How does ΔA=-integral(PdV) relate to the general equation dA=dU-TdS?

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The discussion revolves around the relationship between the Helmholtz free energy (ΔA) and the first law of thermodynamics, particularly in the context of thermodynamic processes. Participants are exploring how the equation ΔA=ΔU-TΔS connects to the expression ΔA=-integral(PdV).

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the meaning of ΔA and its relation to the first law of thermodynamics. Questions are raised about the conditions under which the equations apply, particularly regarding isothermal processes and the implications of integrating the general equation dA=dU-TdS.

Discussion Status

The discussion is active, with participants providing insights into the definitions and conditions of the equations involved. There is an exploration of the assumptions necessary for the equations to hold true, particularly the constancy of temperature in certain scenarios.

Contextual Notes

Some participants note that ΔA=ΔU-TΔS is specifically applicable to isothermal processes, while the general equation dA=dU-TdS is presented as universally valid. There is also mention of the need to integrate under specific assumptions to derive the relationship between ΔA and the work done (integral(PdV)).

sparkle123
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I know ΔA=ΔU-TΔS
how does this lead to ΔA=-integral(PdV)? (which seems like work)
 
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What exactly does 'ΔA' mean? That looks like the first law of thermodynamics just that you replaced W12 with ΔA
 
Are you talking about an isothermal process?
 
I mean Helmholtz free energy.
Thanks!
 
ΔA=ΔU-TΔS is only correct for an isothermal process; the general equation is dA=dU-TdS, which is always true. (Try integrating it; you need to assume that T is constant to obtain the first equation.) Because dU=TdS-PdV for a closed system, ΔA=-integral(PdV). Does this make sense?
 

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