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(a) To derive the Laplace-Young relation (eqn 11.9 ), imagine a spherical object with internal pressure \(p_{i n}\) and external pressure pout. This pressure difference leads to an outward pointing normal force at all points on the membrane, Perform the thought experiment shown in Figure 11.45 in which the spherical shell is cut at the equator and the tension in the membrane acts downward to counterbalance the upward force due the pressure difference. Derive the Laplace-Young relation by balancing the forces acting in the positive z-direction due to the pressure and in the negative z-direction due to the surface tension. Note that the surface tension can be interpreted as a force per unit length acting downward at the equator. Ignore any contribution due to bending energy and focus strictly on the role of the surface tension. (b) Use energy arguments (rather than force balance arguments) to derive the same result. To do so, consider a small change in the radius of the sphere. As a result of this change in radius, we will incur a free-energy change associated with \(p V\) work and a part having to do with YA work. By insisting that the change in free energy for such an excursion be zero obtain the Laplace-Young relation.

Short Answer

Expert verified
The Laplace-Young relation is derived as \(T = R \cdot (p_{in} - p_{out}) / 2\). This obtained result is based on the balance of forces acting on the sphere where the surface tension and pressure difference play a crucial role. The same result is also derived from the energy conservation consideration for small changes in radius of the sphere where a null change in free energy gives back the Laplace-Young relation.

Step by step solution

01

Analysis of forces

Consider a spherical shell of radius \(R\) with internal pressure \(p_{in}\) and external pressure \(p_{out}\), the pressure difference (\(p_{in} - p_{out}\)) results in a force acting on the sphere in the outward direction. This force can be expressed as \(F = p_{in} \cdot A - p_{out} \cdot A\), where \(A\) is the surface area of the sphere.
02

Balancing the forces

Since the force due to the surface tension of the sphere acts downwards, it balances the force acting in the outward direction due to the pressure difference. The force due to the surface tension,\(F_s\), can be calculated by multiplying the surface tension (\(T\)) by the circumference at the equator (\(2 \pi R\)). Therefore, the forces are balanced when \(F = F_s\), giving \(p_{in} \cdot A - p_{out} \cdot A = T \cdot 2 \pi R\).
03

Deriving the Laplace-Young relation

To derive the Laplace-Young relation from the above equation, we replace \(A\) with \(4 \pi R^2\), which is the surface area of a sphere. Then we isolate \(T\) on one side, giving the Laplace-Young relation: \(T = R \cdot (p_{in} - p_{out}) / 2\).
04

Considering energy change

For a small change in the radius of the sphere, there will be a change in free energy associated with \(pV\) work (where \(p\) is the pressure and \(V\) is the volume) and a part having to do with \(T \cdot A\) work. The change in free energy (\(\delta G\)) is thus given by: \(\delta G = p \delta V + T \delta A\).
05

Establishing the zero free-energy condition

If we insist that the change in free energy be zero for such an excursion, energy balance gives: \(0 = p \delta V + T \delta A\).
06

Back to Laplace-Young relation

Considering that \(V = 4/3 \pi R^3\) and \(A = 4 \pi R^2\), we replace these values in the last equation remembering that only small changes in the radius (\(dR\)) are considered. Now, by cancelling identical terms, equating to zero and isolating \(T\), we again get the Laplace-Young relation: \(T = R \cdot (p_{in} - p_{out}) / 2\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Tension
Surface tension is a crucial concept that describes the elastic tendency of a fluid surface. It arises due to the cohesive forces between liquid molecules. At a molecular level, surface tension can be thought of as an "invisible skin" on the liquid's surface. This skin is always trying to compress, minimizing the surface area. This phenomenon results from the imbalance of intermolecular forces experienced by molecules at the surface compared to those in the bulk of the liquid.
For practical understanding, consider how small insects can walk on water. Surface tension supports their weight because the cohesive forces are strong enough to resist external pressure.
In the context of a spherical shell, like a bubble, surface tension plays a pivotal role. It acts along the shell's surface and resists the outward pressure exerted by the trapped gas inside. This tension helps maintain the bubble's shape and ensures its structural stability.
Pressure Difference
Pressure difference refers to the disparity in pressure across two sides of an interface. In our scenario, it's the difference between the internal pressure (\(p_{in}\)) within a spherical shell and the external pressure (\(p_{out}\)) outside of it. This difference is significant as it drives the forces acting on the surface of the shell.
Pressure difference contributes to the net force trying to expand the shell. The internal pressure pushes outwards, while the external pressure pushes inwards against the shell. This interplay of forces creates a balanced state when equilibrium is achieved, primarily involving surface tension to counteract these pressures.
In applications like bubbles or biological membranes, controlling and understanding pressure difference is paramount. It determines the stability and behavior of these structures in various environments, affecting how they grow, shrink, or even burst when influenced by external conditions.
Spherical Shell
A spherical shell is a thin, hollow structure shaped like a sphere. Imagine a soap bubble or a balloon. These are classic examples where we encounter spherical shells in everyday life. The concept of a spherical shell is significant in many scientific and engineering contexts due to its symmetrical properties.
For the Laplace-Young Relation, envision a spherical shell that separates two different pressure zones. The inward and outward forces acting on it due to pressure difference and surface tension are balanced, keeping the shell stable.
Studying spherical shells helps us understand how natural and man-made spheres operate under various conditions. Key features include their uniform stress distribution and resistance to uniform pressure changes. This concept is vital not only in fluid dynamics and material science but also in fields like medicine (e.g., modeling biological cells) and aerospace engineering (e.g., designing pressurized cabins).

