GR | # Login | Register
  GR8677 #15
GREPhysics.NET Official Solution    Alternate Solutions
Verbatim question for GR8677 #15
Statistical Mechanics}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.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
2015-07-22 12:26:56
2015-07-22 12:25:20
testing123 july 22 2015NEC
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.
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?
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}
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!!NEC
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.
2006-01-24 11:27:10
How did you connect P(Ngasatoms) = (1-1E-6)^N to answer C?
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.
2015-07-22 12:31:28
2015-07-22 12:37:21

\langle my \rangle
2015-07-22 12:49:08
\\langle my \\rangle
Answered Question!

Post A Comment!
You are replying to:
testing123 july 22 2015

Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...