Question

A film of stearic acid partially covers the water surface in a container. The work needed to decrease this coverage by 1 cm² is 25.0 × 10⁻⁷ J. The surface tension (in N/m) of the film is .......
(Round off to three decimal places)
(Surface tension of pure water is 0.072 N/m)

Hint:

Surface tension is calculated by dividing the work required to change the surface area by the change in area.

Answer 1

Correct Answer: 0.04

Step 1: Understanding surface tension.
The work required to change the surface area of a liquid is related to the surface tension \( \gamma \) by: \[ W = \gamma \Delta A \] Where \( W \) is the work, \( \Delta A \) is the change in area, and \( \gamma \) is the surface tension.

Step 2: Calculation.
Substitute \( W = 25.0 \times 10^{-7} \, \text{J} \) and \( \Delta A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \): \[ \gamma = \frac{25.0 \times 10^{-7}}{1 \times 10^{-4}} = 0.025 \, \text{N/m} \] Thus, the surface tension is \( \boxed{0.025} \, \text{N/m} \).