15-Phys-203 Summer 2003 Prof. R. A.. Serota Exam 1 Name Formulae and constants: 4Eint = Q W = Z Vf W pdV Vi pV = nRT R = 8.31 J mol K Q = nC T Eint = nCV T f R 2 CV = Cp = CV + R pV = const, = Cp (adiabatic process) CV S = Z i f dQ T 1. Gas within a closed chamber undergoes the cycle shown in the p-V diagram. (a) What is the net change of internal energy in the cycle? (b) What is the net change of entropy? (c) Is the net energy added or extracted from the system in the form of heat? Calculate the heat. Solution Eint = 0, S = 0 Q = W = (50 10) (4 2 2)/2 = 40 J 3. One mole of oxygen O2 is heated at constant pressure starting at 273 K (0 C). How much energy must be added to the gas as heat to double its volume? Hint : f = 5. Solution Q = nCp (Tf but for an isobaric process Ti ) Tf Vf = Ti Vi so that Q = nCp (Tf Ti ) = nCp Ti Vf Vi 1 7 7 = nCpTi = RTi = (8.31) (273) 8 kJ 2 2 since Vf /Vi = 2. 4 4. An ideal diatomic gas, whose molecules are rotating but not oscillating (f = 5), is taken through the following cycle: (p1 ,V1 ,T1) isothermal (p2 ,V2 ,T1 ) → isochoric → (p3 ,V2 ,T3) adiabatic → (p1 ,V1 ,T1) where V2 = 3V1 (a) Sketch the p-V diagram. (b) Determine p2 , p3 , and T3 in terms of p1 and T1. Solution p1 V1 = p2 V2 ⇒ p2 = p1 p3 V2 T1V1 1 = = p1 V1 ⇒ p3 = p1 T3V2 1 ⇒ T3 = T1 V1 p1 = V2 3 V1 V2 V1 V2 = p1 37/5 1 = T1 32/5 or, equivalently, p2 p3 p3 p3 p1 = ⇒ T3 = T1 = T1 T1 T3 p2 p1 p2 1 V1 V1 V1 = T1 = T1 V2 V2 V2 5 1 5. Calculate the change of entropy of 2.00 moles of ideal gas as it is isothermally compressed to half its initial volume. Solution S = Z i f dQ Q = T T where Q = nRT ln so that S = nR ln Vf = Vi Vf Vi (2.00) (8.31) (ln 2) = 6 11.52 J/K

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