Physics Including Human Applications 292 Chapter 14 MOLECULAR MODEL OF MATTER GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: rms velocity surface tension adhesion capillarity cohesion osmosis Molecular Model Explain the properties of fluids using the molecular model. Problems Solve problems that relate the gas laws to the properties of the gas and that involve surface tension, capillarity, and osmosis. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, Chapter 6, Momentum and Impulse, Chapter 10, Temperature and Heat, and Chapter 13, Elastic Properties of Materials.
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Chapter 14 MOLECULAR MODEL OF MATTER GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: rms velocity surface tension adhesion capillarity cohesion osmosis Molecular Model Explain the properties of fluids using the molecular model. Problems Solve problems that relate the gas laws to the properties of the gas and that involve surface tension, capillarity, and osmosis. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, Chapter 6, Momentum and Impulse, Chapter 10, Temperature and Heat, and Chapter 13, Elastic Properties of Materials.
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Chapter 14 MOLECULAR MODEL OF MATTER
14.1 Introduction In our previous discussions of the properties of materials, we have treated them as if they were continuous media. But if materials are assumed to be a uniform gelatinous substance, how can we explain the existence of three different states of matter, solid, liquid, and gas? In a continuous medium what can be the cause of pressure acting upward on the inside of the top of a container holding a confined gas? Even though all of our previous explanations have been based upon a continuous-media model of matter, they seem to explain natural phenomena properly. However, there are some phenomena that are puzzling in our present framework. We have, for example, no way of explaining the fact that evaporation is a cooling process or the fact that one material can diffuse into another. How is a continuous medium made thin enough to be a gas and also thick enough to be a solid? It is clear that our continuous- media model has some serious limitations. With what model of matter shall we replace it? 14.2 The Molecular Model of Matter If matter seems not to behave like a uniform gelatin, then perhaps we can describe it as being discontinuous, made up of individual clumps being more or less far apart. In developing our model here is how we shall proceed.
1. We begin by making some hypotheses, or assumptions, or conjectures, to form a model, or mental construct, to explain a natural phenomenon.
2. We use the laws, conservation principles, and mathematics we know with this model to make predictions and develop explanations.
3. We test these predictions and explanations against experimental results. We cannot show that our model is correct. We can show that the predictions of our model agree with experimental results, but another model that gave the same predictions would be equally good. However, if we can disprove experimentally the predictions that follow from our model, then we know our model is wrong. You will notice that our model building and testing procedure is a powerful way of showing that a model is wrong. It is less useful in proving a model is correct because it is always possible that if you can make a new prediction from our model it might prove to be wrong. Nevertheless, we tend to find comfort in simple models that we can use to provide us with a feeling of understanding our universe. On the basis of our experience with materials let us postulate some basic ingredients of our molecular model of matter. We postulate that all material is composed of small entities called molecules. We may think of molecules as very small, perfectly elastic spheres. The molecules are close together in solids and far apart in gases.
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We postulate that molecules are always in motion. This gives us a logical way of explaining the internal energy term in the first law of thermodynamics. It can be considered the total kinetic energy of all the molecules in the material. We postulate that there are interactions between the molecules. From our experiences and observations of the properties of the three states of matter, we can draw some conclusions about the nature of the interaction between molecules. If you have ever tried to pull a small wire apart, you know that a force is required and that as the diameter of the wire is increased, you will soon find that you cannot pull the wire apart. From this experiment you can predict that on the molecular scale in the wire there is an attractive force. Also, you have observed that very little force is required to separate a liquid into parts. For a gas, it seems that molecules are quite independent from one another. An experiment that suggests a repulsive force comes from attempts to reduce the volume of a solid or liquid. In the chapter on elasticity you learned that large forces are required to compress liquids and solids. These observations suggest that the force between molecules changes as the distance between the molecules change. We expect there exists some equilibrium position where repulsive and attractive forces between molecules cancel. At some very small distances the molecules must repel each other. At larger distances there is an attractive force, and finally at very large distances the molecules feel almost no forces from other molecules. The general shape of the curve that represents the force between molecules as a function of the separation between molecules is shown in Figure 14.1. You will notice that a small displacement from equilibrium results in a force, nearly linear with displacement, which tends to restore the system to equilibrium. At very small separations the repulsive force is strong; so we can treat the molecules as hard, impenetrable particles. Let us apply the postulates of the molecular model of matter to explaining the natural phenomena we have studied.
