Solving Change in Temp. Homework w/Maxwell's & Coeff. of Thermal Exp.

In summary, this conversation discusses an experiment on solid at high pressure, where the resulting change in temperature (\Delta T) of the sample is determined by the increase in pressure (\Delta p), based on the sample's specific heat at constant pressure (cp), density (rho), and volume coefficient of thermal expansion (\alpha). The Maxwell relations are used to derive expressions for the partial derivatives of entropy (S) at constant pressure and temperature, and these can be related to the thermal expansion coefficient. The fundamental thermodynamic relation is also used to further manipulate the equations.
  • #1
tan90
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Homework Statement


This is an experiment done on solid at high pressure. If the pressure is increased by an amount [tex]\Delta[/tex]p, this being done under condition where the sample is thermally insulated and at a sufficiently slow rate that the process can be regarded as quasi-static, what is the resulting change of temperature [tex]\Delta[/tex]T of the sample ? If[tex]\Delta[/tex]p is fairly small, derive an expression for [tex]\Delta[/tex]T in terms of [tex]\Delta[/tex]p, the absolute temperature T of the sample, its specific heat at constant pressure cp (in ergs g-1 deg -1), its density rho (in g/cm3), and its volume coefficient of thermal expansion [tex]\alpha[/tex](in deg -1)

Homework Equations



dv = [tex]\alpha[/tex]V dt

Maxwell's equations

The Attempt at a Solution



I started from an expression for coefficient of thermal expansion and tried to relate it with maxwell's equation. The constant terms in maxwell's equations are very confusing.
 
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  • #2
Well, first of all, what variable are you going to hold constant?

(P.S. They're usually referred to as Maxwell relations to distinguish them from all that B-field and E-field stuff.)
 
  • #3
Since the questions says "If [tex]\Delta[/tex] p is fairly small, " is it ok to consider p as a constant? Other variables [tex]\Delta[/tex]T and [tex]\Delta[/tex]V do change. I am rather confused.

Thank you
 
  • #4
Unfortunately, we can't assume [itex]\Delta P[/itex] is zero because we have to divide [itex]\Delta T[/itex] by [itex]\Delta P[/itex] (that is, we want to find [itex](\partial T/\partial P)_X[/itex], where X is the variable we hold constant). Since we've eliminated pressure, temperature, and volume, what's left? We're blocking heat transfer; besides energy, what flows during heat transfer?
 
  • #5
P,V,T all change. It means that the only entity left in Maxwell's relations is S. So are you suggesting[tex]\Delta[/tex]S is a constant?
 
  • #6
Yes, S is constant, or [itex]\Delta S[/itex] is zero.
 
  • #7
Thank you ! Can you please explain a bit further about what is actually going on in the problem? Its rather hard to picture the problem for me.
 
  • #8
The experiment involves pressurizing a solid. Will the solid heat up or cool down, and by how much for a given increase in pressure? The effect is somewhat related to thermal expansion, in which we observe how the volume of a solid changes for a given temperature increase at constant pressure. You should be able to write the differential term (e.g., [itex](\partial Y/\partial Z)_X[/itex]) for each effect, and manipulate them via Maxwell relations.

It turns out that even the sign of the answer can vary between materials--some materials heat up under pressure, some cool down--so I wouldn't try to intuit the answer through visualization. You have to go through the (admittedly abstract) calculations.
 
  • #9
To find the partial derivative at constant S, you can write:

[tex]dS =\left(\frac{\partial S}{\partial P}\right)_{T}dP + \left(\frac{\partial S}{\partial T}\right)_{P}dT[/tex]

If you equate dS to zero and solve for the ratio dT/dP you got your partial derivative at constant S. Now, you can express the partial derivate of S w.r.t. T at constant P in terms of the heat capacity at constant pressure. But what to make of the partial derivative of S w.r.t. P at constant T? Well, we have the findamental thermodynamic relation:

dE = T dS - P dV

We can write:

T dS = d(TS) - S dT

and

P dV = d(PV) - V dP

This means that:

dE = d[TS - PV] -SdT + V dP ---------->

d[E + PV - TS] = -S dT + V dP

Define G = E + PV - TS. The above relatiomn implies that minus S is the partial derivative of G w.r.t. T and constant P and that V is the partial derivative of G w.r.t. P at constant T. Then the fact that te second derivative of G w.r.t. P and T does not depend on the order of differentiation implies that:

[tex]\left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P} [/tex]

So, the partial derivative of S can be expressed in terms of the thermal expansion coefficient.
 
  • #10
Brilliant !
 

1. What is Maxwell's equation and how is it related to solving changes in temperature?

Maxwell's equation is a set of four equations that describe the relationship between electric and magnetic fields. It is related to solving changes in temperature because it can be used to determine the flow of heat and energy in a system, which is necessary for understanding changes in temperature.

2. How does the coefficient of thermal expansion affect changes in temperature?

The coefficient of thermal expansion is a measure of how much a material expands or contracts when its temperature changes. This coefficient is important in solving changes in temperature because it allows us to predict how much a material will expand or contract when its temperature changes.

3. What role does heat transfer play in solving changes in temperature?

Heat transfer is the movement of thermal energy from one object to another. It plays a crucial role in solving changes in temperature because it is the mechanism by which heat is transferred from a hotter object to a colder one, causing changes in temperature.

4. Can Maxwell's equation and the coefficient of thermal expansion be used to solve all types of temperature changes?

Yes, Maxwell's equation and the coefficient of thermal expansion can be used to solve changes in temperature for all types of materials and systems. However, the specific equations and methods used may vary depending on the specific circumstances and properties of the material.

5. How can the concepts of Maxwell's equation and coefficient of thermal expansion be applied in real-world scenarios?

Maxwell's equation and the coefficient of thermal expansion have numerous applications in real-world scenarios. They are used in industries such as engineering, materials science, and thermodynamics to solve problems related to temperature changes in various systems, including buildings, electronics, and machinery.

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