Is heat capacity (C) extensive or intensive? What about specific heat (c) ? Explain briefly.

Short Answer

Expert verified

Heat capacity is an extensive property

Specific heat is an intensive property.

Step by step solution

01

Given Information

Heat Capacity C and

Specific Heat c.

02

Explanation

Explanation

On the basis of physical properties of matter, it can be classified into two parts

1. Intensive property: It does not depend on the quantity which means the intensive property does not vary when the mass changes.
2.Extensive property: It depends on the quantity of matter which means the extensive property varies when the mass changes.

Specific heat capacity is an intensive property. Specific heat capacity is given by

c=Cm

Where c is specific heat capacity, C is the heat capacity, m is the mass.


Here both C and m are extensive properties.
The ratio of two extensive property is intensive property.
This means Heat capacity is an extensive property.

The amount of heat capacity is given byCv=dUdT
Where, U is the internal energy and T is the temperature.
U is an extensive property and T is an intensive property.

The ratio of extensive to intensive results in extensive property.

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Most popular questions from this chapter

The 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:

NaAlSi3O8NaAlSi2O6+SiO2

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 CPandCV.

(a) With the heat capacity expressions from Problem 3.33 in mind, first considerSto be a function of TandV.Expand dSin terms of the partial derivatives (S/T)Vand (S/V)T. Note that one of these derivatives is related toCV

(b) To bring in CP, considerlocalid="1648430264419" Vto be a function ofTand 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 dP=0and note that you have derived a nontrivial expression for (S/T)P. This derivative is related to CP, so you now have a formula for the difference CP-CV

(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

CP=CV+TVβ2κT

(d) Check that this formula gives the correct value of CP-CVfor an ideal gas.

(e) Use this formula to argue that CPcannot be less than CV.

(f) Use the data in Problem 1.46 to evaluateCP-CVfor water and for mercury at room temperature. By what percentage do the two heat capacities differ?

(g) Figure 1.14 shows measured values of CPfor three elemental solids, compared to predicted values of CV. It turns out that a graph of βvs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CPand CVagree 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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