14.3 Kinetic Theory of Gases We have already studied the properties of gases and used the gas laws to solve many problems in Chapter 10. To use our molecular model to predict the properties of gases, let us make the following assumptions about our idealized gas:
1. The average distance between the molecules is many times greater than the size of the molecules. Therefore we will neglect the volume of the molecules in our calculations.
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2. The force of interaction between the molecules is zero except when they collide. 3. All collisions between molecules are perfectly elastic. 4. The idealized gas consists entirely of molecules of mass m which are moving with
the same speed v. The molecules are assumed to be moving in completely random directions.
Assume we have N molecules in a rectangular box that has sides of length, l1, l2 , and l3. One side is a piston, which has an area of l2 x l 3 and which is parallel to the yz plane. We want to develop an expression for the force exerted on the piston by the molecules of the gas. Consider a molecule such as molecule B which is moving with a velocity v (Figure 14.2). This velocity will have the three rectangular velocity components vx, vy, and vz. The magnitude of the velocity will be related to the individual components by the resultant equation from the section on vector addition in Chapter 3, v2 =vx
2 +vy2 +vz
2 (14.1)
Consider the component vx as the molecule strikes the face of the piston. The molecule will make an elastic collision with the wall. From Chapter 6 we know that the molecule will rebound with its speed unchanged but moving in the opposite direction. This means that the molecule will have a change of momentum in x direction from mvx to -mvx. Hence the change of momentum for each collision of molecule B with the piston is +2mvx. From the impulse-momentum equation (Equation 6.2), each collision will impart an impulse of 2mvx to the piston. Impulse is equal to the product of force times the time, so the force on the piston will be equal to the number of impulses the piston receives each second times the magnitude of each impulse. force = (impulse)(number of impulses per second) (14.2) If we continue to observe the motion of molecule B, it will not strike the piston again until it has traveled from the back wall and returned to hit the piston, a distance of 2l1 in the x direction. The time between impacts of molecule B on the piston is the distance divided by the speed, 2l1/vx. Hence the number of impacts per second of molecule B on
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the piston is 1/(2l1/υx), or υx/2l1. We can substitute these values for impulse and impact rate into Equation 14.2. The force exerted by molecule B on the piston is given by FB = (2mvx) (vx/2l1) =m vx
2/l1 (14.3) The total force of all N particles acting on the piston is equal to the sum of the contribution of each of the N particles, or to the product of the average force per particle mvx
2/ l1, multiplied by the number of particles, total force on piston = N mvx
2/l1 (14.4) But, of course, with all the molecules moving in random directions we cannot know the x-component of velocity of the molecules. We need to relate this to some property of the whole system. Notice that mvx
2 is almost proportional to the kinetic energy of a molecule. Let us define the total kinetic energy of the gas as the number of molecules N times the kinetic energy of the individual molecules, total kinetic energy = N (KE) KE = N (1/2) mv2 =Nmv2/2 (14.5) Because the direction of motion of the molecules is random, we can expect the total kinetic energy to be equally divided among the three components of velocity, so mvx
2= (1/3)mv2 (14.6) We can substitute this value into Equation 14.4 to obtain an equation for the force on the piston in terms of the speed of the molecules, total force on piston = (1/3)Nmv2/l1 (14.7) We can now obtain an expression for the pressure on the piston since the pressure on the piston is equal to the force divided by the area and the area of the piston is l2 l3 , pressure = total force / (l2 l3) P = (1/3)Nmv2/(l1 l2 l3) (14.8) Perhaps as you look at Equation 14.8 it will occur to you that the volume of the container V is equal to the product of the three sides l1 l2 l3. We can replace l1 l2 l3 in Equation 14.8 by the volume V, P = (1/3)Nmv2/V (14.9) or PV = (1/3)Nmv2 (14.10) Look at that expression; PV is a constant of the system by Boyle's law! We can perform some manipulations on Equations 14.9 and 14.10 to obtain some predictions from our model. The density of our gas is given by Nm/V, so Equation 14.9 becomes P = (1/3)ρv2 (14.11) where ρ is the density of the gas. For constant velocities the pressure of a confined gas should be proportional to its density. The kinetic energy of a molecule is given by (½) mv2; so Equation 14.10 becomes PV = (2/3)N(KE) (14.12) where KE is the kinetic energy of a molecule.