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Most popular questions from this chapter

The pipette aspiration experiment described in the chapter can be used to measure the bending modulus \(K_{b}\) as well as the area stretch modulus. Lipid bilayer membranes are constantly jostled about by thermal fluctuations. Even though a flat membrane is the lowest energy state, fluctuations will cause the membrane to spontaneously bend The goal of this problem is to use equilibrium statistic mechanics to predict the nature of bending fluctuations and to use this understanding as the basis of experimental measurement of the bending modulus. (NOTE: This problem is challenging and the reader is asked to con sult the hints on the book website to learn more of our Fourier transform conventions, how to handle the relevant delta functions, the subtleties associated with the limits of integration, etc.). (a) Write the total free energy of the membrane as an integral over the area of the membrane. Your result should have a contribution from membrane bending and a contribution from membrane tension. Write your result using the function \(h(\mathbf{r})\) to characterize the height of the membrane at position \(\mathbf{r}\). (b) The free energy can be rewritten using a decomposition of the membrane profile into Fourier modes. Our Fourier transform convention is $$h(\mathbf{r})=\frac{A}{(2 \pi)^{2}} \int h(\mathbf{q}) \mathrm{e}^{-\mathrm{i} \mathbf{q} \mathbf{r}} \mathrm{d}^{2} \mathbf{q.}$$ where \(A=L^{2}\) is the area of the patch of membrane of inter est. Plug this version of \(h(\mathbf{r})\) into the total energy you derived above (that is, bending energy and the energy related to tension) to convert this energy in real space to an energy in \(q\) -space. Note that the height field in \(q\) -space looks like a sum of harmonic oscillators. (c) Use the equipartition theorem in the form \(\langle E(\mathbf{q})\rangle=\) \(k_{B} T / 2,\) where \(E(\mathbf{q})\) is the energy of the \(q\) th mode and the free energy can now be written as $$F[\hbar(\mathbf{q})]=\frac{A}{(2 \pi)^{2}} \int E(\mathbf{q}) \mathrm{d}^{2} \mathbf{q}.$$ Use this result to solve for \(\left(\left.\mathrm{h}(\mathbf{q})\right|^{2}\right\rangle\) which will be used in the remainder of the problem. (d) We now have all the pieces in place to compute the relation between tension and area and thereby the bending modulus. The difference between the actual area and the projected area is $$A_{a c t}-A=\frac{1}{2} \int(\nabla h(\mathbf{r}))^{2} \mathrm{d}^{2} \mathbf{r}.$$ This result can be rewritten in Fourier space as $$\left(A_{a c t}-A\right)=\frac{1}{2} \frac{A^{2}}{(2 \pi)^{2}} \int_{\pi / \sqrt{A}}^{\pi / \sqrt{\alpha_{0}}} q^{2}\left(\left.\pi \tilde{h}(\mathbf{q})\right|^{2}\right) 2 \pi q d q.$$ Work out the resulting integral which relates the areal strain to the bending modulus, membrane slze, temperature, and tension. The limits of integration are set by the overall size of the membrane (characterized by the area \(A\) ) and the spacing between lipids \(\left(a_{0}\right.\) is the area per lipid), respectively. This result can be directly applied to micropipette experiments to measure the bending modulus. In particular, using characteristic values for the parameters appearing in the problem suggested by Boal (2002) such as \(r=10^{-4} \mathrm{J} / \mathrm{m}^{2}, K_{b} \approx 10^{-1} \mathrm{J}\) and \(a_{0} \approx 10^{-20} \mathrm{m}^{2}\) show that $$\frac{\Delta A}{A} \approx \frac{k_{B} T}{8 \pi K_{B}} \ln \left(\frac{A r}{K_{D} \pi^{2}}\right).$$ and describe how this result can be used to measure the bending modulus. (For an explicit comparison with data, see Rawicz et al., 2000 . For further details on the analysis, see Helfrich and Servuss, \(1984 .\) An excellent account of the entire story covered by this problem can be found in Chapter 6 of 30 al, 2002 .)

The height function used in the chapter can be used to compute the change in area of a membrane when it suffers some deformation. Consider the height function \(h(x, y)\) defined over an \(L \times L\) square in the \(x-y\) plane. The function \(h(x, y)\) tells us the deformed state of the \(L \times L\) square. To compute the change in area we divide the square into small \(\Delta x \times \Delta y\) squares and consider how each of them Is deformed on the surface \(h(x, y) \text { (see Figure } 11.17(\mathrm{A}))\) The function \(h(x, y)\) maps the small square into a tilted parallelogram. (a) Show that the two sides of the parallelogram are given by the vectors $$\Delta x\left(e_{x}+\frac{\partial h}{\partial x} \mathbf{e}_{z}\right), \text { and } \Delta y\left(\mathbf{e}_{y}+\frac{\partial h}{\partial y} \mathbf{e}_{z}\right).$$ where \(\mathbf{e}_{x}, \mathbf{e}_{y},\) and \(\mathbf{e}_{z}\) are unit vectors. (b) The magnitude of the vector product of two vectors Is equal to the area subsumed by them, that is, a \(x \mathbf{b}=\) absin\theta, where \(\theta\) is the angle between vectors a and \(\mathbf{b}\). Using this fact show that the area of the parallelogram is $$a^{\prime}=\sqrt{1+\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}} \Delta x \Delta y.$$ (c) Now assume that the partial derivatives appearing in the above equation are small and expand the squareroot function using \(\sqrt{1+x^{2}}=1+4 x^{2},\) to demonstrate the useful formula for the area change, $$\Delta a=a^{\prime}-a=\frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}\right] \Delta x \Delta y.$$ By integrating this expression over the entire domain, we can find the total area change associated with the surface. Divide both sides by \(a=\Delta x \times \Delta y\) to find a formula for the relative change of area \(\Delta a / a\).

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