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I often read in engineering books that the fundamental theorem of thermodynamics $$dU=TdS-pdV$$ is valid for integration along reversible paths. I think the relationship is applicable along every quasi-static-path. Can someone clarify why there should be a need for reversibility?

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I think the relationship is applicable along every quasi-static-path.

Yes.

It is an abstract relation in the space of equilibrium states. You can see equilibrium states as pairs $(V,S)$ (you can choose mostly any other variable pair), with functions $U$,$T$,$P$... of these variables. Then this is just a relation between these functions:

$$dU=TdS-PdV$$

This differential equation will then be true for any abstract path in the space of equilibrium states, no matter if it corresponds to a real transformation or not.

In particular, you may want to apply it along the path of a real transformation. But this transformation must be represented in the space of equilibrium states. Otherwise, there is no reason during the transformation that $S,V,T,P...$ actually exist. For example, the pressure or temperature may have a gradient inside the system, and thus there is no global pressure or temperature.

Saying we are in the space of equilibrium states is exactly saying the transformation is a sequence of equilibrium states, which exactly means quasistatic in the usual sense. So yes, this relation applies exactly to quasistatic transformations, otherwise $T,P$... may not even exist.

If the system is not undergoing a quasistatic transformation, you can still use this equation on any abstract path $\varphi$ between the initial and final equilibrium states.

$$\Delta U=\int_\varphi TdS-PdV$$

This equation is unrelated to reversibility. Note: reversibility can also be defined as "a sequence of equilibrium states" but in a much stronger sense than quasistaticity. See: Is there a quasistatic process that is not reversible?

A succession of small free expansions (say for an ideal gas) is a good example of irreversible quasistatic process. It consists in moving the piston very fast (no molecule can touch the piston while it moves), by a very small displacement, many times. Between each step, we let the gas recover equilibrium. Unlike moving the piston slowly, a progressive free expansion does not exchange work between the gas and the piston. We have:

$$dU=0=TdS-PdV$$

$$dS=\frac{P}{T}dV=\frac{Nk_B}{V}dV$$

$$S-S_0=Nk_B\log\left(\frac{V}{V_0}\right)$$

We find without surprise that a succession of small free expansions is equivalent to one big free expansion. We could have known it from the start, because for both cases (progressive and non progressive), the initial and final states are the same: $(U_0,V_0)\rightarrow (U_0,V_{final})$. The entropy is a state function, its variation does not depend on the path.

Benoit
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Simply, the implicit assumption of this theorem is that the system is in thermal and mechanical equilibirum with ist surroundings, in particular that $P$=$P_{ext}=P_{sys}$. It can be readily shown that a quasi-static irreversible process cannot both maintain the same differential $dU$ and maintain the mechanical equilibrium condition $P_{ext}=P_{sys}$, so either the integration will not yield the correct result for the irreversible process or the pressure $P$ appearing in the equation is not that of the system.

Proof: for an irreversible process $\delta Q_{irrev} \gt TdS$,so either:

$P_{sys}= P_{ext}$, $\delta W= - P_{ext}dV=-P_{sys}dV$ and $dU_{irrev}>TdS-PdV$

or

$dU_{rev}=dU_{irrev}$ and $\delta W= - P_{ext}dV< -P_{sys}dV$, $P_{sys} \ne P_{ext}$

user8736288
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