HW2 Assn - Christie Donahue 92337669 Homework#2 1 1.11 A...

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Christie Donahue 92337669 Homework #2 September 26, 2007 1) 1.11 A sample of methane of mass 4.5 g (4.5g(1mol/16.04g)=.2804mol) occupies 12.7 L at 310K. a) Calculate work done when the gas expands isothermally (delta T=0) against a constant external pressure of 30.0kPa until its volume is increased by 3.3 L. J v v p W v v p W pdv dw f ex f v v f 99 ) 3 . 3 ( 30 ) ( ) ( 0 0 0 - = - = - - = - = = b) Calculate the work that would be done if the same expansion occurred isothermally and reversibly. J L L K mol v v nRT W molK J i f 9 . 166 7 . 12 16 ln * 310 * 31451 . 8 * 2804 . ln - = - = - = 2) 1.16 A lab animal exercised on a treadmill, which, through pulleys, raised a mass of 200g (.2kg) through 1.55m. At the same time, the animal lost 5.0J of energy as heat (q=-5J). Disregarding all other losses and regarding the animal as a closed system, what is the change in internal energy of the animal? J m kg d F w w q U s m 038 . 3 55 . 1 * 8 . 9 * 2 . * 2 = = = + = The work is negative since the system (the rat) does work on the surroundings. w= -3.038 J J J J U 038 . 8 038 . 3 5 - = - - =
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3) 1.17 The internal energy of a perfect gas does not change when the gas undergoes isothermal expansion (by definition). What is the change in enthalpy? 0 0 0 = = = + = + = m m m m m m H U and T T R U H RT U H In isothermal expansion, delta T=0. Therefore, there is no change in internal energy. 4) 1.18 A sample of serum of mass 25g (.025kg) is cooled from 290K to 275K (deltaT=-15K) at a constant pressure (deltaP=0) by the extraction of 1.2kJ of energy as heat. Calculate q and deltaH and estimate the heat capacity of the sample. q= 1200J The heat is a positive value because the heat is being forcefully extracted as work from the system. Since there is no change in volume, no work is done on or by the serum. K J p p K J T H C J q H 80 15 1200 1200 - = - = = = = We say the heat capacity is 80J/K, though, because it does not make sense to have a negative heat capacity.
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