The question is:

Using the Young-Laplace Equation (if applicable), find the surface tension (dynes/cm) for water at 20 degrees Celsius with 2.5 psi. Round to the nearest tenth.

Well, I didn't use the Young-Laplace equation, not sure if needed though. What I did was use the Eötvös rule and its special case for water to solve the question. The equation is:

$$\gamma = 0.07275\;\frac{N}{m}\;\times\;(1-0.002\times(T-291K))$$

What I did was convert 20 Celsius to Kelvin (293K) and then put it in the equation to get:

$$\gamma = 0.07275\;\frac{N}{m}\;\times\;(1-0.002\times(293K-291K))= 0.072459\frac{N}{m}$$

However, I think I may be wrong as this does not account for pressure at all. Which ends up becoming about $72.46\frac{dynes}{cm}$ Am I right or wrong? And is there a better/correct way of doing this?

  • 4
    $\begingroup$ Why the downvote? This question definitely "shows research effort"... $\endgroup$
    – wchargin
    Commented May 7, 2013 at 3:20
  • $\begingroup$ Is it safe to assume we're dealing with a water droplet? I'd like to model the problem as a spherical bead of water in air whose radius is a function of temperature (and pressure). Also, I'm fairly certain whether or not the Young-Laplace equation is applicable hinges on this point. $\endgroup$
    – David H
    Commented May 7, 2013 at 19:19
  • $\begingroup$ @DavidH, I honestly don't know, this is all the context I'm given, so I can't be certain, but I'm guessing you could. $\endgroup$
    – Rivasa
    Commented May 7, 2013 at 19:26
  • $\begingroup$ @Link On an unrelated note, whoever wrote this question is either evil or an idiot just for using pounds-per-square-inch pressure units but dynes and centimeters for force and distance units. $\endgroup$
    – David H
    Commented May 7, 2013 at 19:33
  • $\begingroup$ @DavidH, agreed :p I actually had to search up dynes to see... And that would be my professor $\endgroup$
    – Rivasa
    Commented May 7, 2013 at 19:35

1 Answer 1


Using the Young-Laplace Equation (if applicable)

Basically a trick question trying to get you to equate the pressure in the question with $\Delta P$ in the Young-Laplace equation.

The actually pressure dependence of water's surface tension is given in On the Evaluation of the Surface Tension-Pressure Coefficient for Pure Liquids

The rate of change of surface tension with respect to pressure is $7 \times 10^{-8} cm$ near atmospheric pressure. So since the question says "Round to the nearest tenth", the pressure effect is insignificant.


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