Answer to Question #179244 in Molecular Physics | Thermodynamics for sunny

Question

Calculate the probability that the speed of a chlorine molecule will lie between 200 m s−1 and 220 m s1 at 300 K. Given that mass of chlorine molecule = 70u, kB = 1.38 × 1023 JK1 and NA = 6.02 × 1026 kmol1

Answer 1

To be given in question

$v_{1}=200meter/sec$

$v_{2}=220meter/sec$

$M_{cl}=70 amu$ $K_{B}=1.38\times10^{-23}Jule/mol$ $T=300K$

To be asked in question

Probability speed

We know that

$P=[\frac{m}{2\pi K_{B}T}]^\frac{1}{2}e^{-\frac{mv^2}{2K_{B}T}}$

$P_{1}=[\frac{m}{2\pi K_{B}T}]^\frac{1}{2}e^{-\frac{mv_{1}^2}{2K_{B}T}}$

Put value

$P_{m}=[\frac{m}{2\pi K_{B}T}]^\frac{1}{2}$

$P_{m}= [\frac{70 \times 1.67\times10^{-27}}{2\times3.14\times1.38\times10^{-23}\times 300}]^\frac{1}{2}$

$P_{m}=2.12\times10^{-3}$

$\frac{mv_{1}^2}{2K_{B}T}=x$

$\frac{mv_{1}^2}{2K_{B}T}=$ $[\frac{70 \times 1.67\times10^{-27}\times200^2}{2\times1.38\times10^{-23}\times 300}]^\frac{1}{2}$

$x=0.564$

$\frac{mv_{2}^2}{2K_{B}T}=y$

$\frac{mv_{2}^2}{2K_{B}T}=$ $y= [\frac{70 \times 1.67\times10^{-27}\times220^2}{2\times1.38\times10^{-23}\times300}]$

$y=0.6824$

First velocity

$P_{1} =P_{m}e^{-x}$

Put value

$P_{1}=1.206\times10^{-03}$

Second velocity

$P_{2} =P_{m}e^{-y}$

Put value

$P_{2}=$ $1.071\times10^{-03}$