Measuring surface tension of a bubbles.

  • #1

Homework Statement


The pressure by the walls of a bubble is 4σ/r (where σ is the surface tension).
The radius r of 2 bubbles is measured. After that you bring the two together by connecting them with a straw (the smaller bubble has higher pressure and gets smaller, while the bigger bubble gets bigger). You now measure the new radius of the new bubble. You do this for different values of r_1 and r_2. Find the value of σ.
(given atmospheric pressure)

Homework Equations


p=4σ/r
(PV=nRT) (I think)

The Attempt at a Solution


Solving the system of equations (PV=nRT for each bubble)
The pressure P=p_walls + p_atmosph.
You get V by r. T is alway the same (therm. equilibrium)
nR adds, when bubbles are fused.
Solve for σ.
Should I add another relation between the different p_walls?
Is this approach ok, or is there something I missed?
 

Answers and Replies

  • #2
21,774
4,951

Homework Statement


The pressure by the walls of a bubble is 4σ/r (where σ is the surface tension).
The radius r of 2 bubbles is measured. After that you bring the two together by connecting them with a straw (the smaller bubble has higher pressure and gets smaller, while the bigger bubble gets bigger). You now measure the new radius of the new bubble. You do this for different values of r_1 and r_2. Find the value of σ.
(given atmospheric pressure)

Homework Equations


p=4σ/r
(PV=nRT) (I think)

The Attempt at a Solution


Solving the system of equations (PV=nRT for each bubble)
The pressure P=p_walls + p_atmosph.
You get V by r. T is alway the same (therm. equilibrium)
nR adds, when bubbles are fused.
Solve for σ.
Should I add another relation between the different p_walls?
Is this approach ok, or is there something I missed?

Your interpretation of the starting equation is incorrect. The gas pressure inside the bubble minus the gas pressure outside the bubble is 4σ/r. You know that the pressure outside the bubble is 1 atm., so, knowing the bubble radius and the surface tension, you can calculate the pressure inside the bubble. From the radius of the bubble, you can get its volume. (I assume air is the gas inside the bubble also). You can then use the ideal gas law to determine the number of moles of air inside each bubble. The number of moles of air in the final bubble is equal to the number of moles in the two initial bubbles. The final pressure inside the coalesced bubble is determined by its radius, and the final volume of the coalesced bubble is also determined by its radius. Use this information and the number of moles in the bubble, along with the ideal gas law, to determine the radius of the final bubble.
 
  • #3
@Chestmiller
I basically did exactly this, but I think there is information missing, when you don't know the values of the number of gas particles.
[itex]
(\frac{4\sigma}{r_1}+1atm)\frac{4\pi}{3}r^3_1=n_1 \cdot R\cdot T\\

(\frac{4\sigma}{r_2}+1atm)\frac{4\pi}{3}r^3_2=n_2 \cdot R\cdot T\\

(\frac{4\sigma}{r_3}+1atm)\frac{4\pi}{3}r^3_3=(n_1+n_2) \cdot R\cdot T\\
[/itex]
3 equations, 4 unknowns ([itex]n_1, n_2, \sigma, T[/itex])

Why doesn't this work?
 
  • #4
Oh, I see:
[itex]
(\frac{4\sigma}{r_1}+1atm)\frac{4\pi}{3}r^3_1 +
(\frac{4\sigma}{r_2}+1atm)\frac{4\pi}{3}r^3_2=
(\frac{4\sigma}{r_3}+1atm)\frac{4\pi}{3}r^3_3
[/itex]
That should be it.
 
Last edited:
  • #5
this is true for any sigma.
:frown:
what is wrong?
 
  • #6
21,774
4,951
this is true for any sigma.
:frown:
what is wrong?

Nothing is wrong. This result is correct. Now, all you need to do is solve the equation explicitly for σ in terms of r1, r2, and r3.
 

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