# Does first law contradicts 2nd law of themodynamics?

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?

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 $$TdS-PdV$$.

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

The second law is actually an inequality,

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

The strict equality holds only for reversible processes.

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