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# 2006 midterm 1 answers - Answer Key Mid term exam 1 Version...

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Answer Key, Mid term exam 1, Version A, 2-1-06, Chem 1B. Corresponding question numbers for version B are shown in parenthesis next to each problem number. 1.( 15 ) In a thermodynamic process a system does 20 J of work and absorbs 40 J of heat. What are the values of the internal energy change ( E), heat (q) and work (w) for this process? a) E = + 60 J, q = + 40 J, w = + 20 J b) E = - 20 J, q = - 40 J, w = +20 J c) E = - 60 J, q = - 40 J, w = -20 J d) E = + 20 J, q = +40 J, w = -20 J e) A process with these values of q and w cannot occur because energy is not conserved. ANSWER: D) E = 20J, q = 40 J, w = -20 J 2.( 16 ) Along pathway A between an initial state and a final state, the system absorbs 300 J of heat and its internal energy decreases by 200 J. Along a pathway B between the same initial and final states, the system absorbs 250 J of heat and does 450 J of work. What is the value of the work along pathway A? a) + 200 J b) - 200 J c) +500 J d) - 700 J e) none of the above ANSWER: E) NONE OF THE ABOVE w = E – q = -200 J – 300 J = -500 J System Surroundings q = + 40 J w = - 20 J E = -20 J + 40 J = +20 J See also quiz question #1

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Initial State Final State Pathway 1. q = 300 J, E = -200 J Pathway 2. q = 250 J, w = -450 w 1 = E – q 1 = -200 J – 300 J w 1 = -500 J Since E is a state function, the value it has on pathway 1, E = - 200 J, must be the same on pathway 2. For Pathway 2: w = E – q = - 200 J – 350 J = -550 J q 1 = 300 J q 2 = 250 J w 2 = -450 J w 1 = ? See also quiz question #2 3. (17) When a gas is cooled at constant volume from an initial temperature of 50 o C to 25 o C it loses 60.0 J of heat. When the same gas is cooled from 50 o C to 25 o C at constant pressure, it loses 100 J of heat. What is the value of the enthalpy change, H, when the gas is heated from 25 o to 50 o at constant volume? H is a state function that is equal to the heat transferred at constant P The gas losses 100 J of heat when it is cooled at constant pressure, q P = -100 J The enthalpy change at constant pressure is the same as the enthalpy change at constant volume. • Therefore, H = -q P = +100 J Constant V 50 o C 25 o C q V = -60 J 50 o C Constant P q P = -100 J 25 o C H = q P See Also Quiz Question #3
4. (18) Which of the following nuclei has the smallest bonding energy per nuclear particle? Nuclear Binding Energies Fission used to produce heat, work Fusion

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