pi k_B T))^(3//2) overbrace(int_(0)^(oo) v^4 e^(-mv^2//2k_BT) dv)^((3pi^(1//2))/8 ((2k_B T)/m)^(5//2))# ,where

#rho = N/V# is the number density, i.e. the number of particles per unit volume. Note that#Nk_B = nR# , where#k_B# is the Boltzmann constant.

With some algebraic simplification, the result is:

#P = N/V k_BT = ul(n/V RT)#

which is what we expect, as we just got

Therefore, if the pressure changes for fixed mols

**As a result, the rate of effusion also depends on whether the pressure was constant. Therefore, at constant pressure AND temperature, it is ONLY dependent on gas speed, and consequently, molar mass.**