GREPhysics.NET
GR | # Login | Register
   
  GR8677 #15
Problem
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!
Comments
ernest21
2019-08-10 03:09:28
A great day and a very explicit information. apex hq
danielsw98667
2019-10-21 05:20:36
With Kinetic Theory and the atomic and molecular explanation of pressure, the answer is c. business for sale cincinnati ohio
NEC
fredluis
2019-08-08 13:02:41
The writing style is somewhat conversational and straightforward, and the problems show a nice range of difficulty. tile contractorNEC
joshuaprice153
2019-08-08 07:04:25
I was suggested this blog by means of my bro. I am not sure whether or not this publish is written by means of him as no one else understand such precise about my problem. You’re wonderful! Thanks! towing serviceNEC
yosun2015tester
2015-07-22 12:26:56
whatNEC
yosun2015tester
2015-07-22 12:25:20
testing123 july 22 2015NEC
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.
NEC
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}
NEC
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!!NEC
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.
NEC
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
testing123
yosun2015tester
2015-07-22 12:37:21
testering12345

\langle my \rangle
yosun2015tester
2015-07-22 12:49:08
\\langle my \\rangle
Answered Question!

Post A Comment!
Username:
Password:
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...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...