Ch. 5: Applications Using Maxwell Relations & Measurable Properties

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Presentation transcript:

Ch. 5: Applications Using Maxwell Relations & Measurable Properties “But ma, it’s not my fault… the universe wants my room like this!”

Some Measureable Properties Heat Capacity at Constant Volume: ∂E Choose T & V as Independent Variables

Some Measureable Properties Heat Capacity at Constant Volume: ∂E Choose T & V as Independent Variables Heat Capacity at Constant Pressure: Choose P & V as Independent Variables

More Measureable Properties Volume Expansion Coefficient: Note!! Reif’s notation for this is α

More Measureable Properties Volume Expansion Coefficient: Note!! Reif’s notation for this is α Isothermal Compressibility: The Bulk Modulus is the Inverse of the Isothermal Compressibility! B  (κ)-1

Some Sometimes Useful Relationships Summary of Results Derivations are in the text and/or are left to the student! Entropy S = S(T,P): Enthalpy H = H(T,P): Gibbs Free Energy G = G(T,P):

in terms of some measureable properties. A Typical Example Given the entropy S as a function of temperature T & volume V, S = S(T,V), find a “convenient expression” for (S/T)P, in terms of some measureable properties.

in terms of some measureable properties. A Typical Example Given the entropy S as a function of temperature T & volume V, S = S(T,V), find a “convenient expression” for (S/T)P, in terms of some measureable properties. Start with the exact differential:

in terms of some measureable properties. A Typical Example Given the entropy S as a function of temperature T & volume V, S = S(T,V), find a “convenient expression” for (S/T)P, in terms of some measureable properties. Start with the exact differential: Use the triple product rule & definitions:

Use a Maxwell Relation:

Use a Maxwell Relation: Combining these expressions gives:

Use a Maxwell Relation: Combining these expressions gives: Converting this result to a partial derivative gives:

This can be rewritten as: The triple product rule is: Substituting gives:

Note again the definitions: Volume Expansion Coefficient β V-1(V/T)p

Note again the definitions: Volume Expansion Coefficient β V-1(V/T)p Isothermal Compressibility κ  -V-1(V/p)T

Note again the definitions: Volume Expansion Coefficient β V-1(V/T)p Isothermal Compressibility κ  -V-1(V/p)T Note again!! Reif’s notation for the Volume Expansion Coefficient is α

A GENERAL RELATIONSHIP Using these in the previous expression finally gives the desired result: Using this result as a starting point, A GENERAL RELATIONSHIP between the Heat Capacity at Constant Volume CV & the Heat Capacity at Constant Pressure Cp can be found as follows:

General Relationship between Cv & Cp Using the definitions of the isothermal compressibility κ & the volume expansion coefficient , this becomes General Relationship between Cv & Cp

Simplest Possible Example: The Ideal Gas For an Ideal Gas, it’s easily shown (Reif) that the Equation of State (relation between pressure P, volume V, temperature T) is (in per mole units!): Pν = RT. ν = (V/n) With this, it is simple to show that the volume expansion coefficient β & the isothermal compressibility κ are: and

So, for an Ideal Gas, the volume expansion coefficient & the isothermal compressibility have the simple forms: and We just found in general that the heat capacities at constant volume & at constant pressure are related as So, for an Ideal Gas, the specific heats per mole have the very simple relationship:

Other, Sometimes Useful, Expressions

More Applications: Using the Combined 1st & 2nd Laws (“The TdS Equations”) Calorimetry Again! Consider Two Identical Objects, each of mass m, & specific heat per kilogram cP. See figure next page. Object 1 is at initial temperature T1. Object 2 is at initial temperature T2. Assume T2 > T1. When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf.

Two Identical Objects, of mass m, & specific heat per kilogram cP Two Identical Objects, of mass m, & specific heat per kilogram cP. Object 1 is at initial temperature T1. Object 2 is at initial temperature T2. T2 > T1. When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf. Q  Heat Flows For some time after initial contact: Object 2 Initially at T2 Object 1 Initially at T1 After a long enough time, the two objects are at the same temperature Tf. Since the 2 objects are identical, for this case,

The Entropy Change ΔS for this process can also be easily calculated: Of course, by the 2nd Law, the entropy change ΔS must be positive!! This requires that the temperatures satisfy:

Some Useful “TdS Equations” NOTE: In the following, various quantities are written in per mole units! Work with the Combined 1st & 2nd Laws: Definitions: υ  Number of moles of a substance. ν  (V/υ)  Volume per mole. u  (U/υ)  Internal energy per mole. h  (H/υ)  Enthalpy per mole. s  (S/υ)  Entropy per mole. cv  (Cv/υ)  const. volume specific heat per mole. cP  (CP/υ)  const. pressure specific heat per mole.

Given these definitions, it can be shown that the Combined 1st & 2nd Laws (TdS) can be written in at least the following ways:

Internal Energy Enthalpy h(T,P): Student exercise to show that, starting with the previous expressions & using the definitions (per mole) of internal energy u & enthalpy h gives: Internal Energy u(T,ν): Enthalpy h(T,P):

Student exercise also to show that similar manipulations give at least the following different expressions for the molar entropy s: Entropy s(T,ν):