5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 13
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16Problem set #4 solutions for 5.112, Fall 2011
Page 13 of 16
( )
2
(
)
( )
( )
v
⎡
⎤
M
2
2
=
exp −
− 4 v
v
( )
⎣ ⎢
⎦ ⎥
2
2RT
2v
( )
⎡
⎤
2
3M v
1
⎢
⎥
=
exp −
⎢
⎥
4
2RT
⎣
⎦
⎡
⎤
⎛
⎞
⎛
⎞
1
3M
8RT
=
exp
⎝ ⎜
⎠ ⎟
⎝ ⎜
⎠ ⎟
⎢
⎥
π M
⎣
⎦
4
2RT
⎡
⎤
1
12
=
exp
⎣ ⎢
⎦ ⎥
π
4
(
)
(
)
= .25000
45.5913
= 11.3978
= 11.4
The ratio of the number of molecules with speed v to that of 2v is 11.4.
(b) Calculate the ratio, to 3 significant figures, of the number of methane molecules that
have the average energy E to the number of molecules that have twice the average
energy, 2 E .
We need to look at the energy distribution equation:
3
⎛
⎞
−E
1
1
2
f (E) = 2 π
⎜
⎟
E
2
e
RT
⎝
⎠
π RT
We can substitute into the equation and look at the ratios:
3
⎛
⎞
−E
1
1
2
2 π
⎜
⎟
E
2
e
RT
⎝
π RT
⎠
f (E)
=
3
f (2E)
⎛
⎞
−2E
1
1
( )
2
2 π
⎜
⎟
2E
e
RT
2
⎝
π RT
⎠
1
⎛
⎞
−E
⎛
⎞
2E
+
E
⎜
⎟
2
=
⎝
⎠
⎜
⎟
RT
RT
e
⎝
⎠
2E
1
⎛
⎞
E
1
2
=
⎜
⎟
e
RT
⎝
⎠
2
1
⎛
⎞
⎛
⎞
⎛
⎞
1
3RT
⎜
⎟
⎜
⎟
1
2
⎝
⎠
⎝
⎠
=
⎜
⎟
RT
2
e
⎝
⎠
2
1
⎛
⎞
3
1
2
=
⎜
⎟
e
2
⎝
⎠
2
(
)
(
)
= 0.70711
4.48169
= 3.1690
= 3.17
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