5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 8
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16Problem set #4 solutions for 5.112, Fall 2011
Page 8 of 16
(c) Calculate the temperature that corresponds to your answer for (b).
1
⎛
⎞
3RT
2
From our lectures #12 notes, we know that: v
=
⎝ ⎜
⎠ ⎟
rms
M
For the equation:
• V
-1
that we calculated in part (b) = 1.8 m s
rms
• R which is the universal gas constant = 8.31447 J K
-1
-1
mol
(Use this value because a Joule is a
-2
Newton.meter, and a Newton = m kg s
which once you cancel units gives:
1
⎛
⎞
3RT
2
⎝ ⎜
⎠ ⎟
v
=
rms
M
1
(
)
( )
⎛
⎞
−1
2
-2
-1
m
kg s
K
mol
K
2
⎜
⎟
=
-1
⎝
⎠
kg mol
1
⎛
⎞
2
m
2
=
⎝ ⎜
⎠ ⎟
2
s
• M which is the molar mass in kg mol
-1
-3
-1
= 85.47 x 10
kg mol
• T is the unknown:
Now we can solve for T:
1
⎛
⎞
3RT
2
⎝ ⎜
⎠ ⎟
v
=
rms
M
1
⎛
⎞
-1
-1
3(8.31447 J K
mol
T
2
-1
1.8 m s
=
⎝ ⎜
⎠ ⎟
-3
85.47 x 10
kg
⎛
⎞
-1
-1
3(8.31447 J K
mol
T
=
2
-2
3.24 m
s
⎝ ⎜
⎠ ⎟
-3
85.47 x 10
kg
T=0.011 K
Question 13 (out of 12 points: 3 for each speed)
Consider Ne gas at T=298 K. For every million molecules traveling at the root mean square
speed, approximately how many travel at 0.1 v rms ; at 1.2 v rms ; at 2 v rms ; at 3 v rms .
This is a problem looking at the Maxwell-Boltzmann speed distribution function;
3
⎛
⎞
⎛
⎞
ΔN
− Mv
2
M
( )
( )
2
( )
⎜ ⎜
⎟ ⎟
= f v
Δv, where f v
= 4 π
⎜
⎟
2
v
exp
⎝
⎠
2 π RT
⎝
⎠
N
2RT
-3
So we can compare the functions for the different speeds remembering that m is 20.180 x 10
kg
-1
-1
-1
mol
, R is 8.315 J mol
K
, T is 298 K and v is going to be v
.
rms
We have to first determine v
rms
-1
-1
3RT
3(8.31451 J mol
K
)(298 K)
-1
=
=
= 606.91 m s
v
rms
-3
M
20.180 x 10
kg
Substituting the first ratios;
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