- #1

asdfTT123

- 9

- 0

## 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!