- #1

Qwurty2.0

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- 0

## Homework Statement

A gas consisting of 15,200 molecules, each of mass 2.00 x 10

^{-26}kg, has the following distribution of speeds, which crudely mimics the Maxwell distribution:

__Number of Molecules__-

__Speed (m/s)__

1600 - 220

4100 - 440

4700 - 660

3100 - 880

1300 - 1100

400 - 1320

(a) Determine

*v*for this distribution of speeds.

_{rms}## Homework Equations

v

_{rms}= √(2 * E

_{k}/m)

E

_{k}= (1/2) * n * M * v

^{2}

## The Attempt at a Solution

__Weighted Average__

((1600 * 220) + (4100 * 440) + (4700 * 660) + (3100 * 880) + (1300 * 1100) + (400 * 1320)) / 15200

= (9944000 m/s) / 15200

= 654.32 m/s

v

_{rms}= √(2 * E

_{k}/m)

E

_{k}= (1/2) * n * M * v

^{2}

n = number of moles

M = molar mass

m = mass of single molecule

n = 15200 molecules / 6.02x10

^{23}molecules/mole

= 2.525x10

^{-20}mole

M = 2.00x10

^{-26}kg / molecule * 6.02x10

^{23}molecules / mole

= 0.01204 kg / mol

E

_{k}= (1/2)(2.525x10

^{-20}moles)(0.01204 kg / mole)(654.21 m/s)

^{2}

= 6.506x10

^{-17}kg⋅m/s

v

_{rms}= √(2 * (6.506x10

^{-17}kg⋅m/s) / 2.00x10

^{-26}kg)

= 80659.78 m/s ...

The correct answer is 710 m/s so obviously I am either grossly overcomplicating this, or I am using the wrong equation.