[cokuardi]

1 a)Suppose that you have a 130.0 liter vessel containing an ideal gas. Suppose that you imagine a smaller volume of 50.0 liters within the vessel. If the vessel contains only 40 molecules of the gas, what is the probability that all 40 gas molecules would be found within the imagined smaller volume rather than spread throughout the vessel? Enter your answer as a number in scientific notation representing the fractional probability.

  b)Suppose that the ideal gas described in the previous question was actually confined to the smaller volume. Now suppose that the gas was allowed to expand isothermally to fill the entire volume of the vessel. Calculate the entropy change for this expansion. Enter your answer in J/K with the correct sign.

I know that w=sites^n, but i'm not sure how the volumes factor in.  Thanks for the help.

[xiankai]

a) think of it in another way; this is a simple probability question.

if i hid 5 balls in 10 boxes, what are the probability that one box will contain the balls i am looking for?

each ball has a probability of 1/10 of being in the one box, and the other balls also have a probability of 1/10 of being in the same box.

1/10 x 1/10 x 1/10 x 1/10 x 1/10 = (1/10)⁵

surely u can relate to your problem now, instead of balls we have molecules. instead of boxes we have volume.

[cokuardi]

Nope.  Still not understanding what to do with the volumes.

[xiankai]

pretend each unit of volume is one box, and each molecule is a ball.

[cokuardi]

so then is it (1/50)^40

[xiankai]

u progress from a volume of 130.0 liters to 50.liters, not from 50 to 1!

[cokuardi]

You really need to give me more than that.  Does that mean (130/50)^40, or what?

[xiankai]

we preffer to help the users figure it out on their own, but since u havent got it, let me clarify things:

the probability for a molecule originally in a volume of 130.0 liters being in a volume of 50.0 liters is effectively 50/130.

since there are 40 molecules, the probability is (50/130)^40.

PS: have u studied probability? this logic is quite simple...

[cokuardi]

You know, there is no need to be rude.  Being polite is quite simple...

[xiankai]

my apologies if i offended you; my main intention is to help u out :/