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Why don't the surface tension of the film of soap solution ( or any other liquid) increases or decreases when we increase the area of the film manually by applying a force ?? Isn't it analogous to the increase in tension of the string on stretching it.

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    $\begingroup$ If the fluid is non-Newtonian then it is possible to have surface elasticity (in addition to surface tension). In that case your analogy with the stretched string will hold. $\endgroup$
    – Deep
    Commented Oct 4, 2020 at 13:55
  • $\begingroup$ @Deep is there any relation between viscosity and surface tention ?? $\endgroup$
    – user270071
    Commented Oct 5, 2020 at 5:46
  • $\begingroup$ None that I know of. However both of them are affected significantly by variations in temperature (among other factors). $\endgroup$
    – Deep
    Commented Oct 5, 2020 at 5:47

2 Answers 2

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There are two ways to look at surface tension. One is as a force per unit length and the other is as an interfacial energy i.e. an energy per unit area associated with the interface. So when we say the surface tension of an air water interface is $0.0728~\textrm{Nm}$ we could also make the equivalent statement that the energy associated with the air-water interface is $0.0728~\mathrm{J/m}^2$. There are various ways of showing this, but you might be interesting in calculating the dimensions of the units of $\mathrm{Nm}$ and $\mathrm{J/m}^2$ and showing that they are the same.

The point of all this is that the interfacial energy is arguably the more fundamental quantity because it depends on the way the water molecules interact at the air-water interface. The interfacial energy per unit area is a property of the interface and does not depend on how much interfacial we have. One square metre of interface has the same energy per unit area as a hundred or a thousand square metres. And since the energy per unit area does not depend on the interfacial area, and since energy per unit area and the surface tension are the same thing, that means the surface tension also does not depend on the interfacial area.

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  • $\begingroup$ There are two ways to increase the area of the interfacial surface , one is when we pour the liquid in a wider container . And other way is the same as i stated in the question i.e. film area expansion . In the former case the no. of molecules can increase or decrease on the interface but in the other case when we increase the film area how do the molecules manage to increase so as to maintain the value of surface tension constant $\endgroup$
    – user270071
    Commented Oct 2, 2020 at 10:06
  • $\begingroup$ @user270071 suppose you have the liquid in a container, then the only way you can increase the area is to make the container wider and that makes the liquid shallower. What happens is that some of the water molecules that were in the bulk of the liquid have now moved to the surface. When those molecules migrate from the bulk to the surface their energy goes up and that is what causes the increase in the interfacial energy. Or another way to look at it is that the interfacial energy per molecule at the interface is constant, and you are increasing the number of molecules at the interface. $\endgroup$ Commented Oct 2, 2020 at 11:12
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It's because surface tension by definition is force per unit length. Or if to be expressed in terms of surface energy :

enter image description here

Then surface tension is surface energy change due to area change :

$$ \gamma = \frac {dE}{dA} $$

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  • $\begingroup$ Can you give the detailed explaination $\endgroup$
    – user270071
    Commented Oct 2, 2020 at 9:53
  • $\begingroup$ There's not much to explain. Just look at Surface tension in terms of energy. Increase of film surface area by force, results in surface potential energy increase. And ratio of these quantities is by definition a surface tension. I think everything is pretty self-explanatory here. $\endgroup$ Commented Oct 2, 2020 at 11:44
  • $\begingroup$ I think you mean per unit area? $\endgroup$ Commented Oct 2, 2020 at 12:57
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    $\begingroup$ @CarlWitthoft Potential energy gained per unit area increase, $dA=Ldx$ $\endgroup$ Commented Oct 2, 2020 at 13:43

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