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From the combined gas laws in Chapter 10 you may recall that the (pressure x volume) product is equal to a constant times the absolute temperature. So we can conclude that the kinetic energy of the molecules is related to the temperature. We can write PV = (2/3)N(KE) = (2/3)N(1/2mv2) = (constant) T (14.13) where T is the Kelvin temperature. The constant is usually expressed as the product of two constants, nR , where n is the number of moles of gas and R is the universal gas constant, which has the value of 8.31 joule/mole-K. The total number of molecules in our gas N is equal to the number of moles times Avogadro's number, N0. So we have PV = (2/3) nNo(1/2)mv2) = nRT (14.14) and the kinetic energy of a molecule KE = (½)mv2 = (3/2) R / No T = (3/2)kT (14.15) where k is R/N0, which has a value of 1.38 x 10-23 joules/molecule-K and is known as Boltzmann's constant, the gas constant per molecule. Notice that the translational kinetic energy of our gas molecule is proportional to the absolute temperature and independent of the kind of gas. This kinetic theory of gases seems quite acceptable. Using only the basic principles of mechanics along with the assumptions of our model, we have been able to derive very general results using no mathematics more difficult than algebra. It is rare that we can develop a fairly acceptable theory from facts and relationships learned in an introductory science course. The development of the kinetic theory of gases is certainly an exception. Now that we know our model can produce interesting results, let us generalize our assumptions a bit. We made the assumption that all of our gas molecules have exactly the same speed v. That is a restrictive assumption not required to obtain the results of Equation 14.15. Instead let us require that the nth molecule have a speed vN, so that molecule 1 has a speed v1, molecule 2 has a speed v2, etc. The system will still have a total kinetic energy given by the sum of the individual kinetic energies, KEtotal = (½) mv1
2 + (½)mv22 +... (½)mvN
2 (14.16) We can use ratio of the total kinetic energy to the total number of molecules to calculate a value of the average kinetic energy per molecule KE, KE = KEtotal/N = ((½) mv1
2 + (½) mv22 + ... + (½) mvN
2) /N (14.17) We can use the value of the average kinetic energy per molecule to obtain a value for the average value of the square of the speeds of the individual molecules v2
rms, KE = (½)m v2
rms = (½)mv12 + (½)mvs
2 +... + (½)mvN2) / N = (3/2)kT
v2rms = (v1
2 + v22 +... + vN
2)/N = 3kT/m (14.18) The square root of vave
2 is called the root mean square (rms) speed. For a collection of molecules with widely different speeds, the root mean square speed v2
rms is that common value of speed of the molecules in the collection that would give each molecule the average kinetic energy of the collection. Thus all of our previous results such as Equations 14.10 and 14.15 are generalized to a collection of molecules with a distribution of speeds by replacing the speed in those equations by the root mean square speed.
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EXAMPLES 1. Find the rms speed of oxygen molecules at 27oC.
(1/2) m v2rms = (3/2) kT
where m = 32 x 1.67 x 10-27 kg and T = 300oK. Therefore v2
rms = 3KT/M = 3 x 1.38 x 10-23 J/molecule-K / (32 x 1.67 x 10-27 kg) = 23 x 104 (m/sec)2 vrms = 4.82 x 102 m/sec = 482 m/sec How many times per second will an oxygen molecule in the air transverse your room?
