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Ch1b2010_week4_bc

# Ch1b2010_week4_bc - Second Lecture Series(Winter 2010...

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Unformatted text preview: Second Lecture Series (Winter 2010): Thermodynamics Chem 1B Jim Heath Lecture Topic #2 Kinetic Theory of Gases & Thermodynamics (Part I) Most (hopefully all) of you will remember from high school chemistry the simple ideal gas law: PV = nRT which relates the pressure (P) and volume (V) of an ideal gas to the number of molecules or Moles (n) and the temperature (T), with a proportionality constant R (= 8.314 J K-1 mol-1 ). In this section, we are going to expand upon this equation and give it a fundamental underpinning based upon the motion of atoms and molecules in the gas phase. We begin by considering the particle within the container shown in Fig. KT.1. This particle (with mass m) has a momentum mv x as it heads toward the container wall. Upon striking the wall, the momentum changes to –mv x . Thus, the total change in momentum Δ p = 2m|v x |. Within a particular time interval Δ t there are a certain number of collisions of molecules with the wall, and we first want to calculate that number. The distance a molecule can travel within this time interval = |v x | Δ t. The area of the wall is A, and so A|v x | Δ t specifies a volume element that contains all of the particles that will strike the wall within the time interval Δ t. The total number of molecules per unit volume is η , and so there are η A|v x | Δ t molecules of interest. It turns out that about ½ of those molecules will be moving away from the wall, while the other half will be moving toward the wall, so, factoring this in, we can now state that the number of collisions within the period Δ t = ½ η A|v x | Δ t. At this point we would like to begin relating some of these calculations to observables, such as pressure. The way to do this is to first put into this expression the total momentum change within a time period Δ t. Total momentum change = ½ η A|v x | Δ t 2m|v x | = η Amv x 2 Δ t. (eq.1) Since, according to Newton, the time derivative of momentum = Force: Force = η Amv x 2 (eq.2) And, since pressure = force/unit area Pressure = P = η mv x 2 (eq. 3) Now we would like to generalize this expression to include the average speed of all molecules in the gas, rather than v x . The detected pressure is a measurement of the average velocity = <v x 2 >. The magnitude of velocity is speed = <v 2 > = <v x 2 > + <v y 2 > + <v z 2 > = 3<v x 2 > = c 2 which we refer to as the mean square speed. Fig KT.1. Particle of mass m striking a wall of area A within a container. Fig KT.1. Particle of mass m striking a wall of area A within a container. So now replacing v x 2 in the Pressure equation above, we get Pressure = P = 1/3 η mc 2 . We can replace η (the number density of particles) by N/V, (V = volume) and then introducing Avogrado’s number through the equality N = nL, where n is the number of moles, and L is Avogrado’s number, we get PV = 1/3nLmc 2 . (eq. 4) Now, to illustrate a point, divide each side by 2 to get: ½ PV = 1/3 nL ½ mc 2 = 1/3 nL ( × the average kinetic energy per particle). the average kinetic energy per particle)....
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Ch1b2010_week4_bc - Second Lecture Series(Winter 2010...

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