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Kinetic Theory of Gases Physics 202 Professor Lee Carkner Lecture 15
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Through which material will there be the most heat transfer via conduction? a) solid iron b) wood c) liquid water d) air e) vacuum
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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
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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
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PAL #14 Heat Transfer Conduction and radiation through a window H = kA(T H -T C )/L = (1)(1X1.5)(20-10)/0.0075 =2000 J/s P = A(T env 4 -T 4 ) P=(5.6703X10 -8 )(1)(1X1.5)(293 4 -283 4 ) = 81.3 J/s Double pane window H = A(T H -T C )/[(L 1 /k 1 )+(L 2 /k 2 )+(L 3 /k 3 )] 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
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Chapter 18, Problem 39 Two 50g ice cubes at -15C dropped into 200g of water at 25 C. Assume no ice melts m i c i T i + m w c w T w = 0 (0.1)(2220)(T f -(-15) + (0.2)(4190)(T f -25) = 0 222T f + 3330 + 838T f – 20950 = 0 T f = 16.6 C Can’t be true (can’t have ice at 16.6 C) Try different assumption
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Chapter 18, Problem 39 (cont) Assume some (but not all) ice melts T f = 0 C, solve for m i m i c i T i + m i L i + m w c w T w = 0 (0.1)(2220)(0-(-15) + m i (333000) + (0.2)(4190)(0-25) = 0 0 + 3330 + 333000m i + 0 – 20950 = 0 m i = 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 T f assuming all ice melts and then warms up to T f
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Chapter 18, Problem 39 (cont) Use one 50g ice cube instead of two We know that it will all melt and warm up m i c i T i + m i L i + m i c w T iw + m w c w T w = 0 (0.05)(2220)(0-(-15) + (0.05)(333000) + (0.05)(4190)(T f -0) + (0.2)(4190)(T f -25) = 0 0 + 1665 + 16650 + 209.5T f - 0 – +838T f - 20950 = 0 T f = 2.5 C
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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
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Mole 1 mol = 6.02 X 10 23 units 6.02 x 10 23 is called Avogadro’s number (N A ) M = mN A Where m is the mass per molecule or atom Gases with heavier atoms have larger molar masses
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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
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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
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Work and the Ideal Gas Law We can use the ideal gas law to solve this equation p=nRT (1/V)
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Isothermal Process If we hold the temperature constant in the work equation: W = nRT(1/V)dV = nRT(lnV f -lnV i ) W = nRT ln(V f /V i )
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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
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Constant Volume or Pressure W=0 W = pdV = p(V f -V i ) W = p V For situations where T, V or P are not constant, we must solve the integral The above equations are not universal
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Random Gas Motions
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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 = (nMv 2 rms )/(3V) Where M is the molar mass (mass contained in 1 mole)
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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 v rms = (3RT/M) ½ For a given type of gas, velocity depends only on temperature
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Translational Kinetic Energy Using the rms speed yields: K ave = ½mv rms 2 K ave = (3/2)kT Where k = (R/N A ) = 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
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Maxwell’s Distribution
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Maxwellian Distribution and the Sun The v rms 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
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Next Time Read: 19.8-19.11 Homework: Ch 19, P: 12, 31, 49, 54
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