2. According to the kinetic theory of gases the pressure exerted by a gas depends upon three factors. What are they? How can each factor be changed? From Equation 14.9 we can identify the three factors as: a. The number of molecules per unit volume (N/V). Changing the total number of
molecules N or the volume of the system will change the pressure. b. The mass of the molecules; so changing the kind of gas would change the
pressure. c. The rms speed of the molecules; so changing the temperature will change the
pressure. 3. If the average speed of the molecules of a gas is doubled, what happens to the kinetic
energy of the gas? Since the kinetic energy is proportional to the square of the speed of the
molecules, the kinetic energy is four times greater. 14.4 Molecular Model of Liquids If we change the assumptions we made for the idealized gas, we can use the molecular model to explain the properties of liquids. For liquids the distances between molecules are assumed to be small, and the forces between the neighboring molecules are strong. The force of attraction between like molecules is called the cohesive force. The cohesive energy for molecules of a liquid is about equal to the kinetic energy of the molecules. We can think of a liquid as a large number of very small, slightly sticky glass beads which are continually moving. This model can be used to explain various properties of liquids - for example, their incompressibility, their ability to assume the shape of a container, and their viscous flow. Molecules at the surface of a liquid will only have cohesive forces acting on them from the interior of the liquid. If a surface molecule has sufficient kinetic energy, it may break away from its neighbors into the gas above the liquid. When this molecule of slightly higher kinetic energy leaves, the average kinetic energy of the liquid is reduced-that is, the temperature of the liquid decreases. This is a molecular explanation of evaporation and its cooling effect.
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14.5 Molecular Model of Solids If we imagine the cohesive forces to become even stronger than they are in liquids so that the cohesive energy is much greater than the kinetic energy of the molecules, then we have a model for a solid. In a solid the molecules are constrained in their motion to a small region around their equilibrium locations. The regular spatial patterns of molecules in crystalline solids is the lowest energy state for a system of molecules that interact with strong cohesive forces. Many of the properties of solids can be explained in a qualitative way by this model. The large values of Young's modulus for most crystalline solids is evidence of the strong cohesive forces in solids. 14.6 Adhesive Forces You know that a piece of adhesive tape will adhere to your skin. Why? Within the framework of the molecular model, the tape sticks to your skin because there is an attractive force between the molecules in your skin and the molecules in the coating of the tape. The attraction of unlike molecules is called adhesion, and the forces of attraction are called adhesive forces. Other applications of adhesive interactions are: wood and glue, solder and copper or brass, paste and paper, cellophane tape and many materials. You know that the cohesive forces of a solid are greatly weakened if a solid is broken apart. If you wish to put two pieces of wood together, you cannot just press the pieces together so that they will cohere. In order to get them together you must use some adhesive material such as glue. The relative magnitude of the cohesive and adhesive forces determine the interaction that results at the surface between substances in contact. You have probably observed that if oil is spilled upon a water surface, there is soon an oil film over a large area of water. Try a simple experiment. Take two pieces of a small-diameter glass tube and insert one in mercury and the other in water. Insert the tube into the liquid to some depth and then raise the tube. Observe the surface contour in each case. A diagram of such an experiment is shown in Figure 14.3.
In the mercury you will observe that the surface is convex, and the column of mercury in the tube is actually depressed. In the water the surface is concave, and inside the tube the water surface is higher than the surface in the main container. The surface for mercury is a typical example of the effect produced when the cohesive force is greater than the adhesive force. The water case is typical of situations in which the adhesive force is greater than the cohesive force. At each boundary of surface as indicated in Figure 14.3, there are three substances in contact. There is a surface effect at
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each surface of separation, and there is a force parallel to each surface. Thus at the point of contact of the three surfaces there are three forces parallel to each of the surfaces - that is, F1 solid-air surface, F2 liquid-solid surface, FS liquid-air surface. In addition there is the fourth force-that of adhesion between the liquid and solid FA. At the point of contact of the three substances the four forces are in equilibrium. The liquid surface will adjust itself so that this condition is satisfied. If the liquid wets the solid, the angle of contact will be less than 90o, as it is for the water-glass interface. If it does not, the angle of contact will be greater than 90 o, as for mercury and glass. 14.7 Surface Tension Some interesting experiments can be carried out with a soap bubble solution. Suppose that you have a light circular wire frame and tie a double thread across the ring as shown in Figure 14.4a.Then dip the ring in the soap bubble solution so that there is a film over the entire ring. The thread will show no well-defined shape. If you puncture the soap film inside the thread, then the thread takes on the circular shape shown in Figure 14.4b. For the thread to take on this shape there must be a force acting on the thread whose direction is perpendicular to the thread.
