# Entropy of Irreversible Process for 2nd Law Proof

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

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Andrew Mason
Homework Helper
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

^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
Homework Helper
^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

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