Question

Calculate the total reversible thermodynamic work required to produce a spray of water of droplet diameter 2000 nm from 40 liters of water at

25C. Take the surface tension of water as 72 mN m size?  How many water drops would be produced, theoretically if the droplets are at all equal size?

this is all the information given in the text book.

Answer 1

Volume of one droplet = 4/3 π r³

= 4/3 × π ×(1000×10^-9m)³ = 4.1888 ×10^-18m³

Number of droplets = total volume / volume of 1 droplet

= 40 ×10^-3 m³/(4.1888×10^-18m³)

Number of water drops would be produced= 9.55×10¹⁵ (answer)

Total work required = surface tension × area

= 72×10^-3N/m × 9.55×10¹⁵×(4×π×(1000×10^-9m)²)

= 8640 N.m

= 8640 J. (Answer)

= 8.64 kJ (answer)