From our molecular model we can explain this surface force as a result of the unbalanced molecular forces that exist at the surface of a liquid. The force per unit length of surface is called the surface tension of a liquid. So if you measure the force F acting over a peripheral length L of surface, the surface tension, S, is given by S = F/L (14.19) The unit of surface tension is unit of force over unit of length or, in the SI system, newtons/meter. However, most tables give the value in dynes/cm. Some values of surface tension are given in Table 14.1. (Note that 1 dyne/cm = 10-3 N/m.)
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14.8 Measurement of Surface Tension
Approximate surface tension measurements can be made in different ways, but accurate measurements are difficult because the surface tension is influenced by impurities and surface dirt, traces of which are extremely hard to remove. An instrument for measuring surface tension is shown in Figure 14.5. A light platinum circular ring is supported on delicate torsion balance. Platinum is used in making the ring as it can be heated until red hot in a flame to remove traces of impurities, particularly greasy materials which are very undesirable substances in these measurements. The ring, held in a horizontal position, is dipped into clean water and is pulled upward by twisting the torsion wire. A film of water clings to both sides of the wire ring. The film from the water becomes vertical as the torsion wire is twisted. The total length of the film then becomes two times the circumference of the ring. The torsion head is twisted until the film breaks. The torsion balance can be calibrated by placing a small known mass on the ring and determining the angle of twist to support the calibrating weight in a horizontal plane.
The force F required to break the film is then given by F = Wcθw/θc (14.20) where Wc is the calibrating weight, θc is the angle of twist for calibrating weight, and θw is angle of twist for the water. The surface tension S is then given by S = F/(2(2πr)) =F/(4πr) (14.21) where r is the radius of the ring, and 2πr is the perimeter of each surface acting on the ring. The surface tension is dependent on the temperature of the liquid, as shown in Table 14.1, and also on any impurities that may be in the liquid. Another way to measure surface tension is to use a light wire frame with one movable side (see Figure 14.6).
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If the frame is dipped into a soap bubble solution so that a film is formed on the inside of the rectangle, there will be a force on the movable side tending to decrease the enclosed area. To hold the movable side of length l in equilibrium a force F will have to be applied to it. This force will be 2l times the surface tension S as there will be a film on each side of the wire. If the wire is slowly moved with the force remaining constant, the area increases, and work is done. The work done when the wire is moved a distance x is: W = Fx = 2lSx (14.22) but the increase in area is lx for each film and for the two films 2lx, so then W = SA or S =W/A (14.23) Thus S, the surface tension, may be regarded as work done per unit area in increasing the film area, and S would be expressed in erg/cm2. In the cgs system this is equivalent to dyne/cm. Any surface that is under tension will always tend to minimize its area for the given boundaries. For example, a droplet tends to take a spherical shape, which makes its surface area a minimum. 14.9 Pressure in Liquid Drops A drop of liquid has an inside pressure due to the surface tension. Consider a spherical droplet cut into two halves by an imaginary plane. The force holding the halves together results from the surface tension around the circumference of the circle. This force is then F = 2πrS . The internal pressure balances this force so that, if ΔP = Pi - P0, where P0 = outside pressure and Pi = inside pressure, then π r2 ΔP = 2πrS or ΔP = 2S/r (14.24) Let us consider a soap bubble. You know that the pressure of the air inside the bubble is greater than that of the air outside because of the process of blowing the bubble. The film of the soap bubble has two sides. The force resulting from surface tension is two times as large in a bubble as in a liquid droplet. Using the same approach as above F = 2(2πr) S This force is balanced by pressure so that with ΔP =Pi - P0 πr2 ΔP =4πrS or ΔP = 4S/r (14.25) EXAMPLE Compare the pressure difference in a soap bubble with that difference for a water drop of same size. The surface tension of soap solution is one-third that of water. Pdrop/Pbubble = (2S/r)/ [(4S/3)/r] = 6/4 = 1.5 14.10 Capillarity Earlier it was mentioned that water will rise in a glass tube of small diameter. This effect is caused by surface tension (Figure 14.3) and is called capillarity. The upward force is produced by the surface tension and is given by the product of vertical component of the surface tension and the length of surface; namely, the circumference of inside of the
