- #1

henryc09

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## Homework Statement

(a) A piston is used to compress an ideal gas quasistatically from volume Vi to volume Vf . If the gas is in thermal contact with a heat bath at temperature T, such that the compression is carried out isothermally, calculate the work done on the gas, the change in entropy of the gas and the change in entropy of the heat bath

(b) The compression is repeated but but non-quasistatically. Are the three calculated quantities higher, lower, or the same as before?

## Homework Equations

## The Attempt at a Solution

For (a) I said that W=-NkT*ln(V

_{f}/V

_{i}), and [tex]\Delta[/tex]E = 0, so [tex]\Delta[/tex]Q = NkT*ln(V

_{f}/V

_{i}).

Then as it's quasistatic [tex]\Delta[/tex]S

_{gas}= [tex]\Delta[/tex]Q/T = Nk*ln(V

_{f}/V

_{i})

I then said as it's quasistatic [tex]\Delta[/tex]S

_{universe}=[tex]\Delta[/tex]S

_{gas}+ [tex]\Delta[/tex]S

_{bath}=0

so [tex]\Delta[/tex]S

_{bath}= -Nk*ln(V

_{f}/V

_{i})

I think this is right but confirmation would be good. For (b) I said that [tex]\Delta[/tex]S

_{gas}will be the same as it's a state variable, but as the entropy of the universe must increase the bath's entropy must increase. The only thing I'm unsure about is whether or not the work increases/decreases/stays the same. Anyone got any ideas about how to approach this?