- #1

MatinSAR

- 593

- 182

- Homework Statement
- What is ##Tds## in this equation : ##dU = Tds - PdV##. Is it the transferred heat in a reversible change or it is equal to transferred heat in any change even irreversible changes?

- Relevant Equations
- _

More explanation :

First law is given by ##dU=dQ+dW##.

For a reversible change we have:

##dQ = Tds##

##dW = - PdV##

So I rewrite first law as :

##dU=Tds - PdV##

As mentioned before this ##Tds ## is the heat transferred in a reversible change. And the ##-PdV## is the work done by system in a reversible change. Then why we use this formula for any change

This question was asked by my professor in class and I should answer it in one page. He said we can use in our proof :

##dQ_r/T >= dQ/T##

My idea :

Because dU in the equation is independent of path it is true for any change or process. But that Tds is not equal to the heat transferred in a irreversible change. Sadly what I've said is not logical because if dU is the same for both reversible and irreversible changes so how their dQ can have different values?

First law is given by ##dU=dQ+dW##.

For a reversible change we have:

##dQ = Tds##

##dW = - PdV##

So I rewrite first law as :

##dU=Tds - PdV##

As mentioned before this ##Tds ## is the heat transferred in a reversible change. And the ##-PdV## is the work done by system in a reversible change. Then why we use this formula for any change

**even**irreversible changes?This question was asked by my professor in class and I should answer it in one page. He said we can use in our proof :

##dQ_r/T >= dQ/T##

My idea :

Because dU in the equation is independent of path it is true for any change or process. But that Tds is not equal to the heat transferred in a irreversible change. Sadly what I've said is not logical because if dU is the same for both reversible and irreversible changes so how their dQ can have different values?