Does first law contradicts 2nd law of themodynamics?

Question

We can write first law of thermodynamics in two forms. $$dU=TdS-pdV$$ and $$dU=dq+dw$$ It is also true that $dw=-pdV$ therefore $TdS=dq$ for every process irrespective of reversibility. What I am missing here?

Answer 1

It is true, for all processes whether reversible or not, that:

$$dU=TdS-PdV=dq+dw$$

as long as two of the four variables ($T, S, V, P$) can be defined for the system.

However, it is not true that $dq = T dS$ always; that equality only holds for reversible processes. Likewise, $dw = -PdV$ is only true for reversible processes. If the processes is irreversible, then $dq \lt T dS$ and $dw \gt -PdV$.

Nevertheless, because of the first law, the sum of $dq$ and $dw$ always equals the change in energy, whether the process is reversible or not:

$$dU=dq_\textrm{rev}+dw_\textrm{rev} = dq_\textrm{irrev}+dw_\textrm{irrev}$$

which also is equal to $TdS-PdV$.

Answer 2

No, dw = -pdV is ONLY for a reversible process and hence you can compare your two equations only if both are written for reversible processes for which TdS = dq. What you are doing wrong is comparing an equation which holds for all processes with an equation that is true only for reversible processes. That is why you are getting a contradiction