3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $TdS-PdV$.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The second law is actually an inequality,

$$ \mathrm{d}S \geq \frac{\delta Q}{T}.$$

The strict equality holds only for reversible processes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think the OP is concerned that according to the first law $\delta q=T\text dS$ always. $\endgroup$ – BioPhysicist May 16 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.