Answer :
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]P = 292 \ bar[/tex]
[tex]P_{psi} = 4235.17 \ psi[/tex]
b
[tex]P = 2.92 \ bar[/tex]
[tex]P_{psi} = 42.3517 \ psi[/tex]
c
[tex]P = 2.92*10^{-5} \ bar[/tex]
[tex]P_{psi} = 0.0423517 \ psi[/tex]
Explanation:
From the question we are told that
[tex]1 \ bar = 10^{5} Pa = 14.504 \ psi[/tex]
[tex]\sigma_s = 0.073 \ N/m \ at \ 20^oC[/tex]
Generally the pressure difference is mathematically represented as
[tex]P = \frac{2 \sigma_s }{r}[/tex]
Considering question a
Here the diameter is [tex]d = 10 nm = 10*10^{-9} \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2} = \frac{10*10^{-9}}{2} = 5*10^{-9} \ m[/tex]
So
[tex]P = \frac{2 * 0.073 }{5*10^{-9}} \ Pa[/tex]
=> [tex]P = 292 *10^{5}\ Pa[/tex]
Converting to bar
[tex]P_{bar} = \frac{292 *10^{5}}{1*10^{5}}[/tex]
[tex]P = 292 \ bar[/tex]
Converting to psi
[tex]P_{psi} = \frac{ 292 *10^{5}}{14.504}[/tex]
[tex]P_{psi} = 4235.17 \ psi[/tex]
Considering question b
Here the diameter is [tex]d = 1\mu m = 1*10^{-6} \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2} = \frac{1*10^{-6}}{2} = 0.5*10^{-6} \ m[/tex]
So
[tex]P = \frac{2 * 0.073 }{ 0.5*10^{-6}} \ Pa[/tex]
=> [tex]P = 292 *10^{3}\ Pa[/tex]
Converting to bar
[tex]P_{bar} = \frac{292 *10^{3}}{1*10^{5}}[/tex]
[tex]P = 2.92 \ bar[/tex]
Converting to psi
[tex]P_{psi} = \frac{ 292 *10^{3}}{14.504}[/tex]
[tex]P_{psi} = 42.3517 \ psi[/tex]
Considering question c
Here the diameter is [tex]d = 1mm = 1*10^{-3} \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3} \ m[/tex]
So
[tex]P = \frac{2 * 0.073 }{ 0.5*10^{-3}} \ Pa[/tex]
=> [tex]P = 292 \ Pa[/tex]
Converting to bar
[tex]P_{bar} = \frac{292 }{1*10^{5}}[/tex]
[tex]P = 2.92*10^{-5} \ bar[/tex]
Converting to psi
[tex]P_{psi} = \frac{ 292 }{14.504}[/tex]
[tex]P_{psi} = 0.0423517 \ psi[/tex]
