A box of length 1 meter and cross-sectionalarea A has a moveable partition inside it.There is some gas on either side of thepartition.The number of particles on the left is Ni =3x102%. The number of particles on the rightis Na = 2x10. The gas on both sides is inthermal equilibrium at the same temperatureT. When the partition settles down to its finalposition, find the lengths Li and La of the leftand right sides of the box.ImL,

Solution:- No.of particles on left N1=3×1023 No.of particles on right…

Answered: A box of length 1 meter and…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A box of length 1 meter and cross-sectional
area A has a moveable partition inside it.
There is some gas on either side of the
partition.
The number of particles on the left is Ni =
3x102%. The number of particles on the right
is Na = 2x10. The gas on both sides is in
thermal equilibrium at the same temperature
T. When the partition settles down to its final
position, find the lengths Li and La of the left
and right sides of the box.
Im
L,

Expert Solution

Step 1

$Solution : - No . of particles on left N_{1} = 3 \times 10^{23} No . of particles on right N_{2} = 2 \times 10^{23}$

$At equilibrium condition the box inside is$

$Then, we can write its as PV = nRT \frac{V}{n} = \frac{RT}{P} constant \frac{V}{n} = constant \underset{}{\to} (1) so, \frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}} \frac{L_{1} A}{3 \times 10^{23}} = \frac{L_{2} A}{2 \times 10^{23}} 2 L_{1} = 3 L_{2} \underset{}{\to} (2)$

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