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 r_{0} < r < 2.5 r_{0}),
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 r_{0} < r < 2.5 r_{0} find the
pressure difference ΔP = P_{i} - P_{o} 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

F_{P}= 2πr^{2}∫_{0}^{π/2}sinθ dθ ΔP cosθ = 2πr^{2}ΔP ∫_{0}^{1}x dx = πr^{2}ΔP.

The net force due to the surface tension points to the left. It has magnitude

F_{γ}= -2*γ2πr = -4γπr.

For equilibrium we need F_{P}+ F_{γ}= 0

πr^{2}ΔP = 4γπr. ΔP = 4γ/r.

(b) In equilibrium we have P_{total}(r) = -P_{atm}- 4γ/r + 3nRT/(4πr^{3}) = -a - b/r + c/r^{3}= 0, with a, b, and c positive constant.

P_{total}(r + Δr) = dP/dr|_{r}Δr = (b/r^{2}- 3c/r^{4})Δr = (1/r) (b/r + a - a - c/r^{3}+ c/r^{3}- 3c/r^{4})Δr

= (1/r)(-P_{total}(r) - a - 2c/r^{3})Δr = -(1/r)( a + 2c/r^{3})Δr = -constant*Δr.

We have a restoring force per unit area, the balloon is stable.