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First law of Thermodynamics and the definition of Internal Energy

Physics Asked on June 12, 2021

From the Wikipedia page on internal energy I get the following definition

$$U=TS-PV+summu_iN_i$$

Hence,
$$dU=TdS+SdT-VdP-PdV+summu_idN_i$$

For constant pressure and temperature and when there is no transfer of matter,
$$dU=TdS-PdV$$

which is the 1st law of thermodynamics. My question is: For isothermal processes (constant temperature) $dU=0$. Then how will I get to the 1st law of thermodynamics from this definition?

One Answer

You should carefully identify the validity range of each statement. A more general statement can contain a special case, but it is not possible to obtain the more general from special cases.

First principle of thermodynamics is actually $$ Delta U = q + w, $$ where $q$ and $w$ are heat and work exchanged by the system with the environment.

For the special case of a reversible transformation $$ dU = TdS -PdV.~~~~~~~~~~~~~~~~~[1] $$ In general, $U$ depends on $S$ and $V$, and such a dependence can be transformed into a dependence on $T$ and $V$. It is only for the special case of a perfect gas that $U$ depends on $T$ only. Therefore, from the special case of the isothermal behavior of a perfect gas it is impossible to obtain a relation like $[1]$ which is valid for all phases and for all systems.

Correct answer by GiorgioP on June 12, 2021

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