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All Solutions of Type: Thermodynamics
 0 Click here to jump to the problem! GR8677 #14 Thermodynamics}Exact differentials The key equation is: , and its players, , are terms one should be able to guess. (A) True, according to the ideal gas law. (This is also the final step in deriving Mayer's Equation, as shown below.) (B) This translates into the statement .The problem gives away the fact that for an ideal gas . B can't be right. (C) According to the ideal gas law, the volume might change. (D) False. An ideal gas's internal energy is dependent only on temperature. More elegantly, . (E) Heat needed for what? If one is interested in the formal proof of the relation , read on about Mayer's equation: For thermo, in general, there's an old slacker's pride line that goes like, When in doubt, write a bunch of equations of states and mindlessly begin taking exact differentials. Without exerting much brainpower, one will quickly arrive at a brilliant result." Doing this, Plugging in the first law of thermodynamics into the equation of state, one gets , where the last simplification is made by remembering the fact that the internal energy of an ideal gas depends only on temperature. (Taking the derivative with respect to T at constant volume, one gets .) Plugging in the simplified result for into the third equation of state, the ideal gas equation, one gets: . Taking the derivative at constant pressure, one gets: So, one sees that it is the ideal gas equation that makes the final difference. The work of an ideal gas changes when temperature is varied. Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #66 Thermodynamics}First Law Recall that , where work done by the system is positive and heat input into the system is positive. For constant volume, the equation becomes . . The reciprocol gives the right answer. Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #95 Thermodynamics}Carnot Cycle The Carnot cycle is the cycle of the most efficient engine, which does NOT have (unless ), but rather, ---this means that entropy stays constant. Choice (C) is thus false. The efficiency of the Carnot cycle is dependent on the temperature of the hot and cold reservoir. The hot reservoir has decreasing entropy because it gets cooler as the cycle proceeds. From writing down the thermodynamic relations for isothermal and adiabatic paths and matching P-V boundary conditions, one can determine that . The efficiency is thus . Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #15 Thermodynamics}Phase Diagram Recall that for an ideal gas and . Don't forget the first law of thermodynamics, . For , , since the temperature is constant. Thus, . For , . , and thus . For , , , thus . Add up all the Q's from above, cancel the term, to get , as in choice (E).s Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #16 Thermodynamics}Mean Free Path Air is obviously less dense than the atomic radius , thus choices (C), (D), and (E) are out. Air is not dilute enough that the distance between particles is actually within human visible range, as in (A)! Thus, the answer must be (B). (Note how this problem exemplifies the usefulness of common sense.) Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #73 Thermodynamics}Adiabatic Work One should recall the expression for work done by an ideal gas in an adiabatic process. But, if not, one can easily derive it from the condition given in the problem, viz., . Recall that the definition of work is , which when one plugs in the endpoint limits, becomes choice (C). Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #74 Thermodynamics}Entropy Recall the definition of entropy to be . The heat is defined here as , and thus . One is given two bodies of the same mass. One mass is at and the other is at before they're placed next to each other. When they're put next to each other, one has the net heat transferred being 0, thus . The entropy is thus , as in choice (B). Click here to jump to the problem!
 9 Click here to jump to the problem! GR8677 #75 Thermodynamics}Fourier's Law Recall Fourier's Law , where is the heat flux vector (rate of heat flowing through a unit area) and is the temperature and is the thermal conductivity. (One can also derive it from dimensional analysis, knowing that the energy flux has dimensions of ) Fourier's Law implies the following simplification: The problem wants the ratio of heat flows , as in choice (D). (The problem gives , , and , .) Click here to jump to the problem!
 10 Click here to jump to the problem! GR8677 #91 Thermodynamics}Second Law The Second Law of thermodynamics has to do with entropy; that entropy can never decrease in the universe. One form of it states that from hot to cold things flow. A cooler body can thus never heat a hotter body. Since the oven is at a much lower temperature than the wanted sample temperature, the oven can only heat the sample to a maximum of 600K without violating the Second Law. (This solution is due to David Latchman.) (Also, since the exam is presumably written by theorists, one can narrow down the choices to either (D) or (E), since the typical theorist's stereotype of experimenters usually involves experimenters attempting to violate existing laws of physics---usually due to naivity.) Click here to jump to the problem!
 11 Click here to jump to the problem! GR8677 #5 Thermodynamics}Degree of Freedom Acording to the Equipartition Theorem, there is a $kt/2$ contribution to the energy from each degree of quadratic freedom in the Hamiltonian. In equation form, the average total energy is , where s is the degrees of freedom. For a n-dimensional 1-particle system, the Hamiltonian is , where () refer, respectively, to the component of momentum (position). Thus, for a 3-dimensional 1-particle system, one has 6 quadratic terms in the Hamiltonian, as in . Plugging in s=6, one finds that the average energy is 3kT. (This revised solution is due to the GREPhysics.NET user kolndom.) Click here to jump to the problem!