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### 15kineticgas2s

Course: PH 202, Fall 2009
School: Augustana
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Theory Kinetic 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? 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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Theory Kinetic 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? a) solid iron and wood b) wood and liquid water c) liquid water and air d) vacuum and solid iron e) vacuum and air 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 PAL #14 Heat Transfer Conduction and radiation through a window H = kA(THTC)/L = (1)(1X1.5)(2010)/0.0075 =2000 J/s P = A(Tenv4T4) P=(5.6703X108)(1)(1X1.5)(29342834) = 81.3 J/s H = A(THTC)/[(L1/k1)+(L2/k2)+(L3/k3)] H = (1X1.5)(2010)/[(.0025/1)+(0.0025/0.026)+(0.0025/1)] H = 148.3 J/s Double pane window Conduction dominates over radiation and doublepane windows are about 13 times better Chapter 18, Problem 39 Two 50g ice cubes at 15C dropped into 200g of water at 25 C. Assume no ice melts miciTi + mwcwTw = 0 (0.1)(2220)(Tf(15) + (0.2)(4190)(Tf25) = 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 Chapter 18, Problem 39 (cont) Assume some (but not all) ice melts miciTi + miLi + mwcwTw = 0 (0.1)(2220)(0(15) + mi(333000) + (0.2)(4190)(025) = 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 Tf = 0 C, solve for mi Chapter 18, Problem 39 (cont) Use one 50g ice cube instead of two We know that it will all melt and warm up miciTi + miLi + micwTiw + mwcwTw = 0 (0.05)(2220)(0(15) + (0.05)(333000) + (0.05) (4190)(Tf0) + (0.2)(4190)(Tf25) = 0 0 + 1665 + 16650 + 209.5Tf 0 +838Tf 20950 = 0 Tf = 2.5 C 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 Mole 1 mol = 6.02 X 1023 units x 6.02 1023 is called Avogadro's number (NA) M = mNA Gases with heavier atoms have larger molar masses Where m is the mass per molecule or atom 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 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 Work and the Ideal Gas Law W= pdV Vf Vi We can use the ideal gas law to solve this equation p=nRT (1/V) 1 W= nRT dV V Vf Vi Isothermal Process If we hold the temperature constant in the work equation: W = nRT(1/V)dV = nRT(lnVflnVi) W = nRT ln(Vf/Vi) Isotherms PV = nRT T = PV/nR For an isothermal process temperature is constant so: Can plot this on a PV diagram as isotherms One distinct line for each temperature If P goes up, V must go down Constant Volume or Pressure W=0 W = pdV = p(VfVi) W = p V For situations where T, V or P are not constant, we must solve the integral The above equations are not universal Random Gas Motions 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 trans...

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