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

[sparkle123]

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

 

[rock.freak667]

What exactly does 'ΔA' mean? That looks like the first law of thermodynamics just that you replaced W₁₂ with ΔA

 

[danago]

Are you talking about an isothermal process?

 

[sparkle123]

I mean Helmholtz free energy.
Thanks!

 

[Mapes]

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?