Chapter 5: Q. 5.21 (page 163)
Is heat capacity (C) extensive or intensive? What about specific heat (c) ? Explain briefly.
Short Answer
Heat capacity is an extensive property
Specific heat is an intensive property.
Chapter 5: Q. 5.21 (page 163)
Is heat capacity (C) extensive or intensive? What about specific heat (c) ? Explain briefly.
Heat capacity is an extensive property
Specific heat is an intensive property.
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Get started for freeThe methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:
Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both S and V.
Figure 5.35 (left) shows the free energy curves at one particular temperature for a two-component system that has three possible solid phases (crystal structures), one of essentially pure A, one of essentially pure B, and one of intermediate composition. Draw tangent lines to determine which phases are present at which values of x. To determine qualitatively what happens at other temperatures, you can simply shift the liquid free energy curve up or down (since the entropy of the liquid is larger than that of any solid). Do so, and construct a qualitative phase diagram for this system. You should find two eutectic points. Examples of systems with this behaviour include water + ethylene glycol and tin - magnesium.
How can diamond ever be more stable than graphite, when it has
less entropy? Explain how at high pressures the conversion of graphite to diamond
can increase the total entropy of the carbon plus its environment.
The partial-derivative relations derived in Problems 1.46,3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between and.
(a) With the heat capacity expressions from Problem 3.33 in mind, first considerto be a function of andExpand in terms of the partial derivatives and . Note that one of these derivatives is related to
(b) To bring in , considerlocalid="1648430264419" to be a function ofand P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set and note that you have derived a nontrivial expression for . This derivative is related to , so you now have a formula for the difference
(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be
(d) Check that this formula gives the correct value of for an ideal gas.
(e) Use this formula to argue that cannot be less than .
(f) Use the data in Problem 1.46 to evaluatefor water and for mercury at room temperature. By what percentage do the two heat capacities differ?
(g) Figure 1.14 shows measured values of for three elemental solids, compared to predicted values of . It turns out that a graph of vs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why and agree at low temperatures but diverge in the way they do at higher temperatures.
When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables, Rewrite the van der Waals equation in terms of these variables, and notice that the constants a and b disappear.
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