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tube. So Fupward = (S cosθ )(2πr) where θ is the angle of contact between the liquid surface and the tube and r is the radius of the tube. The equilibrium occurs when the upward force is equal to the weight of the liquid in the tube above the surface of the liquid outside of the tube. The weight of the liquid = mg = ρVg, where ρ is the density of the liquid and V is the volume of the elevated liquid, mg = pπr2hg = (S cosθ) (2πr) where h is the height of the liquid in the tube above the outside level. We can solve this equation to obtain h: h = 2Scosθ/(ρgr) (14.26) The height h is inversely proportional to the radius of the tube. EXAMPLE It is noted that water has a contact angle of 0o in a capillary with a radius of 0.20 mm. The water rises 7.4 cm in the capillary; this means ΔP =ρgh. Find the surface tension of the water. S = ρghr/2cosθ = (1.00 g/cm3 x 980 cm/sec2 x 7.4 cm x 2.0 x 10-2 cm)/(2 x 1.0) S = 73 dyne/cm 14.11 Osmosis Have you observed what happens if a prune is soaked in water? The dried fruit soaks up water and swells to an enlarged size. In living systems there are similar cases that involve the movement of water through biological membranes. This transport process is called osmosis. The net transport of the host liquid across the membrane is always in the direction that tends to equalize the concentrations of dissolved materials on each side of the membrane. For example, if two compartments of sugar and water are separated by a parchment membrane, water will flow through the membrane from the lower concentration of sugar compartment to the higher concentration of sugar compartment until the two concentrations are equalized or until the pressure in the more concentrated solution prevents further flow. In such osmotic processes the membrane is permeable only to the host liquid, the solvent Figure 14.7 .
Figure 14.7 The principle of osmosis. (a) Originally the sugar solution is contained in the inner container, which is a semi-permeable membrane immersed in distilled water. The arrows indicate that the water passes through the membrane but the sugar molecules do not. As water molecules go into the inner tube, the volume of the sugar solution increases, as indicated by the increase in height in the pressure tube. Finally (b) an equilibrium condition is reached in which the number of molecules of water going in each direction through the semi-permeable membrane is the same. The pressure exerted by the liquid in the small tube is called the osmotic pressure. It is also the pressure that will stop the osmotic process.
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By applying pressure in the compartment of the more concentrated solution, the flow of the solvent can be slowed, stopped, or reversed. The applied pressure that is necessary to stop the flow of the solvent from the lower concentration to the higher concentration solution through the membrane is called the osmotic pressure. In the 1880s, W. Pfeffer performed experiments to determine the factors that influence osmotic pressure. First he carried out a series of experiments at constant temperature and varied the concentration of the dissolved materials, or solute. The results indicated that the osmotic pressure was directly proportional to the concentration of solute. In the second series of experiments the concentration of the solute was held fixed and the osmotic pressure was measured as a function of temperature. It was found that the value of the osmotic pressure divided by the Kelvin temperature is constant; that is, the osmotic pressure is directly proportional to the Kelvin temperature. A Dutch chemist named van't Hoff combined these observed relationships to obtain an empirical relationship for the osmotic pressure ΔPos, ΔPos = ΔcRT (14.27) where Δc is the number of moles of solute per unit volume of solvent, R is the universal gas constant andT is the temperature in degrees Kelvin. The osmotic current density (kg of solvent/ m2-sec) is proportional to the concentration difference across the membrane. This can be expressed in equation form as follows: Jos = -k Δc = -k/(RT ΔPos) (14.28) where k is the permeability coefficient of the membrane. EXAMPLE Given that k/RT = 6 x 10-10 kg/N-sec for a parchment membrance at room temperature. Find the osmotic pressure that will result in 6 x 10-4 kg/m2-sec of water current density from one side of the parchment to the other side. (ΔP)os = |Jos/(k/RT) | = 6 x 10-4/6 x 10-10 = 106 N/m2 ≈ 10 atm SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the number of the section where you can find the related content material. Definitions 1. The average kinetic energy of the molecules of a confined gas is used to determine the
_________________ , which is equal to ________________. 2. Solids have large internal forces _______________, i.e., forces of _______________
between ___________ molecules. 3. Glues are _______________ materials which have large _____________ forces, i.e.,
force of ______________ between ______________ molecules.
