Solved thermodynamics exercise

The work needed to stretch a soap film

Exercise 13 · Lesson 4 — The First Law of Thermodynamics

surface tension
generalized work
interface
soap film

Statement

A flat soap film is stretched across a rectangular frame, one side of which is a bar of length $ℓ$ that can slide without friction. The distance $x$ between this bar and the opposite side of the frame increases from $x_{A}$ to $x_{B}$ . The film has two air–liquid interfaces, each with surface tension $γ$ , assumed constant. Edge effects, weight and the work of pressure forces are neglected; the volume of liquid is constant and no liquid is exchanged. The displacement is slow enough for mechanical equilibrium to be maintained. For the creation of interface, the work received is $δ W = γ d A$ , where $A$ is the total area of the interfaces.

Draw the frame and the bar, and show $x$ , $ℓ$ and the direction of the external pulling force. Express $A (x)$ , then $d A$ .
Determine the external force required and calculate the work received when $x$ goes from $x_{A}$ to $x_{B}$ .
Calculate this force and this work for $γ = 30 mN m^{- 1}$ , $ℓ = 5, 0 cm$ and $x_{B} - x_{A} = 2, 0 cm$ . What error would result from forgetting one face?
The film is kept at constant temperature by a thermal reservoir. Is knowledge of $γ$ and of the change in area alone enough to determine $Δ U$ and $Q$ ?

Hint

Each face has area

ℓ x

. Equate the work

F_{ext} d x

with the work of creating the two interfaces. For the last question, distinguish the work received from the energy stored.

Detailed solution

Solution

Question 1. The frame is drawn in the plane of the film, with the movable bar perpendicular to the

x

direction and the external pulling force pointing towards increasing

x

. Each face has area

ℓ x

. The total area and its change are therefore

A = 2 ℓ x, d A = 2 ℓ d x,

since $ℓ$ is constant. The two interfaces are distinct, even though they overlap in a front view.

Question 2. The work received can be written in two ways:

δ W = F_{ext} d x = γ d A = 2 γ ℓ d x .

Equating them gives $F_{ext} = 2 γ ℓ$ . The force exerted by the film opposes this pull. Since $γ$ and $ℓ$ are constant,

W_{A \to B} = \int_{x_{A}}^{x_{B}} 2 γ ℓ d x = 2 γ ℓ [x]_{x_{A}}^{x_{B}} = 2 γ ℓ (x_{B} - x_{A}) > 0.

Increasing the area of the film requires an input of work.

Question 3. In SI units,

γ = 30 \times 1 0^{- 3} N m^{- 1}

ℓ = 5, 0 \times 1 0^{- 2} m

and

x_{B} - x_{A} = 2, 0 \times 1 0^{- 2} m

. Thus

F_{ext} = 2 (30 \times 1 0^{- 3}) (5, 0 \times 1 0^{- 2}) = 3, 0 \times 1 0^{- 3} N, W_{A \to B} = (3, 0 \times 1 0^{- 3}) (2, 0 \times 1 0^{- 2}) = 6, 0 \times 1 0^{- 5} J = 60 μ J .

The units are consistent: $γ ℓ$ is a force and $γ Δ A$ an energy. Forgetting one face would wrongly halve both the force and the work.

Question 4. The first law only requires

Δ U = Q + γ Δ A .

The reservoir may supply or receive heat to keep the temperature constant. Without a model of the internal energy of the film, neither $Δ U$ nor $Q$ is known. One therefore cannot automatically identify $γ Δ A$ with a change in internal energy, nor deduce $Q = 0$ from the fact that the process is isothermal.