Physics 52 Prof Test 3s-key_2

Physics 52 Prof Test 3s-key_2 - ___________ 6.67...

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Physics 52 Proficiency Test 3(Sample) (Time: 10 minutes) 25 points By Todd Sauke Name ________ KEY _____________ Section # __ KEY ___ One liter of an ideal gas, initially at one atmosphere pressure and 20 °C, is isothermally compressed to a volume of 0.15 liters. What is the final temperature of the gas, in both degrees Celsius and kelvins? [The gas constant, R is 8.314 J /(mol K) = 0.0821 L atm/(mol K)] (Isothermal means constant temperature.) _______ _ 20 _ __________ °C ________ _ 293.15 _ _______ °K n = p V / (R T) = 1*1 /(0.0821 * 293.15) = 0.0415 How many moles of gas are being compressed? ______ _ 0.0415 _________ mol The internal energy of an ideal gas depends only on temperature. By how much does the internal energy of the gas change? ________ _ 0 __________ J What is the final pressure of the gas? p 1 V 1 = p 2 V 2 p 2 = p 1 V 1 / V 2 = 1.0 * 1.0 / 0.15 = 6.67
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Unformatted text preview: ___________ 6.67 _____________ atm How much work is done by the gas during this compression? (Note units!) W = p(V) dV = n R T dV / V = n R T ln(V 2 /V 1 ) W = - n R T ln(V 1 /V 2 ) = - .0415 (8.314) (293.15) 1.897 = -192 Note sign! I may vary the question to work done on the gas . _________-192 _____________ J How much heat enters the gas during this process? Δ U = Q – W Q = W = -192 Note sign! I may vary the question to heat leaving the gas. ________-192 ______________ J If the compression were to be done isobarically, how much work would be done on the gas during compression from 1 liter to 0.15 liters? W by the gas = p dV = p dV = p Δ V = 1 * (V 2 – V 1 ) = 1 * (0.15 – 1) = -0.85 L atm = -0.85 * (8.314/.0821) J = -86.1 W on the gas = - W by the gas _________ 86.1 ______________ J...
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This note was uploaded on 09/08/2010 for the course PHYS 52 at San Jose State.

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