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

$$U=TS-PV+\sum\mu_iN_i$$

Hence, $$dU=TdS+SdT-VdP-PdV+\sum\mu_idN_i$$

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?

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  • $\begingroup$ For an isothermal process, dU is zero only for an ideal gas. Your reference gives dU for the more general case of a real gas when U also depends on V (if you look for it). $\endgroup$
    – Chet Miller
    Commented Aug 17, 2020 at 14:55

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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.

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  • $\begingroup$ Oh...Now I got it. Thanks! $\endgroup$
    – Noob
    Commented Aug 17, 2020 at 14:47

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