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Chapter-9-HW - Homework#1 Chapter 9 Energy Enthalpy and...

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1 Homework #1 Chapter 9 Energy, Enthalpy, and Thermochemistry 1. When systems are in thermal contact (touching) they will eventually come to thermal equilibrium (be the same temperature). This happens because heat (energy) flows from the system with greater thermal energy (hot system) to the system with lower thermal energy (cold system). Metal feels colder than plastic, at the same temperature, because the metal has a lower specific heat capacity then plastic. This means that the metal can transfer heat faster than plastic. 10. By convention if work is negative then the system does work, or energy flows from the system to the surroundings. If work is positive then the surroundings does work, or work flows from the surroundings to the system. For the system to do work the volume must increase; therefore, V f -V i =+number. If the negative sign was not in the equation then it would be opposite of convention. 11. C P is the heat capacity of constant pressure and C V is the heat capacity of constant volume. If we are at constant volume no work is done by or on the system, therefore, w=0. The energy, E, is stored in kinetic energy. For a monoatomic ideal gas the change in average kinetic energy is: T nR KE ave 2 3 Since work is 0, all of the energy that is lost/gained goes into heat T nC q V Therefore, R C T nC T nR V V 2 3 2 3 If we are at constant pressure, then the kinetic energy must be distributed between heat and work. V P T nC w q T nR P 2 3 For an ideal gas T nR V P Therefore, R C T nR T nC T nR P P 2 5 2 3 This causes C P to be larger than C V .
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2 16. Plot (a) represents an exothermic reaction. The lower the energy the more energetically stable the system is and the stronger the bonds in the system. In plot (a) the products are at a lower energy, therefore, have stronger bonds. In plot (b) (endothermic reaction) in order for reactants to go to products heat must be introduced into the system. 18. Path (1 2) Step 1 P= 2.00 atm  101.325 1 1 2.00 30.0 10.0 40.0 4050 J ext L atm w P V atm L L L atm J         Step 2 The volume is constant therefore w 2 = 0 J 1 2 4050 0 4050 Tot w w w J J J     Path (3 4) Step 3 ext w P V   The volume is constant therefore w 3 = 0 J Step 4 P= 1.00 atm  101.325 1 1 1.00 30.0 10.0 20.0 2030 J ext L atm w P V atm L L L atm J         3 4 2020 0 2020 Tot w w w J J J     The 2 numbers are different; therefore, work is not a state function. 20. a) When the gas is compressed, work is done by the gas and the sign of work is +. When the system releases heat the sign of heat is -. 23 100. 77 E q w J J J   b) E q w  101.325 1 1.90 2.80 8.30 10.45 1060 351 1060 1410 J ext L atm w P V atm L L L atm J E q w J J J     c) E q w  101.325 1 1.00 29.1 11.2 17.9 1810 1037 1810 773 J ext L atm w P V
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