5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 9
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16Problem set #4 solutions for 5.112, Fall 2011
Page 9 of 16
(
)
⎡
⎤
⎛
⎞
(
)
-3
-1
−20.180 x 10
2
kg mol
1, 000, 000
⎛
⎞
v
⎢
⎥
⎡
(
)
(
)
⎤
rms
⎜
⎟
2
2
rms
=
− 60.691
(
)
⎝ ⎜
⎠ ⎟
exp
606.91
⎢
⎣
⎦
⎥
(
)
ΔN
⎜
⎟
-1
-1
0.1v
0.1 v
rms
2 8.315 J mol
K
298 K
⎝
⎠
rms
⎢
⎥
⎣
⎦
(
)
(
)
1, 000, 000
⎡
⎤
-6
5
rms
= 100 exp (−4.07205 x 10
) 3.64656 x 10
⎣
⎦
ΔN
0.1 v
rms
(
)
1, 000, 000
(
)
(
)
rms
= 100
exp(−1.48490) = 100
(.22653)
ΔN
0.1 v
rms
(
)
1, 000, 000
rms
= 22.653
ΔN
0.1 v
rms
ΔN
= 44, 144
0.1 v
rms
For every 1,000,000 molecules traveling at v
, 40 thousand molecules travel at 0.1v
.
rms
rms
We can now do the same calculation for 1.2v
:
rms
(
)
⎡
⎤
⎛
⎞
(
)
-3
-1
−20.180 x 10
2
kg mol
1, 000, 000
⎛
⎞
v
⎢
⎥
⎡
(
)
(
)
⎤
rms
⎜
⎟
2
2
rms
=
− 728.292
(
)
⎝ ⎜
⎠ ⎟
exp
606.91
⎢
⎥
⎣
⎦
(
)
ΔN
⎜
⎟
-1
-1
1.2v
1.2 v
rms
2 8.315 J mol
K
298 K
⎝
⎠
rms
⎢
⎥
⎣
⎦
(
)
(
)
1, 000, 000
⎡
⎤
-6
5
rms
= 0.6944 exp (−4.07205 x 10
) −1.6207 x 10
⎣
⎦
ΔN
1.2 v
rms
(
)
1, 000, 000
(
)
(
)
rms
= 0.6944 exp 0.659955
= 0.6944
(1.93470) = 1.34346
ΔN
1.2 v
rms
5
ΔN
= 7.44345 x 10
1.2 v
rms
5
= 7.4 x 10
5
For every 1,000,000 molecules traveling at v
, 7.4 x 10
molecules travel at 1.2v
.
rms
rms
We can now do the same calculation for 2v
:
rms
(
)
⎡
⎤
⎛
⎞
(
)
-3
-1
−20.180 x 10
2
kg mol
⎛
⎞
1, 000, 000
v
⎢
⎥
⎡
(
)
(
)
⎤
rms
⎜
⎟
2
2
rms
=
− 1213.82
(
)
⎝ ⎜
⎠ ⎟
exp
606.91
⎢
⎥
⎣
⎦
(
)
ΔN
⎜
⎟
-1
-1
2v
2v
rms
2 8.315 J mol
K
298 K
⎝
⎠
rms
⎢
⎥
⎣
⎦
(
)
(
)
1, 000, 000
⎡
⎤
-6
6
rms
= 0.25 exp (−4.07205 x 10
) −1.10502 x 10
⎣
⎦
ΔN
2v
rms
(
)
1, 000, 000
(
)
(
)
(
)
rms
= 0.25
= 0.25
(89.9896) = 22.49
exp 4.49969
ΔN
2v
rms
ΔN
= 44, 450
2v
rms
For every 1 million molecules traveling at v
, there are 40 thousand molecules traveling at twice
rms
the v
.
rms
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