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# Lecture 13 ideal gas. kinetic model of a gas.

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Lecture 13 ideal gas. kinetic model of a gas.

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### Lecture 13 ideal gas. kinetic model of a gas.

1. 1. Lecture 13 The ideal gas. Kinetic model of a gas.
2. 2. ACT: Coin and mulch A piece of wood and a coin initially at 10°C are left out in the sun until they are both at the outside temperature (35°C). As you know, if you pick them up and hold them in your hand, the coin feels warmer. Why is that? A. Because the specific heat of the metal is larger than that of wood. B. Because the thermal conductivity of the metal is larger than that of wood. C. Because the metal absorbed more heat than the wood. cwood > cmetal It takes more heat to raise the temperature of wood. But this has nothing to do with how warm something feels… Something feels warm when heat is transferring to your skin. kwood < kmetal Heat transfer rate is higher for metal. → More heat is transferred to your skin in a given Δt for metals.
3. 3. State variables State variables for a system: they characterize the system macroscopically at a given time. • • • • Pressure Temperature Volume Mass or moles Important: a system is in a thermal state that can be described by these variables when ALL the system has uniform pressure, temperature, etc. Example: Gas in a tube whose ends are kept at different temperature is NOT in a thermal state.
4. 4. Equation of state The equation of state relates the state variables. Can be obtained: • Empirically • From statistical mechanics
5. 5. The ideal gas An ideal gas is one whose molecules • are very small (point-like) • do not interact with one another This is very true for any gas that is not in an “extreme” situation. The gas is NOT ideal if we have: • high density • high pressure • near transition to liquid (or solid) • molecules are very large
6. 6. The ideal gas state equation n = number of moles pV = nRT J mol × K liter ×atm = 0.082 mol × K R = 8.31 Ideal gas constant Special cases: Constant volume and mass (or moles) → Charles’ law m = nM Total mass pV = Molar mass m RT M p = constant T pM = ρRT (in terms of density)
7. 7. Example: Volume of ideal gas at STP STP = Standard Temperature and Pressure = 0°C and 1 atm What is the volume of one mole of an ideal gas at STP?  J  1 mole )  8.31 ( ÷( 273.15 K ) mol K  nRT  V = = = 0.00224 m3 = 22.4 liters p 1.01 × 105 Pa For any ideal gas
8. 8. Partial pressure If we have a mix of two gases A and B in a volume V, we can still use the ideal gas equation as long as they are both dilute. They both have the sameT and occupy the entire V pXV = nXRT for x = A, B Just like nall = nA + nB , pX = partial pressure of gas X pall = pA + pB pB pA = nB nA  RT  = ÷ V   nX pX = pall nall It makes sense: pressure comes from collisions with the wall. nX is the fraction of collisions with molecules of gas X. nall
9. 9. The Van der Waals state equation First attempt to correct idealization of ideal gas. This model includes: • molecule interaction (attraction) → reduces pressure • molecule size → reduces volume to move  n2   p + a 2 ÷(V − nb ) = nRT  V ÷   a, b are determined empirically. Different for each gas This model is used for more extreme conditions. If the gas is diluted (ie, n/V is very small), the corrections are negligible.
10. 10. Kinetic model of a monoatomic ideal gas All this was the ideal gas from the macroscopic point of view. Let us try to study it from the microscopic point of view.
11. 11. Model assumptions • container with volume V, a large number of molecules (N ), each molecule has mass m • molecules behave as point particles, occupy no volume and do not interact with one another. • molecules are in constant motion, obey Newton’s Laws and collide elastically with walls. • container walls are perfectly rigid and do not move.
12. 12. Strategy 1. Calculate number of collisions (per unit time) of molecules hitting wall (area A) 2. Calculate the momentum change of the molecules due to these collisions 3. Find the force exerted by the wall on gas that causes this momentum change 4. Use Newton 3rd law to get force exerted by gas on the wall 5. Pressure = (force exerted by gas on wall)/area 6. Use this to get/understand pV =nRT (Microscopic expression for p, T )
13. 13. Step 1: Number of collisions |vx| = magnitude of x-component of molecule’s velocity (for now assume all molecules have the same |vx| If a molecule is going to hit the wall in the time interval dt –it must be within a distance |vx|dt from the wall ie, within cylinder of volume A v x dt –within this distance, half of molecules going toward wall, half moving away 1 ( volume ) ( density of molecules ) 2 N  1 = A v x dt  ÷ 2 V  number molecules hitting wall in dt = ( )
14. 14. Step 2: Change in momentum Elastic collisions with an unmovable wall: The x-component of molecule’s velocity changes from vx to –vx ( m = mass of a molecule ) ∆px = 2m v x For all the molecules impacting area A of wall in dt: N  1 dPx = A v x dt  ÷2m v x 2 V  ( ) Step 3: Force exerted by wall Fx , on wall dPx 2 NAmv x = = dt V 2 NAmv x = dt V Step 4: Force exerted on wall Fx , by wall = Fx , on wall ( Newton's 3rd )
15. 15. Step 5: Pressure p= Fx , on wall A 2 Nmv x = V Step 6: Link to ideal gas equation pV = Nmv x2 macroscopic microscopic Problem: Who knows what is vx for each molecule??
16. 16. Additional step: Averages pV = Nmv 3D : 2 x 2 v 2 = v x + v y2 + v z2 2 pV = Nm v x 2 v 2 = v x + v y2 + v z2 2 2 2 All directions are the same, so v x = v y = v z = Thus, v2 3 1 pV = Nm v 2 3  2 1 = N  m v2 ÷ 3 2  2 = N KE 3 pV = 2 total kinetic energy of molecules 3
17. 17. Microscopic interpretation of temperature pV = nRT 2 pV = N 3 n= 1  m v2 ÷  2  N NA Define 3 1 nRT = N m v 2 2 2 where NA = 6.022 × 1023 Avogadro’s number k = R = 1.38 × 10 −23 J/K NA 3 1 kT = m v 2 2 2 Boltzmann’s constant Average kinetic energy of one molecule (for monoatomic ideal gas)
18. 18. ACT: Molecular speeds Two separate containers of gas are in thermal equilibrium with each other. One contains He and the other contains Ar. Which of the following statements is correct? A. 2 2 vHe = v Ar B. 2 2 vHe > v Ar C. 2 2 vHe < v Ar 3 K = kT 2 1 3 m v 2 = kT 2 2 They have the same average kinetic energy. The heavier mass must have slower speeds.
19. 19. Root-mean-square velocities The key quantity is <v2> Take the square-root v rms = → → mean of squared velocities back to the units of velocity v2 1 3 2 mvrms = kT 2 2 Example: 3 molecules at 400,500,600 ms-1 average = rms = 400 + 500 + 600 = 500 m/s 3 4002 + 5002 + 6002 = 507 m/s 3 Not the same!
20. 20. Mean free path Assume molecules are rigid spheres of radius r. Only one moves, with speed v. It will collide with another molecule when their centers are at distance 2 r. ie, in dt it collides if there are other molecules with centers inside this cylinder: 2r v vdt dN = π ( 2r ) vdt 2 N V Collision per unit time: dN N = 4π r 2v dt V dN N = 4π r 2v 2 If all molecules are moving, a similar calculation gives dt V 1 V = Average time between collisions: tmean = dN 4π 2r 2vN dt Average displacement between collisions V λ = vtmean = 4π 2r 2N (mean free path)