Question

) Determine the pressure difference (in bar and in psi) between the inside and outside of an ultra-small air droplet near the surface of water given that the droplet diameter is

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

`P = 292 \ bar`

`P_(psi) = 4235.17 \ psi`

b

`P = 2.92 \ bar`

`P_(psi) = 42.3517 \ psi`

c

`P = 2.92*10^(-5) \ bar`

`P_(psi) = 0.0423517 \ psi`

Step-by-step explanation:

From the question we are told that

`1 \ bar = 10^(5) Pa = 14.504 \ psi`

`\sigma_s = 0.073 \ N/m \ at \ 20^oC`

Generally the pressure difference is mathematically represented as

`P = (2 \sigma_s )/(r)`

Considering question a

Here the diameter is
`d = 10 nm = 10*10^(-9) \ m`

Generally the radius is mathematically represented as

`r = (d)/(2) = (10*10^(-9))/(2) = 5*10^(-9) \ m`

So

`P = (2 * 0.073 )/(5*10^(-9)) \ Pa`

=>
`P = 292 *10^(5)\ Pa`

Converting to bar

`P_(bar) = (292 *10^(5))/(1*10^(5))`

`P = 292 \ bar`

Converting to psi

`P_(psi) = ( 292 *10^(5))/(14.504)`

`P_(psi) = 4235.17 \ psi`

Considering question b

Here the diameter is
`d = 1\mu m = 1*10^(-6) \ m`

Generally the radius is mathematically represented as

`r = (d)/(2) = (1*10^(-6))/(2) = 0.5*10^(-6) \ m`

So

`P = (2 * 0.073 )/( 0.5*10^(-6)) \ Pa`

=>
`P = 292 *10^(3)\ Pa`

Converting to bar

`P_(bar) = (292 *10^(3))/(1*10^(5))`

`P = 2.92 \ bar`

Converting to psi

`P_(psi) = ( 292 *10^(3))/(14.504)`

`P_(psi) = 42.3517 \ psi`

Considering question c

Here the diameter is
`d = 1mm = 1*10^(-3) \ m`

Generally the radius is mathematically represented as

`r = (d)/(2) = (1*10^(-3))/(2) = 0.5*10^(-3) \ m`

So

`P = (2 * 0.073 )/( 0.5*10^(-3)) \ Pa`

=>
`P = 292 \ Pa`

Converting to bar

`P_(bar) = (292 )/(1*10^(5))`

`P = 2.92*10^(-5) \ bar`

Converting to psi

`P_(psi) = ( 292 )/(14.504)`

`P_(psi) = 0.0423517 \ psi`