Question

Calculate the amount of work done in spraying a drop of a liquid of radius 1 mm into million identical droplets under isothermal conditions. [surface tension of liquid $=550\times10^{-3}\text{Nm}^{-1}$]

• A. $6.84\times10^{-4} \text{ J}$
• B. $5.50\times10^{-4} \text{ J}$
• C. $7.25\times10^{-4} \text{ J}$
• D. $3.42\times10^{-4} \text{ J}$

Hint:

Shortcut formula for splitting drops: $W = 4\pi R^2 T (n^{1/3} - 1)$. For $10^6$ drops, $n^{1/3} = 100$, making the term $(100 - 1) = 99$.

Answer 1

The Correct Option is A

Step 1: Work equals extra surface energy.
Breaking one drop into many makes much more surface area. The work done is $W = T\,\Delta A$, the surface tension times the area increase.

Step 2: Volume stays the same.
The total liquid does not change, so $\tfrac{4}{3}\pi R^{3} = n\cdot\tfrac{4}{3}\pi r^{3}$, giving $r = R\,n^{-1/3}$.

Step 3: Find the small radius.
With $R = 10^{-3}$ m and $n = 10^{6}$: \[ r = 10^{-3}\times(10^{6})^{-1/3} = 10^{-3}\times 10^{-2} = 10^{-5} \text{ m} \]
Step 4: Work out the area change.
\[ \Delta A = 4\pi(n r^{2} - R^{2}) = 4\pi(10^{6}\cdot 10^{-10} - 10^{-6}) \]\[ = 4\pi(10^{-4} - 10^{-6}) = 4\pi\times 99\times 10^{-6} \text{ m}^{2} \]
Step 5: Multiply by surface tension.
With $T = 550\times 10^{-3}$ N m$^{-1}$: \[ W = 550\times 10^{-3}\times 4\pi\times 99\times 10^{-6} \]
Step 6: Get the number.
\[ W \approx 6.84\times 10^{-4} \text{ J} \]This is option 1.
\[ \boxed{W \approx 6.84\times 10^{-4} \text{ J}} \]