(a) Express the second law of thermodynamics as a mathematical equation. (Select all that apply.)?

(a) Express the second law of thermodynamics as a mathematical equation. (Select all that apply.)


For a reversible process, Suniverse %26gt; 0.


For a spontaneous process, Suniverse = 0.


For a reversible process, Suniverse %26lt; 0.


For a spontaneous process, Suniverse %26gt; 0.


For a spontaneous process, Suniverse %26lt; 0.


For a reversible process, Suniverse = 0.





(b) In a particular spontaneous process, the entropy of the system decreases. What can you conclude about the sign and magnitude of Ssurr?


[_] The magnitude of Ssurr is greater than the magnitude of Ssystem.


[x] The sign of Ssurr is negative.


[x] Ssurr is less than the magnitude of Ssystem.


[_] The sign of Ssurr is positive.








(c) During a certain reversible process the surroundings undergo an entropy change S = -67 J/K. What is the entropy change of the system for this process?


J/K

(a) Express the second law of thermodynamics as a mathematical equation. (Select all that apply.)?
There are several ways to express the second law of thermo. I think what you are looking for is


S (surround)+S(process) =/%26gt; 0 for any process.


In (a): in a reversible process, S universe=0


The others should be straight-forward from the above.
Reply:a.) First, when you say "Suniverse", I thinkyou mean the change in the entropy of the universe. I will call this deltaSuniverse.





For a reversible process, deltaSuniverse = 0, so the first answer is NOT correct.





For a spontaneous process, deltaSuniverse %26gt; 0, so the second answer is NOT correct.





For a reversible process, deltaSuniverse = 0, so the 3rd answer is NOT correct.





For a spontaneous process, deltaSuniverse %26gt; 0, so the 4th answer IS correct.





For a spontaneous process, deltaSuniverse %26gt; 0, so the 5th answer is NOT correct.





For a reversible process, deltaSuniverse = 0, so the 6th answer IS correct.





b.) For a spontaneous process, deltaSuniverse %26gt; 0. But, deltaSuniverse = deltaSsystem + deltaSsurroundings. Since deltaSsystem %26lt; 0 in this process we can conclude that deltaSsurroundings %26gt;0 and /deltaSsurroundings/ %26gt; /deltaSsystem / . Answers 1 and 4 are correct, assuming our symbols are related by: Ssurr = deltaSsurroundings and Ssys = deltaSsystem





c.) For a reversible process, deltaSuniverse = 0


I assume you mean deltaSsystem = -67 J/K.


deltaSuniverse = deltaSsystem + deltaSsurroundings.


Therefore, we can conclude that deltaSsurroundings = +67 J/K.

columbine