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Topic: Exercise about contact angles.  (Read 819 times)

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Offline arnakon

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Exercise about contact angles.
« on: October 25, 2020, 01:34:26 PM »
Hi, there is a problem that is giving me a hard time. The exercise states as follows: " If the thermodynamic contact angle made by water (saturated with benzene) against a given solid is θW = 70°, and that by benzene (saturated with water) against the same solid is θB = 30°, calculate the contact angle of a water drop on the solid surface immersed in benzene. The surface and interfacial tensions under the given conditions (i.e., mutually saturated liquids) are: σW = 62.2 mN/m; σB = 28.8 mN/m, and σB/W = 35.0 mN/m. The solid surface is perfectly smooth and homogeneous. Also, explain why one must consider mutually saturated liquids to solve the problem. "
I thought to just use young's equation:

But I'm not sure to be right. The angle I obtain is around 130°C, but I didn't use the values of contact angles provided in the text. I don't think they are a red herring, so I'm not sure of my reasoning. I don't even know the numerical answer sadly. Anyone could help? Thank you in advance to everyone.

Offline Corribus

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Re: Exercise about contact angles.
« Reply #1 on: October 26, 2020, 10:54:55 AM »
What values are you using when you solve the young equation?

You are basically given information (contact angles and tensions) for two solid-liquid-air systems (water-air-solid and benzene-air-solid, air implied) and asked to use that information to solve for the water-benzene-solid system. So, you'll need to use the contact angles given for the air systems to solve for the missing information in the air systems, and use that missing information to solve for the contact angle in the liquid-liquid-solid system. If that makes sense.

Do you understand the bit about using mutually saturated liquids?

What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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