Relatively simple question regarding Thermo (entropy and heating)

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Problem 1:

Homework Statement


An ideal gas is heated reversibly at constant volume from T1 to T2. Calculate the entropy change
of the system, the surroundings, and the universe.


Homework Equations


dS = dq/T (reversible change)
and dq = CvdT (constant V)

The Attempt at a Solution


I know how to do the system:
dS = dq/T, so you can substitute the heat capacity for dq to get
dS = CvdT/T and integrate to get dS of the system to be Cv * ln(T2/T1)

Here is where I'm not sure...
Wouldn't the entropy of the surroundings just be the negative of that? I figure because it's a reversible process the entropy change of the universe is 0...correct? Please correct me on that. Also, is there a better expression or more intuitive solution for the entropy of the surroundings?

Problem 2:

Homework Statement


For a particular change of state, why is qrev>qirr? That is, why is heat flow greater for a reversible
change of state rather than when the same change of state is brought about irreversibly?


Homework Equations


None that I know of.


The Attempt at a Solution


Is it because in an irreversible process, heat is often dissipated away in other forms such as work, noise, etc...? Because the system is not in equilibrium for an irr process the heat flow is not always conserved. Is this correct and is there a better explanation for this problem?

Thanks!
 

Answers and Replies

  • #2
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I think you are right in both problems.
For problem 2, you might be able to add some more quantitative arguments in terms of the second law.
 

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