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CU-Boulder PHYS 1110 - Lecture 43

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Physics 1110 Fall 2004L43 - 1Lecture 43 6 December 2004Announcements• No Reading Assignment, review on Wednesday• Final Exam on Tuesday, December 14, 4:30-7pm in the CoorsEvent Center. One sheet of notes allowed.• Exam is cumulative, but emphasis on material after Ch 11, allmultiple choice questions• Q&A on Monday, Dec 13, 3-5pm, ECCR 1B40• Online Survey worth 50 clicker pointsPressure and the Ideal Gas LawOne common everyday force we have not discussed is that due topressure from gasses and liquids. Today, we’ll look at the causesbehind these forces, which we can now understand at the atomiclevel, just using the physics we have studied over the last semester.Whenever is in contact with a fluid or gas it feels a force. This force isproportional to a quantity called pressure p, which is simply definedas the force per unit area, € pressure = Force surface area . The force frompressure always has a direction normal to the surface that is incontact with the gas (or fluid). Thus, while the force from pressurehas a direction and magnitude, the pressure itself does not have adirection, but it can cause a force in any direction.The units of pressure are N/m2 which is given the special name ofPascal, that is, 1 Pascal (Pa) = 1 N/m2. It turns out this is a very smallunit. In the US, the most common unit is p.s.i., which stands forpounds per square inch, lb/in2. The pressure from the Earth’satmosphere at sea level is about 15 p.s.i. = 103,460 Pa.Pressure arises from two main causes: thermal motion of the atomsand gravitational forces on the atoms. In your homework, you’ll derivethe simple formula for the change of pressure in a fluid as you godeeper: € p = psurface+ρgd, where psurface is the pressure at the top surfaceof the liquid, ρ is the density (mass/volume) of the liquid and d is thedepth. Here we will focus on the thermal causes.Physics 1110 Fall 2004L43 - 2Kinetic Model of an Ideal GasOur model of a gas will be the simplest imaginable. The atoms will betreated as little elastic of mass m, which bounce against the walls andeach other elastically (no energy loss) and only have kinetic energy,that is, each atom has energy € E =12mv2. The velocity € r v as usual hasthree components, x, y, and z.Since pressure is a force from many atoms, we will start by analyzingthe force from a single atom with a wall. The wall is one of the wallsof a box of volume V, which holds the atom. To keep the geometryclear, we’ll make the wall perpendicular to the x direction, with area A(in the y and z directions clearly). The figure below shows a beforeand after picture of an atom bouncing elastically on the wall. Sincethe collision is elastic, we know that the energy is conserved, and sothe magnitude of the velocity is unchanged by the collision, eventhough the direction is changed.In this particular case, all that happens is that the x component wasreversed by the collision, whereas the other components remainedunchanged:Physics 1110 Fall 2004L43 - 3€ vfx= −vixvfy= viyvfz= vizSince we don’t know the actual forces in the collision (they areactually electromagnetic), we’ll use impulse and momentum. € Δr p =r F dt =r F ave∫Δt.We can easily calculate the change in momentum over the time of thecollision Δt, so we find that the average force € r F ave during the collisionis € r F ave=Δr p Δt=mΔr v Δt=−2mvxΔt.This is the force on the atom, so by using Newton’s third law, weknow the average force from the collision the wall is just the opposite: € r F ave,wall=2mvxΔt.Now that we have found the force from one atom, all we need to do isfind the force from N atoms held in the box (where N could beAvogadro’s number). If we just multiply by N, this is too much, sinceall the atoms don’t hit the wall at the same time. This results from areasonable and important assumption that the motion of the atoms iscompletely random. So what we want to find is how many atoms Nccollide with the wall during the time Δt. If you consider the situation awhile, then you’ll realize that only those atoms within a certaindistance can get to the wall within this time. This distance d is thatwhich they can travel with speed vx in the time Δt: d = vxΔt. If weassume the atoms are uniformly spread out in the box, then thenumber of atoms within d of the wall can be found by proportions.We’ll assume that the number of atoms within a volume d*A willcollide with the wall, and this number Nc is to the total number ofatoms N, as the volume d*A is to the total volume V:€ NcN=dAV.Solving gives € Nc= NdA /V. So now the total force from the atoms is€ Ftotal= NcFave,wall=NdAV2mvxΔtPhysics 1110 Fall 2004L43 - 4However, this is still too many; only half of these atoms will happen tobe moving towards the wall during Δt, instead of away from it, so we’lldivide this by two. We’ll also substitute vxΔt for d to get€ Ftotal=12N (vxΔt )AV2mvxΔt=NVmvx2AIf we want to find the pressure, we divide by the area A of the wall, sowe get€ p =FtotalA=NVmvx2.This is pretty simple, but it still refers just to one wall; we can againuse the randomness of the gas atoms to provide an average quantityfor the x component (squared). If the gas is random, we assume it isequally like to have speed in the x, y, and z directions. So we’llassume that the average x, y, and z components are all the same€ vx2ave= vy2ave= vz2ave,and since the total velocity squared is the sum of these threecomponents, then on average each one has a third of the total:€ vx2ave=13v2ave.So finally we get a formula for the pressure without reference to aparticular wall:€ p =13NVmv2ave.You might notice the mv2 term in the formula; this is just twice theaverage kinetic energy of the atom:€ p =23NV(KE)aveThis is a pretty striking result! The pressure is directly proportional tothe average atomic kinetic energy. It is also proportional to thenumber of atoms per volume.Physics 1110 Fall 2004L43 - 5Most of you have probably studied the ideal gas law in chemistry, thefamous equation, which was discovered empirically by studying gaschanges when heated, is€ pV = NkBT.The constant kB is a number known as Boltzmann’s constant, and Tis the temperature on the Kelvin scale. Room temperature on theKelvin scale is about 300.If we take our bouncing atom formula and multiply it by V we get€ pV =23N(KE)aveSince these formulas describe the same quantities, we canimmediately say that in our theory the


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