GREPhysics.NET: GR8677 #15

GR8677 #15

Problem

GREPhysics.NET Official Solution

Statistical Mechanics $\Rightarrow$ }Probability

Probability is mostly common sense and adhering to definitions.

The probability that a gas molecule or atom is in the small cube is $P(in)=1E-6$ . The probability that it's not in that small cube is $P(not)=1-1E-6$ . Assuming independent gas molecules or atoms, i.e., the usual assumption of randomness in Stat Mech, one gets, $P(N gas atoms)=\left(1-1E-6\right)^N$ .

The answer is thus C.

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Comments

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yosun2015tester
2015-07-22 12:26:56

what

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yosun2015tester
2015-07-22 12:25:20

testing123 july 22 2015

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mpdude8
2012-04-15 20:09:22

For this, I just looked at it logically. Think about N = a small number, like 2. A and E are out -- there's always at least a chance that neither one will be in that subset, or one of them will. Also, when N = 2, the probability that both particles are outside of such a small subset of the total volume should be almost 1. B is out, as even for N = 2, the probability is extremely small, and races towards zero as you increase N slightly.

D is also out, as it is the exact opposite effect of what you want, logically. Probability, as you add more particles, should tend to 0. C is the only choice that has the correct tendency as N grows, and at a reasonable rate of increase.

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livieratos
2011-11-07 03:54:30

so the probability of one atom to be in the given small volume is the ratio of the small volume to the entire volume? P(yes) = 1E-6/1 = 1E-6?

Ajith
2015-10-14 09:30:40

I guess $P(an atom in 10^{-6})=\\frac{No. of atoms in that volume}{tolal number of atoms}$ \r\nHere, No. of atoms in $1 m^3$ = N\r\nNo. of atoms in $10^{-6} m^3$ = No. of atoms in $1 m^3 * 10^{-6}$ = $N *10^{-6}$ \r\nThus, $P(an atom in 10^{-6})=\\frac{N*10^{-6}}{N}$ = $10^{-6}$

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ubaraj
2007-11-01 01:15:13

the volume is 1. There can be no He atoms in the volume 10E-6. So all N atoms have to be within the volume (1-10E-6), and so the probability becomes (1-10E-6)^N. The choice is C!!

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Furious
2007-08-17 15:40:58

You really just need to look at the limiting factors in this problem.

When N is 0 the probability should be 1. That eliminates A and D.

When N is extremely high, the probability should go to 0. That eliminates E, (not that any of us thought it was E)

So that leaves us with B and C. Which both match the limiting cases, but just think about it when N is 1. When N is one there should be a very high chance that it is not in the specific 1e-6 area.

For B, the probability is 1e-6, not very high at all.
For C, the probability is 0.99999, pretty close to 1.

This helped me since, I get super confused whenever I think about probability.

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wishIwasaphysicist
2006-01-24 11:27:10

How did you connect P(Ngasatoms) = (1-1E-6)^N to answer C?

yosun
2006-02-01 22:02:05

hi wishIwasaphysicist, because each of the N molecules are independent, the probability would be multiplicative. Thus, if the particle for each particle is P, the probability for N particles would be $P^N$ .

yosun2015tester
2015-07-22 12:31:28

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yosun2015tester
2015-07-22 12:37:21

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yosun2015tester
2015-07-22 12:49:08

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For this, I just looked at it logically. Think about N = a small number, like 2. A and E are out -- there's always at least a chance that neither one will be in that subset, or one of them will. Also, when N = 2, the probability that both particles are outside of such a small subset of the total volume should be almost 1. B is out, as even for N = 2, the probability is extremely small, and races towards zero as you increase N slightly. D is also out, as it is the exact opposite effect of what you want, logically. Probability, as you add more particles, should tend to 0. C is the only choice that has the correct tendency as N grows, and at a reasonable rate of increase.

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