Chapter-10-HW

# Chapter-10-HW - Homework#2 Chapter 10 Spontaneity Entropy...

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1 Homework #2 Chapter 10 Spontaneity, Entropy and Free Energy 7. The larger the positional probability (the more available microstates) the larger the entropy 1. Entropy increases from solid to liquid to gas corresponding to an increase in positional probability. 2. Entropy increases when you dissolve a solid in liquid corresponding to an increase in positional probability. 3. The higher the temperature the greater the entropy because the wider the range of possible velocities. 4. The larger the volume the larger the positional probability and the greater the entropy. 5. The larger the pressure the smaller the positional probability and the lower the entropy a) (l) (g), ΔS=+, (1) b) (l) (s), ΔS=-, (1) c) As the gas is compressed the number of available microstates decreases, ΔS=-, (4 and 5) d) ΔS=+, (3) e) (s) (aq), ΔS=+, (2) 12. a) Spontaneous Process: A process that occurs without outside intervention. b) Entropy: Is a value that shows how energy is distributed among energy levels in the “particles” that constitute a given system. Entropy is closely associated with probability, where the most probable arrangement (state) is the highest entropy state. c) Positional Probability: A type of probability that depends on the number of arrangements in space that yield a particular state. d) System: That part of the universe on which attention is to be focused. e) Surroundings: Everything in the universe surrounding the thermodynamic system. f) Universe: The system and the surroundings. 17. a) As you increase the volume of a gas there are more possible locations for the particle to occupy. Or the positional probability of the system has increased. b) As you increase the temperature of a gas the positional probability remains the same because the number of locations that the particle can occupy is the same. Although the positional probability is the same increasing the temperature increases the entropy because, there is a wider range of possible velocities for any given gas particle. c) As the pressure of a gas is increased the particles are forced closer together, decreasing the possible positions of particles. This corresponds to a decrease in positional probability.

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2 21. Case 1 They want you to calculate the energy required to change the temperature. This means that they want you to calculate q (heat) T nC q V 26 44.60 73.5 25.0 48.5 48.5 1 1000 1.00 33.2 1 30.08 J V mol K C T C C C K mol C H g n kg mol C H kg g C H               33.2 44.60 48.5 71800 71.8 J V mol K q nC T mol K J kJ   Case 2 The second part of the problem wants you to calculate the heat again but at constant P. You can assume that C 2 H 6 is an ideal gas and that C P and C V are related by. C P =C V +R T nC q P 44.60 8.3145 52.91 J J J PV mol K mol K mol K C C R       33.2 52.91 48.5 85200 85.2 J P K mol q nC T mol K J kJ   The third part of the problem wants you to calculate the internal energy (ΔE) for case 1.
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## This note was uploaded on 01/18/2012 for the course CHEM 1B taught by Professor Watts during the Winter '08 term at UCSB.

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Chapter-10-HW - Homework#2 Chapter 10 Spontaneity Entropy...

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