6
$\begingroup$

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?

$\endgroup$
6
  • 4
    $\begingroup$ Why the downvote? This question definitely "shows research effort"... $\endgroup$
    – wchargin
    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
    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
    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
    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
    May 7, 2013 at 19:35

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.