Entropy of Irreversible Process for 2nd Law Proof

[zmike]

My textbook tries to prove that for Entropy of Irreversible Process > 0 by using the example of irrev expasion into a vacuum which doesn't make sense to me.

If work is = 0 then no heat is exchanged between the sys and surrounding so qsurr = 0

s(uni) = s(sys) + 0

since S(sys)=qirr/t and no heat is being exchanged then S(sys) = 0

then S(uni) SHOULD = 0 why would does my textbook put S(uni)> 0 by using the nr*ln(T2/T1) equation for S(sys) when no work was done and q was suppose to be 0? how can I prove to that S(uni) is > 0 for irreverisble process?

Thank you

 

[Andrew Mason]

My textbook tries to prove that for Entropy of Irreversible Process > 0 by using the example of irrev expasion into a vacuum which doesn't make sense to me.

If work is = 0 then no heat is exchanged between the sys and surrounding so qsurr = 0

s(uni) = s(sys) + 0

since S(sys)=qirr/t and no heat is being exchanged then S(sys) = 0

then S(uni) SHOULD = 0 why would does my textbook put S(uni)> 0 by using the nr*ln(T2/T1) equation for S(sys) when no work was done and q was suppose to be 0? how can I prove to that S(uni) is > 0 for irreverisble process?

The change of entropy is defined as $\int ds = \int dQ/T$ over the reversible path between two states.

To calculate the change in entropy in a free expansion (allowing a gas to expand into a vacuum) you have to determine a reversible path between the intial and final states. The reversible path between those two states requires a heat flow into the gas.

AM

 

[zmike]

^Thanks, I understood why for reversible. The issue I am having is with IRREVERSIBLE, how could it have a > 0 entropy proved using the equations?

thanks

 

[Andrew Mason]

^Thanks, I understood why for reversible. The issue I am having is with IRREVERSIBLE, how could it have a > 0 entropy proved using the equations?

thanks

Entropy is a state function. It does not depend on the path between two states. That is very important to understand. ENTROPY IS PATH INDEPENDENT.

So, when you speak of a change in entropy in a process you determine the initial state and the final state. Then you determine the reversible path between those two states and determine the change in entropy over that reversible path. You do this EVEN IF THE ACTUAL PATH TAKEN BETWEEN THOSE TWO STATES WAS NOT REVERSIBLE.

For example, you have a gas at state (P, V, T) and you do a free expansion to (P/2, 2V, T).

Step 1: in order to determine the difference in entropy between those two states you determine the reversible path between those two states. That path will be a reversible isothermal expansion.

Step 2: apply the first law to determine dQ. In this case dU=0 (isothermal) so, dQ = PdV

Step 3: determine the change in entropy, the integral of dQ/T, over this reversible path:

$$\Delta S = \int_{V}^{2V} dQ/T = \int_{V}^{2V} PdV/T = nR\int_{V}^{2V} dV/V = nR\ln{2} > 0$$

AM

 

[sardar]

for your question. The concept of entropy and the second law of thermodynamics can be difficult to understand, so it is important to carefully consider the textbook's explanation and any potential misunderstandings.

First, it is important to note that the equation for entropy change in an irreversible process, S(sys) = qirr/T, is only applicable when there is heat exchange between the system and surroundings. In the example of an irreversible expansion into a vacuum, there is no heat exchange, so this equation cannot be used.

Second, the equation S(uni) = S(sys) + S(surr) does not apply in this case because there is no heat exchange with the surroundings. This equation assumes that the system and surroundings are in thermal equilibrium, which is not the case in an irreversible process.

So, how can we prove that the entropy change for an irreversible process is greater than zero? One way to do this is by considering the definition of entropy as a measure of the disorder or randomness of a system. In an irreversible process, there is a net increase in disorder or randomness, which corresponds to an increase in entropy.

Another way to think about it is through the concept of reversibility. In a reversible process, the system can be returned to its initial state by reversing the steps of the process. This means that the entropy change for a reversible process is zero, as there is no net increase in disorder. In contrast, an irreversible process cannot be reversed, and therefore there is a net increase in disorder and a corresponding increase in entropy.

In summary, the textbook's explanation may not be clear, but it is important to understand the underlying concepts of entropy, heat exchange, and reversibility in order to fully grasp the second law of thermodynamics and the concept of entropy change in irreversible processes. I hope this helps to clarify the issue for you.