Elastic balloon of volume V in vacuum

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Homework Statement


Take an elastic balloon of volume V in vacuum. The surface of the balloon has tension T. Find the pressure inside the balloon in terms of V and T, then combine this to the ideal gas law to find an expression for V.

Homework Equations


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The Attempt at a Solution


I think one should use something like dU=-pdV, but you need to add a term with T. From dimensional analysis you get TdA. So you have dU=-pdV+TdA. But I'm confused on what happens then, I get something like

p=-(dU/dV)+T(dA/dV)

But then how does this help if I want V?
 
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  • #2
Quantum Defect
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Homework Statement


Take an elastic balloon of volume V in vacuum. The surface of the balloon has tension T. Find the pressure inside the balloon in terms of V and T, then combine this to the ideal gas law to find an expression for V.

Homework Equations


See below

The Attempt at a Solution


I think one should use something like dU=-pdV, but you need to add a term with T. From dimensional analysis you get TdA. So you have dU=-pdV+TdA. But I'm confused on what happens then, I get something like

p=-(dU/dV)+K(dA/dV)

But then how does this help if I want V?
You have a static equilibrium. Pressure inside the balloon (force pushing out) is exactly compensated for by the tension in the balloon (force decreasing the size of the balloon).

How can you relate the increase in area of the balloon to the increase in volume?
 
  • #3
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maybe like this: dA/dV should go like 1/dr where r is the radius. If you do it for the whole balloon this is A/V=3/r. So effectively you get p = 3T/r = 3T (4π/3)1/3 V-1/3

then you plug this in the ideal gas law (I call the temperature τ) and you get pV=3T (4π/3)1/3 V-1/3 V = Nkτ and solving for V you get V = (Nkτ/3T)3/2 (4π/3)-1/2

I'm not sure about the 3 from A/V
 
  • #4
haruspex
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I think one should use something like dU=-pdV
Finding the pressure is much easier than that.
Think of the sphere in two halves. What is the force pushing them apart? What is the force holding them together?
 
  • #5
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Ok, one half gets F=pπr2 because the area is effectively just the one of the equator (right?). The force that holds together the two halves is the circle 2πr times the tension T. So you get p=2T/r. Then you have exactly the same calculation, but with a 2 instead of a 3: V = (Nkτ/2T)3/2(4π/3)-1/2

Btw I found this http://en.wikipedia.org/wiki/Surface_tension#Thermodynamics_of_soap_bubbles where they get the same using dA/dV!
 

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