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

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ΔA is defined as the change in Helmholtz free energy, expressed as ΔA=ΔU-TΔS, which applies specifically to isothermal processes. The general relationship dA=dU-TdS holds true under all conditions. By integrating the equation dU=TdS-PdV for a closed system, one arrives at ΔA=-integral(PdV), linking free energy to work done. This relationship highlights the connection between thermodynamic potentials and work in physical systems. Understanding these equations is crucial for analyzing energy transformations in thermodynamics.
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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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