Question

In: Advanced Math

An important practice is to check the validity of any data set that you analyze. One...

An important practice is to check the validity of any data set that you analyze. One goal is to detect typos in the data, and another would be to detect faulty measurements. Recall that outliers are observations with values outside the “normal” range of values of the rest of the observations.

Specify a large population that you might want to study and describe the type numeric measurement that you will collect (examples: a count of things, the height of people, a score on a survey, the weight of something). What would you do if you found a couple outliers in a sample of size 100? What would you do if you found two values that were twice as big as the next highest value?

You may use examples from your area of interest, such as monthly sales levels of a product, file transfer times to different computer on a network, characteristics of people (height, time to run the 100 meter dash, statistics grades, etc.), trading volume on a stock exchange, or other such things.

Solutions

Expert Solution

Yes. To prove Le Chatelier's principle doesn't require a general theory of chemical kinetics, just an understanding of (thermodynamic!) fluctuations and response. As usual, Callen's Thermodynamics and an Introduction to Thermostatistics is a good reference. Ch. 8.5 is relevant, and I shall simply recapitulate Callen's development below.

(Incidentally, I think this principle is closer to Newton's third law of action and reaction, but in general thermodynamic laws have a very different flavor from mechanical laws because of their underlying structure. Thermodynamics deals with nebulous relationships; mechanics with definite equations of motion.)

In fact, we will prove an extra statement, the Le Chatelier-Braun principle. This principle states that the secondary effects induced by a perturbation also serve to curtail it. This augments the Le Chatelier principle, that the primary effect induced by a perturbation serves to curtail it.

I will steal Callen's example of a system immersed within a pressure and temperature reservoir, with diathermal walls and a movable domain wall with which to control its volume. The wall is moved slightly outward, causing a positive volume change dV. The primary effect is the decrease in pressure of the system, which leads to a driving force to decrease the system's volume. A secondary effect is the change in temperature dT resulting from this change in volume,

dT=(∂T∂V)SdV=−TαNcvκTdV.

The prefactor is unimportant; we care only about the sign of the result. All variables are positive except for α, which is of variable sign. For now, we can assume it positive. The reduction of temperature in the system then drives heat flow into it, which will itself affect the pressure of the system:

dP=(∂P∂S)VdQT=αNT2cvκTdQ.

This change in pressure is positive and diminishes the effect of the original perturbation. The same result is obtained if one takes α negative.

Let us consider a thermodynamic system with first law

dU=f1dX1+f2dX2.

It is coupled to a reservoir, itself with first law

dU′=f′1dX′1+f′2dX′2=−f′1dX1−f′2dX2,

noting that dX′i=−dXi, because Xi+X′i is assumed fixed. A perturbation dXpert1 drives fluctuations

dffluc1=∂f1∂X1dXpert1anddffluc2=∂f2∂X1dXpert1.

These fluctuations in intensive quantities themselves lead to responses dXresp1 and dXresp2. The signs of these responses can be determined by minimizing the total energy of the system and reservoir given the initial perturbation,

d(U+U′)=(f1−f′1)dXresp1+(f2−f′2)dXresp2=dffluc1dXresp1+dffluc2dXresp2≤0.

Because X1 and X2 are independent variables, each term in the sum must be negative, and we have

dffluc1dXresp1≤0anddffluc2dXresp2≤0.

The first equation yields

0≥dffluc1dXresp1⟺0≥dffluc1∂f1∂X1dXresp1=dffluc1dfresp(1)1,

where we have multiplied both sides of the inequality by ∂f1/∂X1, which must be positive by stability. dfresp(1)1 is the response of f1 to the fluctuation due to X1 only, and so we have the Le Chatelier principle. The second equation yields

0≥dffluc2dXresp2=∂f2∂X1dXpert1dXresp2,

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