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

This preview shows pages 1–5. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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? a) +60.0 J b) - 60.0 J c) + 100 J d) - 100 J e) - 40 J ANSWER: C) H = q P = - (-100 J) = + 100 J 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. •The re fo , 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? a) lead, 207 Pb b) deuterium, 2 H c) iron, 56 Fe d) uranium, 235 U f) gold, 197 Au ANSWER: B) DEUTERIUM Nuclear Binding Energies Fission used to produce heat, work Fusion 5. (19) Who is credited with building the first successful nuclear reactor? a) Erwin Oppenheimer b) Edward Teller c) Enrico Fermi d) Albert Einstein e) Julius Mayer ANSWER: C) ENRICO FERMI

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chicago Pile 1, December 2, 1942, First Sucessful Nuclear Reactor Enrico Fermi, Nobel Prize, 1938 6. (1) What method is used to confine matter when lasers are used to heat the matter to ignition temperatures for fusion? a) magnetic confinement b) inertial confinement c) gravitational confinement d) thermal confinement e) strong force confinement ANSWER: B) INERTIAL CONFINEMENT Principles of Inertial Confinement 7. (2) The source of the heat released in the fission of 235 U is a) conversion of electrons into heat b) conversion of protons into heat
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/07/2008 for the course CHEM 1B taught by Professor Watts during the Winter '08 term at UCSB.

### Page1 / 16

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online