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Empirical temperature problem

  • #1

Homework Statement



In almost every thermodynamics book, I see that they define empirical temperature in terms of parameters of a constant volume gas thermometers.
Like: t=(t_3)[(lim P_3 tends to 0) ((P/P_3)_v)].Here 3 corresponds to triple point.Why?If they are to define empirical temperature and they are to refer to triple point then why they do not take thermocouples or resistance thermometers?
Is it because that for constant volume gas thermometers the exposure to the triple point temperature is readily accessible?
 

Answers and Replies

  • #2
Hello,I found one.Please check if I am correct.In every book the concept has been developed starting from (theta_2/theta_1)=(X_2/X_1) where X is the value of thermometric property and theta is the empirical temperature.
I used the letter t instead theta in last post.
The question was why X should be taken as pressure of a constant volume gas thermometer insted resistence of a resistence thermometer or thermo-emf of a thermocouple.
In the latter cases,X depends on theta square,theta, as well as on constants.
Extracting (theta_2/theta_1) from RHS,we see that the rest also depends on thetas.So,effectively we get 1=f(theta_1,theta_2) which may be correct only for a limited range of tempetatures.
For the constant volume gas ther mpmeters,this complication does not arises.
 

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