Respuesta :
Explanation:
As it is given that [tex]\Delta P[/tex] = 3 atm = [tex]3.04 \times 10^{5} N/m^{2}[/tex]
r = 73 mN/m
[tex]\Theta[/tex] = [tex]110^{o}C[/tex]
As it is known that free energy of a wetting liquid in a solid-liquid interface is lower than the solid-gas interface free energy.
As a result, there will be spontaneous filling of pores but we need a non-reactive gas under the pressure so that the liquid can be forced out of the pores.
The pressure difference will be represented as [tex]\Delta P[/tex]. Hence, work done by the gas to increase interface free energy is as follows.
[tex]\Delta P = (rcos \Theta) (\frac{dS}{d\nu})[/tex]
where, [tex]d\nu[/tex] = incremental volume of gas in the pore
dS = incremental solid-gas interface area due to [tex]d\nu[/tex].
So, [tex](\frac{dS}{d\nu})_{pore}[/tex] = [tex](\frac{dS}{d\nu})_{cylinder}[/tex]
Therefore, formula to calculate pore size is as follows.
D = [tex]\frac{4r cos \Theta}{\Delta P}[/tex]
Putting the given values into the above formula as follows.
D = [tex]\frac{4r cos \Theta}{\Delta P}[/tex]
= [tex]\frac{4 \times 73 mN/m \times cos(110^{o})}{3.04 \times 10^{5}N/m^{2}}[/tex]
= [tex]0.33 m^{2}[/tex]
Thus, we can conclude that pore size is [tex]0.33 m^{2}[/tex].
