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Chapter 5: Q. 5.21 (page 163)

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

Suppose that a hydrogen fuel cell, as described in the text, is to be operated at 75°Cand atmospheric pressure. We wish to estimate the maximum electrical work done by the cell, using only the room temperature data at the back of this book. It is convenient to first establish a zero-point for each of the three substances, H2,O2,andH2O. Let us take Gfor both H2andO2to be zero at 25°C, so that G for a mole of H2Ois -237KJat 25°C.

(a) Using these conventions, estimate the Gibbs free energy of a mole of H2at 75°C. Repeat for O2andH2O.

(b) Using the results of part (a), calculate the maximum electrical work done by the cell at 75°C, for one mole of hydrogen fuel. Compare to the ideal performance of the cell at25°C.

Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,

∂∂V∂U∂S=∂∂S∂U∂V

where each ∂/∂Vis taken with Sfixed, each ∂/∂Sis taken with Vfixed, and Nis always held fixed. From the thermodynamic identity (forU) you can evaluate the partial derivatives in parentheses to obtain

∂T∂VS=-∂P∂SV

a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H,F,andG ). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you've done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems.

Use a Maxwell relation from the previous problem and the third law of thermodynamics to prove that the thermal expansion coefficient β(defined in Problem 1.7) must be zero at T=0.

Problem 5.35. The Clausius-Clapeyron relation 5.47 is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ∆V depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ∆V to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:

P= (constant) x e-L/RT

This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.

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.
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