Calculate the probability that the speed of a chlorine molecule will lie between200 m s^-1 and 220 ms^-1 at 300 K. Given that mass of chlorine molecule = 70u,kb = 1.38 x 10^-23 JK^-1 and NA = 6.02 x 10^26 kmol^-!.
Answers
Answer:
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Explanation:
To be given in question
v_{1}=200meter/secv
1
=200meter/sec
v_{2}=220meter/secv
2
=220meter/sec
M_{cl}=70 amuM
cl
=70amu K_{B}=1.38\times10^{-23}Jule/molK
B
=1.38×10
−23
Jule/mol T=300KT=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=[
2πK
B
T
m
]
2
1
e
−
2K
B
T
mv
2
P_{1}=[\frac{m}{2\pi K_{B}T}]^\frac{1}{2}e^{-\frac{mv_{1}^2}{2K_{B}T}}P
1
=[
2πK
B
T
m
]
2
1
e
−
2K
B
T
mv
1
2
Put value
P_{m}=[\frac{m}{2\pi K_{B}T}]^\frac{1}{2}P
m
=[
2πK
B
T
m
]
2
1
P_{m}= [\frac{70 \times 1.67\times10^{-27}}{2\times3.14\times1.38\times10^{-23}\times 300}]^\frac{1}{2}P
m
=[
2×3.14×1.38×10
−23
×300
70×1.67×10
−27
]
2
1
P_{m}=2.12\times10^{-3}P
m
=2.12×10
−3
\frac{mv_{1}^2}{2K_{B}T}=x
2K
B
T
mv
1
2
=x
\frac{mv_{1}^2}{2K_{B}T}=
2K
B
T
mv
1
2
= [\frac{70 \times 1.67\times10^{-27}\times200^2}{2\times1.38\times10^{-23}\times 300}]^\frac{1}{2}[
2×1.38×10
−23
×300
70×1.67×10
−27
×200
2
]
2
1
x=0.564x=0.564
\frac{mv_{2}^2}{2K_{B}T}=y
2K
B
T
mv
2
2
=y
\frac{mv_{2}^2}{2K_{B}T}=
2K
B
T
mv
2
2
= y= [\frac{70 \times 1.67\times10^{-27}\times220^2}{2\times1.38\times10^{-23}\times300}]y=[
2×1.38×10
−23
×300
70×1.67×10
−27
×220
2
]
y=0.6824y=0.6824
First velocity
P_{1} =P_{m}e^{-x}P
1
=P
m
e
−x
Put value
P_{1}=1.206\times10^{-03}P
1
=1.206×10
−03
Second velocity
P_{2} =P_{m}e^{-y}P
2
=P
m
e
−y
Put value
P_{2}=P
2
= 1.071\times10^{-03}1.071×10
−03