Is the Joule-Thompson experiment an isenthalpic process?

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Homework Help Overview

The discussion revolves around the Joule-Thompson experiment and whether it can be classified as an isenthalpic process, specifically focusing on the condition where the change in enthalpy (ΔH) is zero. Participants are exploring the relationship between temperature and pressure in this context, particularly through the use of partial derivatives.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster expresses confusion about how to start the problem, particularly regarding the total differential of enthalpy (H) and the implications of the Joule-Thompson experiment being isenthalpic. Some participants inquire about the total differential of H and provide its expression, while others question the correctness of this expression.

Discussion Status

The discussion is ongoing, with participants attempting to clarify concepts and definitions related to the problem. Some guidance has been offered regarding the total differential of H, and there is a recognition that if the process is indeed isenthalpic, then the change in enthalpy should be zero. However, there is no explicit consensus on the correctness of the provided expressions or the next steps.

Contextual Notes

One participant notes a lack of familiarity with calculus concepts, specifically partial derivatives, which may be impacting their understanding of the problem. This suggests that there may be varying levels of mathematical background among participants.

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Homework Statement



Show that the Joule-Thompson experiment is an isenthalpic process. (ΔH=0). The experimentally determined quantity is (Partial derivative T)/(Partial derivative P)H . Starting with the total differential for H, show that:

(Partial deriv H)/(Partial deriv P)T = - Cp(Partial deriv T)/(Partial deriv P)H

does anybody know even where to start ??, iam just confused with this question
 
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Do you know what the total differential of H is?
 
no i dont, since i never took calc 3 and still having some hard time partial derivatives
 
The total differential of H is:

[tex]dH=C_p dT+ \left[ v - T \left( \frac{\partial v}{\partial T} \right) _p \right] dp[/tex]
 
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those this even seem right ??
 

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If its isoenthalpic the dH is going to be zero. You're almost there.
 

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