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4. The rise of a liquid in a narrow tube is called ______________ and is caused by the ______________ of the liquid and the _____________ between the liquid and the tube.
5. The force that acts upon the _______________ of a liquid to make it have the minimum possible surface area is called the ______________.
6. The passage of the solvent through a ______________ from a region of _____________ solute concentration to a region of _______________ solute concentration is called ____________ .
Molecular Model 7. Use the molecular model to provide an explanation of surface tension (make a
sketch). Problems 8. If you had an open glass tube between a soap bubble of 2-cm radius and one of 8-cm
radius, what would happen? Why? 9. A home remedy for the removal of candle wax from clothing is to cover the spot with
blotting paper and then pass a hot iron over the paper. Explain. 10. Describe an ideal gas. 11. a. What is the change of momentum per elastic collision per molecule?
b. How many collisions does each molecule make per second on a given face? 12. Compute the root mean square velocity of oxygen molecules at pressure of 1.01 x 105
N/m2 and a temperature of 0oC. The density of oxygen is 1.43 gm/liter. 13. How high will 15oC water rise in a capillary tube of 0.5-mm radius? Answers
1. rms velocity, SQR RT [Σi=1N vi
/N ] (Section 14.3) 2. cohesive, attraction, like (Section 14.5) 3. adhesive, adhesive, attraction, unlike (Section 14.6) 4. capillarity, surface tension, adhesion (Section 14.10) 5. surface, surface tension (Section 14.7) 6. membrane, low, high, osmosis (Section 14.11) 7. (Section 14.11) 8. small bubble shrinks (Section 14.9) 9. adhesion of liquid candle wax is low to cloth, high to blotting paper (Section 14.6) 10. See the list of assumptions (Sections 14.2, 14.3) 11. a. 2mv b. v/2l , where l is the length of the container and v is the velocity parallel to that length (Section 14.3) 12. 460 m/sec (Section 14.3) 13. 3 cm (Section 14.10)
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ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single-concept problems. Equations PV = (1/3)Nmv2 (14.10) P = (1/3)ρv2 (14.11) PV = (2/3) nNo(1/2)mv2) = nRT (14.14) KE = (½)mv2 = (3/2) R / No T = (3/2)kT (14.15) v2
rms = (v12 + v2
2 +... + vN2)/N = 3kT/m (14.18)
S = F/L (14.19) π r2 ΔP = 2πrS or ΔP = 2S/r (14.24) ΔP = 4S/r (14.25) h = 2Scosθ/(ρgr) (14.26) ΔPos = ΔcRT (14.27) Jos = -k Δc = -k/(RT ΔPos) (14.28) Problems
1. What force additional to its weight is required to pull a ring of 4.00-cm circumference from a clean water surface at 15oC?
