solutions 4

solutions 4 - PHYS 253 Assignment 4 Solutions October 20...

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PHYS 253 Assignment 4 Solutions October 20 2010 Total: 12 marks 1) – 2 marks Let the larger bulb be denoted as bulb 1 and the smaller one as bulb 2. V 1 =3V 2 Because there is no heat transfer through the tube: T 2i =T 2f =T 0 Because of the tube pressure is equalized: P i1 =P i2 and P f1 =P f2 We approximate the air as an ideal gas: PV=nRT Initially: RT 0 P 0 = V 1 n 1i = V 2 n 2i Thus: n 1i = n 2i V 1 V 2 = 3n 2i And: n 1i n 2i = n 1f n 2f = 4n 2i When the pressure in both bulbs is doubled: P 1f = P 2f = 2P 0 = n 1f RT 1f V 1 = n 2f RT 0 V 2 T 1f = V 1 V 2  n 2f n 1f T 0 n 2f = 2 P 0 V 2 RT 0 n 1f = 4n 2i n 2f n 1f = 4 P 0 V 2 RT 0 2 P 0 V 2 RT 0 = n 2f T 1f = 3T 0 2) 4– marks a) V 1 V 2
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P 1 LA = nRT P 2 L x A = nRT = P 1 LA F 2 L x = F 1 L k = F x x = P x = F 1 L L x 1 x = F 1 L x = P 0 A L x For small x: K = P 0 A L b) When adiabatic PV γ =Constant P 0 V 0 = P f V f P = P f P i = P 0 [ V 0 V f 1
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This note was uploaded on 12/15/2010 for the course PHYS 253 taught by Professor Petergrutter during the Fall '10 term at McGill.

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solutions 4 - PHYS 253 Assignment 4 Solutions October 20...

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