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I read that $U=q+w$ and $H=u+PV$ so aren't $PV$ and $w$ same? If they are, can't we write $H=q+2w$? Also, $dU=$ change in internal energy wrt change in temperature keeping $V$ constant times $dT+$change in internal energy wrt change in volume keeping $T$ constant times $dV$ Here is the second term $=w$? And, $dH=$ change in enthalpy wrt change in temperature keeping $P$ constant times $dT+$ change in enthalpy wrt change in pressure keeping $T$ constant times $dP$. Here is the second term $=PV$?

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    $\begingroup$ Possible duplicate of Explain internal energy and enthalpy $\endgroup$
    – ACuriousMind
    Sep 30, 2016 at 10:59
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    $\begingroup$ My query is quite different $\endgroup$
    – Kshitij
    Sep 30, 2016 at 11:11
  • $\begingroup$ For starters, if $U=q+w$, then by the same definition, $PV=-w$. Which definition of $w$ are you using? In physics, commonly the definition $U=q-w$ is used, then we have $PV=w$. $\endgroup$ Sep 30, 2016 at 13:02
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    $\begingroup$ @FreezingFire The work w is not PV or -PV. It is equal to the integral of PdV or -PdV, at least for a reversible change. $\endgroup$ Sep 30, 2016 at 14:40

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According to the sign convention you are using, $$\Delta U=q+w$$where, for a reversible expansion or compression, $$w=-\int{PdV}$$So, $$\Delta U=q-\int{PdV}$$and$$\Delta H=\Delta U+\Delta (PV)=q-\int{PdV}+\Delta (PV)$$Integrating by parts, we get$$\Delta H=q+\int{VdP}$$

The quantity $$\int{\left(\frac{\partial U}{\partial V}\right)_TdV}$$ is not equal to the work. In fact, for an ideal gas, this quantity is always equal to zero. Similarly for $$\int{\left(\frac{\partial H}{\partial P}\right)_TdP}$$

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