Question

7. Calculate the probability that the speed of an oxygen molecule will lie between 100
m/s and 101 m/s at 200 K. The mass of oxygen molecule is 32 u. Take
kg = 1.38 x 10'J/k and NA = 6 x 1026 kmol-1
A 12.57 x 10'
B. 6.64 x 10-5
C. 4.47 x 10-8
D. 5.35 x 10-6​

Answer 1

Answer:

D one is the correct one 100%

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Answer 2

Answer:

we can get the probability which is

$P = D. 5.35 x 10^-6$

Explanation:

The probability that the speed of an oxygen molecule will lie between 100 m/s and 101 m/s at 200 K can be calculated using the Maxwell-Boltzmann distribution.

The probability density function (pdf) for the Maxwell-Boltzmann distribution is given by:

$f(v) = (4πm/kT)^(-3/2) * v^2 * e^(-mv^2/2kT)$

Where:

m = mass of oxygen molecule (32 u)

v = speed of the oxygen molecule

k = Boltzmann constant ($1.38 x 10^-23 J/K$)

T = temperature (200 K)

The probability that the speed of the oxygen molecule will be between v1 and v2 is given by the integral of the pdf from v1 to v2.

So, To find the probability that the speed of an oxygen molecule will lie between 100 m/s and 101 m/s, we need to integrate the above pdf from 100 to 101:

$P = \int\limits^a_b {x} \, dx 100-101 f(v) dv$

By solving the above integral we can get the probability which is

$P = D. 5.35 x 10^-6$

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