# Stupid question on black holes and thermodynamics

I read in a book that since the area of a black hole increases, the entropy increases, so black holes are consistent w/ the second law of thermodynamics.

So...I have a really stupid question on black holes and the second law of thermodynamics: Consider an ideal gas of non-interacting particles that is spherical in shape. According to GR, once it contracts to less than r=2M (M is the mass), it will get smaller and smaller. However, the entropy of this ideal gas is, of course:

S~log(V)+(terms involving energy)

Where V is the volume. So...wouldn't this mean that the entropy decreases? Or is there some sort of extra energy added to the system to "make up" for the loss of entropy resulting from decreasing volume? Is there some sort of modification in curved spacetime to the equation above to make it OK? Is there a modified version of the second law?

Sorry, this is a really dumb question, I realize.

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It's not a dumb question at all.
The entropy of an ideal gas is related to the product of the volume and temperature, by:
$S \propto \textrm{ln}(V T^a)$
for some power 'a'.

When a gas compresses, what happens to the temperature?

Oh, so the thermal energy increases and "makes up" for the loss of volume?

Exactly!

All right, thank you!