# Explain difference between internal energy and enthalpy

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 $=PV$?

• Possible duplicate of Explain internal energy and enthalpy Sep 30, 2016 at 10:59
• My query is quite different Sep 30, 2016 at 11:11
• 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 $PV=w$. Sep 30, 2016 at 13:02
• @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. Sep 30, 2016 at 14:40

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 $$\int{\left(\frac{\partial H}{\partial P}\right)_TdP}$$