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View Full Document50utA TI OAJ3 I tJO Pe?l~ tr To-I-P J Midterm #1 / Chemistry 4511 / Physical Chemistry for Engineers Helpful Information: PV= nRT 760 Torr = 1 atm R = 0.08206 atm dm3K1mOr1 R = 8.3145 J K1 mor1 k= 1.381 x 10-23J K1 dm = O.lm dm3 = 1000 cm3 = 1 Liter 1 J = 1 kg m2 S-2 1 amu = 1.66 x 10-27kg 1 amu x Avogadro's Number = Mass of mole in grams Avogadro's Number = 6.022 x 1023mor1 ~U = q (heat absorbed by system) + w (work done on system) ~U=q-P~V H=U +PV ~H = ~U + ~(PV) MPV) = ~nRT qP= ~H ~U = 0 for isothermal process w = nRT In (V t/V 2) for work done on the system for isothermal reversible process ~U = Cy(T 2- T1)for adiabatic reversible process ~S = qfT Efficiency = 1 - TefT h G = H- TS ~G=~H-T~S ~G < 0 is spontaneous ~Go= ~Ho-T~So ~Go= -RT In Kpo q= mC~T Heat Capacity of Water, C(H2O) = 4.186 JfgOC F=mg (resulting from gravity) Potential Energy = mgh (resulting from gravity) g= 9.8 mls2 1 lb. = 454 g 1 ft. = 0.3048 m (13 Ph) ( 12-P+V 1. ( 25" Pcn '" rs ) In the 1800s, James Joule measured the conversion between work and heat in his lab in Manchester, England. In his most famous experiment, he dropped a weight that was connected by pulleys to a paddle wheel in a tank of water. The turning of the paddle wheel led to a rise of the water temperature.... View Full Document
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View Full DocumentMidterm3-2005-Solutions
FinalExamSolutions
class 4
Fall 2013 Exam 2
HW#1
Chem notes gen
Chapter 13, Lecture 5
Chapter 14, Lecture 1
Chapter 14, Lecture 3 Problems
Chapter 17, Lecture 2
Chapter 17 Problem Solution
Chapter 17 problems
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