5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 10
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16Problem set #4 solutions for 5.112, Fall 2011
Page 10 of 16
We can now do the same calculation for 3v
:
rms
(
)
⎡
⎤
⎛
⎞
(
)
-3
-1
−20.180 x 10
2
kg mol
⎛
⎞
1, 000, 000
v
⎢
⎥
⎡
(
)
(
)
⎤
rms
⎜
⎟
2
2
rms
=
− 1820.73
(
)
⎝ ⎜
⎠ ⎟
exp
606.91
⎢
⎥
⎣
⎦
(
)
ΔN
⎜
⎟
-1
-1
3v
3v
rms
2 8.315 J mol
K
298 K
⎝
⎠
rms
⎢
⎥
⎣
⎦
(
)
(
)
1, 000, 000
⎡
⎤
-6
6
rms
= 0.1111 exp (−4.07205 x 10
) −2.94672 x 10
⎣
⎦
ΔN
3v
rms
(
)
1, 000, 000
(
)
(
)
rms
= 0.1111 exp 11.9992
= 0.1111
(162622) = 18067.3
ΔN
3v
rms
ΔN
= 55.3
3v
rms
For every 1 million molecules traveling at v
, there are 55 molecules traveling at three times the
rms
v
.
rms
Here is another way to set up the problem using symbols:
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