State and explain Graham's law of Diffusion.

Explanation

Graham'Law of Diffusion: At constant temperature and pressure. The rate of diffusion of a gas is inversely proportional to the square root of its density (or) Vapour pressure (or) molecular weight.
`rprop1/sqrtd:rprop1/sqrt(VD):rprop1/sqrtm(or)`
`r=Ktimes1/sqrtd:r=K times1/sqrt(VD):r=K times1/sqrtm`
IF `r_1` and `r_2` are the rates of diffusion of two gases and `d_1,d_2` are their densities then
`r_1/r_2=sqrt(d_2/d_1)`..........(1)
IF `r_1` and `r_2` are the rates of diffusion of two gases and `VD_1,VD_2` are their vapour pressures , then
`r_1/r_2=sqrt((VD_2)/(VD_1))`..........(2)
IF `r_1` and `r_2` are the rates of diffusion of two gases and `m_1,m_2` are their molecular weights then `r_1/r_2=sqrt((m_2)/(m_1))`
`thereforer_1/r_2=sqrt((d_2)/(d_1))=sqrt((VD_2)/(VD_1))=sqrt(m_2/m_1)`
`therefore` But rate of diffusion
`r=(V(volume))/(t(tim es of dif fusion))` Then `r_1/r_2=V_1/t_1timest_2/V_2`
`thereforer_1/r_2=(V_1t_2)/(V_2t_1)=sqrt(d_2/d_1)=sqrt((VD_2)/(VD_1))=sqrt(m_2/m_1)`
Case:1 If the times pf diffusion are equal i.e., `t_1=t_2`. then we can write
`thereforer_1/r_2=(V_1)/(V_2)=sqrt(d_2/d_1)=sqrt((VD_2)/(VD_1))=sqrt(m_2/m_1)`
Case.2: If the volumes of the two gases are the same (i.e.,) `V_1-V_2`
then `thereforer_1/r_2=(t_1)/(t_2)=sqrt(d_2/d_1)=sqrt((VD_2)/(VD_1))=sqrt(m_2/m_1)`