Chapter 8, Problem 8.26NP

(a)

Interpretation Introduction

Interpretation:

The difference is pressure across the liquid-air interface for a mercury droplet of radius 125 nm needs to be calculated

Concept Introduction:

Surface tension (γ) is the energy required to increase the surface area of a liquid by 1 m²

For a droplet of radius r, the surface tension across a pressure gradient ΔP is given as:

  $\begin{array}{l}\text{ΔP = }\frac{\text{2γ}}{\text{r}}\text{ -----(1)}\\ \Delta {\text{P = P}}_{\text{inner}}-{\text{P}}_{\text{outer}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The difference is pressure across the liquid-air interface for a methanol droplet of radius 125 nm needs to be calculated

Concept Introduction:

Surface tension (γ) is the energy required to increase the surface area of a liquid by 1 m²

For a droplet of radius r, the surface tension across a pressure gradient ΔP is given as:

  $\begin{array}{l}\text{ΔP = }\frac{\text{2γ}}{\text{r}}\text{ -----(1)}\\ \Delta {\text{P = P}}_{\text{inner}}-{\text{P}}_{\text{outer}}\end{array}$