Why Is Empirical Temperature Defined Using Constant Volume Gas Thermometers?

  • Thread starter Thread starter Kolahal Bhattacharya
  • Start date Start date
  • Tags Tags
    Temperature
Click For Summary
Empirical temperature is defined using constant volume gas thermometers due to their ability to provide reliable measurements at the triple point, which is a standard reference point in thermodynamics. The relationship between pressure and temperature in these thermometers simplifies the calculations, avoiding complications seen in other methods like thermocouples or resistance thermometers. In those cases, the thermometric properties depend on temperature in a more complex manner, making them less straightforward. The discussion highlights that constant volume gas thermometers offer a consistent and accessible means of defining temperature without the limitations of other devices. Overall, this choice is rooted in the desire for accuracy and simplicity in temperature measurement.
Kolahal Bhattacharya
Messages
133
Reaction score
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?
 
Physics news on Phys.org
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
Replies
16
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 29 ·
Replies
29
Views
5K
Replies
14
Views
3K