Thermodynamics question. Could be simpler than it looks, but I am really stuck.

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Homework Help Overview

The discussion revolves around a thermodynamics equation involving the gas constant and enthalpy of formation. The original poster expresses uncertainty about what the equation allows them to calculate, suspecting a connection to entropy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the notation and clarify the equation's form. The original poster seeks to understand the implications of the equation, while others suggest rewriting and manipulating the expression to gain insights.

Discussion Status

Some guidance has been offered regarding rewriting the expression and exponentiating it, which may lead to a clearer understanding. The conversation reflects a collaborative effort to explore the equation's meaning without reaching a definitive conclusion.

Contextual Notes

There is ambiguity in the notation used, and the original poster is working within the constraints of their understanding of the equation's components and their relationships.

Jasbraman
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Precisely what does the following equation, in which R is the gas constant, allow one to calculate?



ln xs = -dHf/R(T0-T/TT0)



I have no idea what this could be. I know what each part of the equation is (gas constant, change in enthalpy of formation, Freezing point depression etc.) I just don't know what it allows me to calculate...except that its the natural log of something. I think it has something to do with entropy, but I'm not sure. Any help would be much appreciated.
 
Last edited:
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Notation is ambiguous. Did you mean

[tex] \ln x_s = \frac{-dH_f}{R}\left(\frac{T_0-T}{TT_0}\right)[/tex]

in which case [itex]x_s[/itex] is dimensionless?
 
That is exactly what I meant, yes. I did not know you could put in equations like that. Anyway, I have to find out precisely what it allows me to calculate, but I have no idea.
 
Jasbraman said:
That is exactly what I meant, yes. I did not know you could put in equations like that. Anyway, I have to find out precisely what it allows me to calculate, but I have no idea.

Try this:
1. the expression with T can be rewritten as two terms
2. exponentiate
 
Last edited:
That should give you something that looks a lot like something else you're familiar with :-/
 
Oohhhhh yeeeah! Thanks for that!
 

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