# Change in Temperature

1. Apr 4, 2009

### tan90

1. The problem statement, all variables and given/known data
This is an experiment done on solid at high pressure. If the pressure is increased by an amount $$\Delta$$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 $$\Delta$$T of the sample ? If$$\Delta$$p is fairly small, derive an expression for $$\Delta$$T in terms of $$\Delta$$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 $$\alpha$$(in deg -1)

2. Relevant equations

dv = $$\alpha$$V dt

Maxwell's equations

3. 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.

Last edited: Apr 4, 2009
2. Apr 5, 2009

### Mapes

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. Apr 5, 2009

### tan90

Since the questions says "If $$\Delta$$ p is fairly small, " is it ok to consider p as a constant? Other variables $$\Delta$$T and $$\Delta$$V do change. I am rather confused.

Thank you

4. Apr 5, 2009

### Mapes

Unfortunately, we can't assume $\Delta P$ is zero because we have to divide $\Delta T$ by $\Delta P$ (that is, we want to find $(\partial T/\partial P)_X$, 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. Apr 5, 2009

### tan90

P,V,T all change. It means that the only entity left in Maxwell's relations is S. So are you suggesting$$\Delta$$S is a constant?

6. Apr 5, 2009

### Mapes

Yes, S is constant, or $\Delta S$ is zero.

7. Apr 5, 2009

### tan90

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. Apr 5, 2009

### Mapes

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., $(\partial Y/\partial Z)_X$) 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. Apr 5, 2009

### Count Iblis

To find the partial derivative at constant S, you can write:

$$dS =\left(\frac{\partial S}{\partial P}\right)_{T}dP + \left(\frac{\partial S}{\partial T}\right)_{P}dT$$

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:

$$\left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P}$$

So, the partial derivative of S can be expressed in terms of the thermal expansion coefficient.

10. Apr 5, 2009

### tan90

Brilliant !!!