I know ΔA=ΔU-TΔS
how does this lead to ΔA=-integral(PdV)? (which seems like work)
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.
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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