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Kinetic Theory of Gases

Kinetic Theory of Gases. Physics 202 Professor Lee Carkner Lecture 15. Through which material will there be the most heat transfer via conduction? a) solid iron b) wood c) liquid water d) air e) vacuum. Through which 2 materials will there be the most heat transfer via radiation?

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Kinetic Theory of Gases

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  1. Kinetic Theory of Gases Physics 202 Professor Lee Carkner Lecture 15

  2. Through which material will there be the most heat transfer via conduction? a) solid iron b) wood c) liquid water d) air e) vacuum

  3. Through which 2 materials will there be the most heat transfer via radiation? a) solid iron and wood b) wood and liquid water c) liquid water and air d) vacuum and solid iron e) vacuum and air

  4. Through which 2 materials will there be the most heat transfer via convection? a) solid iron and wood b) wood and liquid water c) liquid water and air d) vacuum and solid iron e) vacuum and air

  5. PAL #14 Heat Transfer • Conduction and radiation through a window • H = kA(TH-TC)/L = (1)(1X1.5)(20-10)/0.0075 =2000 J/s • P = seA(Tenv4-T4) • P=(5.6703X10-8)(1)(1X1.5)(2934-2834) = 81.3 J/s • Double pane window • H = A(TH-TC)/[(L1/k1)+(L2/k2)+(L3/k3)] • H = (1X1.5)(20-10)/[(.0025/1)+(0.0025/0.026)+(0.0025/1)] • H = 148.3 J/s • Conduction dominates over radiation and double-pane windows are about 13 times better

  6. Chapter 18, Problem 39 • Two 50g ice cubes at -15C dropped into 200g of water at 25 C. • Assume no ice melts • miciDTi + mwcwDTw = 0 • (0.1)(2220)(Tf-(-15) + (0.2)(4190)(Tf-25) = 0 • 222Tf + 3330 + 838Tf – 20950 = 0 • Tf = 16.6 C • Can’t be true (can’t have ice at 16.6 C) • Try different assumption

  7. Chapter 18, Problem 39 (cont) • Assume some (but not all) ice melts • Tf = 0 C, solve for mi • miciDTi + miLi + mwcwDTw = 0 • (0.1)(2220)(0-(-15) + mi(333000) + (0.2)(4190)(0-25) = 0 • 0 + 3330 + 333000mi + 0 – 20950 = 0 • mi = 0.053 kg • This works, so final temp is zero and not all ice melts • If we got a number larger than 0.1 kg we would know that all the ice melted and we could try again and solve for Tf assuming all ice melts and then warms up to Tf

  8. Chapter 18, Problem 39 (cont) • Use one 50g ice cube instead of two • We know that it will all melt and warm up • miciDTi + miLi + micwDTiw + mwcwDTw = 0 • (0.05)(2220)(0-(-15) + (0.05)(333000) + (0.05)(4190)(Tf-0) + (0.2)(4190)(Tf-25) = 0 • 0 + 1665 + 16650 + 209.5Tf - 0 – +838Tf -20950 = 0 • Tf = 2.5 C

  9. What is a Gas? • A gas is made up of molecules (or atoms) • The temperature is a measure of the random motions of the molecules • We need to know something about the microscopic properties of a gas to understand its behavior

  10. Mole 1 mol = 6.02 X 1023 units • 6.02 x 1023 is called Avogadro’s number (NA) M = mNA • Where m is the mass per molecule or atom • Gases with heavier atoms have larger molar masses

  11. Ideal Gas • Specifically 1 mole of any gas held at constant temperature and constant volume will have the almost the same pressure • Gases that obey this relation are called ideal gases • A fairly good approximation to real gases

  12. Ideal Gas Law • The temperature, pressure and volume of an ideal gas is given by: pV = nRT • Where: • R is the gas constant 8.31 J/mol K

  13. Work and the Ideal Gas Law • We can use the ideal gas law to solve this equation p=nRT (1/V)

  14. Isothermal Process • If we hold the temperature constant in the work equation: W = nRT(1/V)dV = nRT(lnVf-lnVi) W = nRT ln(Vf/Vi)

  15. Isotherms PV = nRT T = PV/nR • For an isothermal process temperature is constant so: • If P goes up, V must go down • Can plot this on a PV diagram as isotherms • One distinct line for each temperature

  16. Constant Volume or Pressure W=0 W = pdV = p(Vf-Vi) W = pDV • For situations where T, V or P are not constant, we must solve the integral • The above equations are not universal

  17. Random Gas Motions

  18. Gas Speed • The molecules bounce around inside a box and exert a pressure on the walls via collisions • How are p, v and V related? • The rate of momentum transfer depends on volume • The final result is: p = (nMv2rms)/(3V) • Where M is the molar mass (mass contained in 1 mole)

  19. RMS Speed • There is a range of velocities given by the Maxwellian velocity distribution • It is the square root of the sum of the squares of all of the velocities vrms = (3RT/M)½ • For a given type of gas, velocity depends only on temperature

  20. Translational Kinetic Energy • Using the rms speed yields: Kave = ½mvrms2 Kave = (3/2)kT • Where k = (R/NA) = 1.38 X 10-23 J/K and is called the Boltzmann constant • Temperature is a measure of the average kinetic energy of a gas

  21. Maxwell’sDistribution

  22. Maxwellian Distribution and the Sun • The vrms of protons is not large enough for them to combine in hydrogen fusion • There are enough protons in the high-speed tail of the distribution for fusion to occur

  23. Next Time • Read: 19.8-19.11 • Homework: Ch 19, P: 12, 31, 49, 54

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