HW8-10Sol

# HW8-10Sol - Oregon State University Physics 202 , Winter...

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Unformatted text preview: Oregon State University Physics 202 , Winter 2010 HW8-10 (due Mar. 12 at 5:00 p.m.) Page 1 Oregon State University Physics 202 Winter Term, 2010 HW8-10 Solutions 1. Helium (He), a monatomic gas, f lls a 0.031-m 3 container. The pressure oF the gas is 5.6 x 10 5 Pa. How long would a 0.333-hp engine have to run (1 hp = 746 W) to produce an amount oF energy equal to the thermal energy oF this gas? For a monatomic ideal gas, E th = ( 3 / 2 ) NkT . Or, since PV = NkT , E th = ( 3 / 2 ) PV . We need to equate this much energy (Joules) to the energy generated by the engine. The energy generated by the engine is its power multiplied by the time it operates: E = ( Power ) t . (We spell out Power here to distinguish it from the gas pressure, P.) Therefore: ( 3 / 2 ) PV = ( Power ) t Solving for the time, we have: ( 3 / 2 ) PV / Power = t Plug in the numbers (converting to SI units): ( 3 / 2 )(5.6 x 10 5 )(0.031)/[(0.333)(746)] = t t = 105 s The thermal energy oF the helium is equivalent to running a .333-hp engine For 105 seconds. 2. The pressure oF a certain tank oF sulFur dioxide (SO 2 ) is 3.21 x 10 4 Pa. There are 214 moles oF this gas in a volume oF 40.0 m 3 . ind the translational rms speed oF these gas molecules. Note: Although this gas is not monatomic (and would thus have some signif cant kinetic energy due to rotation oF its molecules) , we can still approximate its translational kinetic energy (and thereFore its translational rms speed) using the monatomic ideal gas model. For a monatomic ideal gas: K T.a vg = ( 1 / 2 ) m p v rms 2 = ( 3 / 2 ) kT And: T = PV /( / nR ) Therefore: ( 1 / 2 ) m p v rms 2 = ( 3 / 2 ) kPV k / ( nR ) Solve this for v rms . Multiply both sides by 2/ m p : v rms 2 = 3 kPV / ( nRm p ) Take the square root of both sides: v rms = [3 kPV k / ( nRm p )] 1/2 Note that m p = m mole / N A : v rms = [3 kPV k N A /( nRm mole )] 1/2 Plug in the numbers (using SI units): v rms = {3(1.38 x 10-23 )(3.21 x 10 4 )(40)(6.02 x 10 23 )/[(214)(8.31)(0.064)]} 1/2 v rms = 530 m/s The translational rms speed oF SO 2 gas under these conditions is approximately 530 m/s. Oregon State University Physics 202 , Winter 2010 HW8-10 (due Mar. 12 at 5:00 p.m.) Page 2 3. A heat engine produces a total of 425 J of work by using an ideal gas in these three processes: Begin- ning at a pressure of 1.32 x 10 5 Pa and a volume of 2.20 x 10-3 m 3 , the pressure is increased isochorically to 5.70 x 10 5 Pa. Next, the gas expands adiabatically back to its original pressure. Finally, its volume is reduced isobarically back to its original volume. Calculate the ef ciency of this engine. The efF ciency, e , for a heat engine is deF ned as e = W / Q H In this case, we know W (and thats the magnitude of the work for the whole cycle)its given: 425 J. The question is, whats Q H ?...
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## This note was uploaded on 03/27/2010 for the course PH 202 taught by Professor Staff during the Spring '08 term at Oregon State.

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HW8-10Sol - Oregon State University Physics 202 , Winter...

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