Timeline for Temperature scale from equation of state
Current License: CC BY-SA 4.0
15 events
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| Nov 9, 2022 at 15:26 | comment | added | Themis | What is an "observational" temperature? Any monotonic function $c(T)$ of the absolute temperature $T$ can serve as observational scale. The Celsius and Fahrenheit scales are such examples and so is the temperature of Galileo's thermometer or the $c(T)$ in your expression. The only way to obtain true absolute $T$ via an observational approach is through the ideal-gas law, $T = PV/\text{const}$ when $P$ is low enough. And the only way to convince ourselves that this is the true absolute temperature is put the ideal gas law in a Carnot cycle and show it satisfies the isentropic condition. | |
| Nov 9, 2022 at 13:05 | comment | added | scmartin | I think @JohnDoty 's point is that the question I'm asking is clearly not trying to define temperature from the fundamental equation, but to come up with a scale for temperature based on an equation of state (some experimentally observable relationship between system properties). | |
| Nov 9, 2022 at 0:11 | comment | added | Themis | Let us continue this discussion in chat. | |
| Nov 9, 2022 at 0:00 | comment | added | John Doty | @Themis I have a Galilean thermometer on my desk. Galileo knew nothing of thermodynamics. | |
| Nov 8, 2022 at 23:58 | comment | added | John Doty | @Themis Temperature was measured long before it had a clear physical definition. The phenomena come first, and then the theory is crafted to capture them. | |
| Nov 8, 2022 at 23:52 | comment | added | Themis | @JohnDoty Of course not. But measuring something that has no definition is pretty useless. Historically, temperature obtained its thermodynamic meaning first through the ideal gas law,then through entropy and the Carnot cycle, again applied to an ideal gas. | |
| Nov 8, 2022 at 23:07 | comment | added | John Doty | @Themis That doesn't seem to be how metrologists figure out temperature scales. See bipm.org/documents/20126/41489682/SI-App2-kelvin.pdf/… | |
| Nov 8, 2022 at 21:13 | comment | added | Themis | @JohnDoty In principle $$S(E,V,N) = k \ln\Omega(E,V,N)$$ where $\Omega$ is the number of microstates. In principle (again) this can be calculated from a molecular level model of matter. Easier said than done, but molecular simulations today have reached a pretty sophisticated level, so it is indeed possible. | |
| Nov 8, 2022 at 19:31 | comment | added | John Doty | "The main point is that the fundamental equation is the relationship between 𝑆, 𝐸, 𝑉 and 𝑁". But how do you know what that is? | |
| Nov 8, 2022 at 19:03 | history | edited | Themis | CC BY-SA 4.0 |
Corrected typos in Eqs 1 and 2
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| Nov 8, 2022 at 19:02 | comment | added | Themis | @scmartin oops.... of course you're right. I corrected the typos. | |
| Nov 8, 2022 at 16:57 | comment | added | scmartin | Aren't the more standard definitions of $T$ and $p$ given by $(\partial S / \partial E)_{V,N} = 1/T $ and $(\partial S / \partial V)_{E,N} = p/T $? | |
| Nov 8, 2022 at 12:44 | comment | added | Themis | You can always define an operational temperature scale, but whether such scale makes thermodynamic sense, you need to check against the fundamental equation. If the operational temperature is not linked to entropy, it will not work. It works for the ideal gas because the ideal-gas equation is consistent with the fundamental equation, i.e., it can be derived from it assuming that molecules are point masses that do not interact. If you can obtain the entropy from your equation of state, you can recover temperature. | |
| Nov 8, 2022 at 11:57 | comment | added | scmartin | I know how to define the thermodynamic temperature in terms of the fundamental equation. I believe the question from the University Physics book that prompted my question is asking for an operational way to define a temperature scale. | |
| Nov 8, 2022 at 10:49 | history | answered | Themis | CC BY-SA 4.0 |