### Surface tension

#### Problem:

The surface tension γ is defined as the force along a line of unit length, γ = F/L (N/m).  The force is parallel to the surface and perpendicular to the line.   For an inflated, (1.5 r0 < r < 2.5 r0), spherical rubber balloon γ is nearly constant. The surface tension balances the outward force due to the pressure difference between the inside and the outside air.
(a)  For a balloon inflated to a radius r in the range 1.5 r0 < r < 2.5 r0 find the pressure difference ΔP = Pi - Po as a function of r and γ.
(b)  Show that is in a stable equilibrium.

Solution

• Concepts:
Surface tension
• Reasoning:
The balloon skin has an inner and an outer surface.  The net force per unit length holding the skin together is 2γ.
• Details of the calculation:

(a)  Imagine a cut through the balloon as shown.  Consider the right half of the balloon.
The net force due to the pressure difference points to the right.  It has magnitude
FP = 2πr20π/2sinθ dθ ΔP cosθ = 2πr2 ΔP ∫01x dx = πr2 ΔP.
The net force due to the surface tension points to the left.  It has magnitude
Fγ = -2*γ2πr = -4γπr.
For equilibrium we need FP + Fγ = 0
πr2 ΔP = 4γπr.  ΔP = 4γ/r.

(b)  In equilibrium we have Ptotal(r) = -Patm - 4γ/r + 3nRT/(4πr3) = -a - b/r +  c/r3 = 0, with a, b, and c positive constant.
Ptotal(r + Δr) = dP/dr|rΔr = (b/r2 - 3c/r4)Δr = (1/r) (b/r + a - a - c/r3 +  c/r3 -  3c/r4)Δr
= (1/r)(-Ptotal(r) - a - 2c/r3)Δr = -(1/r)( a + 2c/r3)Δr = -constant*Δr.
We have a restoring force per unit area, the balloon is stable.