2. What is the pressure within a mercury droplet (20oC) of 4.00-mm diameter? 3. What is the pressure within a soap bubble (20oC) whose radius is 4 cm if the
surface tension of the solution is 50 dynes/cm? 4. Assume the angle of contact between the water and glass tube is zero. How high
will water rise in a glass tube of radius 0.6 mm if the temperature is 20oC? 5. What is the kinetic energy of a room temperature, 20oC, air molecule? 6. How many moles of air are there in a typical physics classroom (7.2 m x 9.4 m x
3.4 m) at 20oC and at a pressure of 1.01 x 105 N/m2? 7. The density of air at 1 atm pressure and 0oC is 1.29 g/liter. What is the rms speed
per molecule? 8. If in problem 7 the pressure remains constant, but the temperature is raised to
100oC, what is the rms speed at 100oC? Answers
1. 588 dynes 2. 4.65 x 103 dynes/cm2 + p0 3. 50 dynes/cm2 + p0 4. 2.48 cm "
EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate, the numerical answer is given in brackets at the end of the exercise. Section 14.3
1. Ten moving particles have the following speeds (m/s): 100, 120, 110, 140, 180, 150, 130, 170, 165, 135, find the rms speed for these 10 particles and compare it with the average speed of the particles. [avg = 140 m/s; rms = 142 m/s]
2. If the internal energy U of a gas of N particles (in volume V) is defined as the total kinetic energy of the particles, show that the pressure equals (2/3)U/V .
3. According to the kinetic theory of gases, the pressure exerted by a gas depends upon three factors. With two factors remaining constant, how would the pressure change if a. the number of particles per unit volume is increased by a factor of 2? b. the gas is changed from hydrogen to oxygen? c. the rms speed is doubled? [a. pressure doubled; b. ratio ρO2/ρH2; c. pressure
increased by factor of 4] Section 14.8
4. When using an apparatus like that shown in Figure 14.6, we find that the equilibrium force required is 500 dynes for a moving wire 2.00 cm long. Find the surface tension of the liquid. [125 dyne/cm]
Section 14.9 5. Using the data from Table 14.1 , find the ratio of pressure at 10oC and at 100oC
when the surface tension is 59 dyne/cm. [P100/P10 = 0.80] Section 14.10
6. Water at 15oC rises in a glass capillary of 0.5-mm radius to a height of 3.00 cm. Find the surface tension of the 15oC water. [73.5 dyne/cm]
Section 14.11 7. Given that k/RT = 6 x 10-10 kg/N- sec for a membrane, find the necessary pressure
difference that will produce an osmotic current density of 60 micrograms/cm2 -sec of water. [106 N/m2]
Physics Including Human Applications
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PROBLEMS Each problem may involve more than one physical concept. The numerical answer to
each problem is given in brackets at the end of the problem. 8. Compute the number of molecules per cubic centimeter in a vacuum tube in which
the pressure is 1.00 x 104 mm Hg. Assume temperature is 300oK and molecular mass = 3.20 x 10- 27 kg. [3.22 x 10 12 molecules/cm3]
9. The horizontal length of wire frame in Figure 14.6 is 5 cm. Compute the value of surface tension from following data: a. With clean water a force of 730 dynes was required to pull the frame away from the water. b. With water containing a trace of soap the force was 500 dynes. [a. 73 dynes/cm; b. 50 dynes/cm]
10. Mercury is used in a barometer made of glass tubing having an inside diameter of 2 mm. If the mercury stands at 75 cm, what allowance should be made for capillary action, and what should the corrected height be? Angle of contact for mercury is 128o. [add 0.4 cm]
11. Find the ratio of the work necessary to make a soap bubble of radius 1 cm to that required to make a bubble with 2-cm radius. [W1/W2 =1/4]
12. Compare the rms velocity of hydrogen and oxygen molecules at 300oK. Both are diatomic molecules; does this affect your answer? [vH2/ uO2 rms = 4]
13. A vertical solid glass rod 0.6 cm in diameter stands partly submerged in water. What is the downward pull on the rod due to the surface tension? [274 dynes]
14. A 0.5-liter flask contains 1.34 x 1022 molecules each of mass 5.31 x 10-23 g with an rms speed 4.50 x 104 cm/sec. Compute the pressure in the flask in atmospheres. [9.61 x 104 N/m2 = .96 atm]
15. A needle is 5 cm long. Assuming the needle is not wetted, how heavy can it be and still float on the water? [730 dynes]
16. The surface tension of a soap solution is 40 dynes/cm. How much work must be done in blowing a soap bubble 6 cm in diameter? [905 ergs]
