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- What would be the greatest effect on the ideal gas law if there is a slight attractive force between the molecules? A) At low densities, the pressure would be less than that predicted by the ideal gas law. B) At high densities, the pressure would be less than that predicted by the ideal gas law. C) At high densities, the pressure would be greater than that predicted by the ideal gas law. D) At low densities, the pressure would be higher than that predicted by the ideal gas law. E) There is no effect. Midterm Exam Problem 10 Example of using van der Waals equation P = RT /( V - b ) - a / V 2 for n = 1 mole At T = 273K, applying the ideal gas law to 1 mole of CO 2 in V = 22.4 L we get P = 1 atm (STP). Using the van der Waals equation, we find P = 0.995 atm, so indeed P is slightly less. But the situation reverses at high density, where the RT /( V - b ) term becomes more important. For example, if we compress the CO 2 from V = 22.4 L to V = 0.05 L, the ideal gas law gives P = 448 atm while the van der Waals equation gives 1620 atm, so at high density the pressure P is greater than that predicted by the ideal gas law. However, I apologise that problem 10 was ambiguous, and I will be happy to give 5 points credit for answer B as well as A on problem 10 if you turn your Midterm in again by next weekās lecture. Note: The Final Exam is Wed Dec 11 4 to 7 pm, not 5 to 8 pm as the Syllabus said.
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Understanding Entropy
Physics 5D - Lecture 7 - Nov 25, 2013
The Carnot cycle consists of:
- Isothermal expansion ab
- Adiabatic expansion bc
- Isothermal compression cd
- Adiabatic compression da The change in entropy S when an amount of heat Q is added in a reversible process at constant T is ā S = Q / T.^ For a Carnot cycle, we proved that^ Q H/ T H = Q L/ T L , so it follows that ā S = Q H/ T H ā^ Q L/ T L = 0. ā Sin = Q H/ T H ā Sout = Q L/ T L
Analogy: Entropy and Volume
The fact that entropy S is a property that depends uniquely on the state of a system makes it useful. Like internal energy U , S is a property that is not obvious, but needs to be calculated from other properties of the system. If volume V were such a non- obvious property, how could we discover it? Consider as the system an ideal gas in a cylinder confined by a piston. We would do reversible experiments and look for something that just depends on the change of state of the system.
in the P-T diagram at the right. We might notice that, although Q and W each have different values for each path, Looking for more regularities, we might try plotting T vs. Q, P vs. Q, T vs. W, P vs. W, etc. Getting more desperate, we next try plotting reciprocals, for example 1/ P vs. W. Plotting 1/ P vs. W actually turns out to be really interesting, as the next slide shows. Q-W = ā U is always the same. Here U is the internal energy of the gas, and this equation is just energy conservation, i.e. the 1st Law of Thermodynamics. from H. C. van Ness, Understanding Thermodynamics For example, we might take the system through many reversible paths from to
Extensive vs. Intensive Variables
The volume, internal energy, and entropy have the property that if we double a system, each of them also doubles. Such a variable is called āextensiveā. In contrast, if we double a system, the pressure and temperature are unchanged. Variables like these are called āintensiveā. Thus writing the 1st Law as with each product on the rhs (intensive)x(extensive), is more symmetrical than, for example, writing it as Perhaps one reason that entropy seems more abstract than energy is that there are many forms of energy (heat, chemical, electical, etc.) but only āinternalā entropy. d U = d Q - d W = T d S - P d V. d U = d Q - d W = C V d T -P d V.
The Carnot cycle consists of:
- Isothermal expansion ab
- Adiabatic expansion bc
- Isothermal compression cd
- Adiabatic compression da We can alternatively draw the Carnot cycle in a T-S diagram. The area in the T-S diagram is the heat Q transferred since d Q = T d S. T S a b d c The area enclosed in the P-V diagram is the work W done in the cycle since d W = P d V.
Q = Q H - Q L T H T L Q H Q L It is also the work done W, by energy conservation.
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20-6 Entropy and the Second Law of
Thermodynamics
The fact that after every interaction the entropy of the system plus the environment increases is another way of putting the second law of thermodynamics:
The entropy of an isolated system never
decreases. It either stays constant (reversible
processes) or increases (irreversible
processes).
Copyright Ā© 2009 Pearson Education, Inc. Entropy is a measure of the disorder of a system. This gives us yet another statement of the second
law: Natural processes tend to move toward a
state of greater disorder. This gives an arrow of
time. Example: If you put a drop of dye in a beaker of water, it will spread outā but case (c) will never revert to case (a) spontaneously!
20-7 Order to Disorder
Thermal equilibrium is a similar processāthe uniform final state has more disorder than the separate temperatures in the initial state.
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Which has higher entropy ā a mole of ideal
gas at 20 ĀŗC occupying 10 liters or a mole
of the same gas at 20 ĀŗC occupying 100
liters?
A. The gas in 10 liters
B. the gas in 100 liters
C. No difference
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How much higher entropy does a mole of
ideal gas at 20 ĀŗC occupying 100 liters
have compared to a mole of gas at 20 ĀŗC
occupying 10 liters?
Copyright Ā© 2009 Pearson Education, Inc. Another consequence of the second law:
In any irreversible process, some energy becomes
unavailable to do useful work.
If we look at the universe as a whole, it seems inevitable that, as more and more energy is converted to unavailable forms, the ability to do work anywhere will gradually vanish. This is called the āheat death of the universeā. We will come back to this in the last lecture, next week.
20-8 Unavailability of Energy; Heat
Death
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20-9 Statistical Interpretation of
Entropy and the Second Law
Microstate: a particular configuration of atoms Macrostate: a particular set of macroscopic variables This example uses coin tosses:
Copyright Ā© 2009 Pearson Education, Inc. The more coin tosses, the more probable it is that the number of heads is about half. Similarly, for gas in a box, the most probable configuration is to have half the gas in each half of the box. Least probable: all in one half! The most probable state is the one with the greatest disorder, or the greatest entropy. With k being Boltzmannās constant and W the number of microstates, Boltzmann showed that the entropy is Text Boltzmannās tombstone, Vienna
Copyright Ā© 2009 Pearson Education, Inc. Statistical determination of entropy: Determine the change in entropy for the adiabatic free expansion of one mole of an ideal gas as its volume increases by a factor of 10. Assume W, the number of microstates for each macrostate, is the number of possible positions of the gas molecules.
