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GR8677 #15
Problem
 GREPhysics.NET Official Solution Alternate Solutions

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, $P(N gas atoms)=\left(1-1E-6\right)^N$.

Alternate Solutions
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ernest21
2019-08-10 03:09:28
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 danielsw986672019-10-21 05:20:36 With Kinetic Theory and the atomic and molecular explanation of pressure, the answer is c. business for sale cincinnati ohio
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yosun2015tester
2015-07-22 12:26:56
what
yosun2015tester
2015-07-22 12:25:20
testing123 july 22 2015
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.
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?
 Ajith2015-10-14 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, $P(an atom in 10^{-6})=\\frac{N*10^{-6}}{N}$=$10^{-6}$
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!!
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.
wishIwasaphysicist
2006-01-24 11:27:10
How did you connect P(Ngasatoms) = (1-1E-6)^N to answer C?
 yosun2006-02-01 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 $P^N$.
 yosun2015tester2015-07-22 12:31:28 testing123
 yosun2015tester2015-07-22 12:37:21 testering12345 $langle my rangle$
 yosun2015tester2015-07-22 12:49:08 $\\langle my \\rangle$

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$