5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 5
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16Problem set #4 solutions for 5.112, Fall 2011
Page 5 of 16
⎛
⎞
⎛
⎞
18
1.5 J
1 photon
4.30 x 10
photons
=
⎝ ⎜
⎠ ⎟
x
⎝ ⎜
⎠ ⎟
-19
1 s
3.49 x 10
J
second
As developed in Section 4-13, the total force equals the change in momentum per collision
multiplied by the number of collisions per second:
(
)
(
)
-27
-1
18
-1
-8
-2
= 1.00 x 10
F= 2.33 x 10
kg m s
x 4.30 x 10
s
kg m s
Pressure is defined as force divided by area. The area of the circular wall is πr
2
where r is the
radius. Hence:
-8
-2
F
F
1.00 x 10
kg m s
P =
=
=
(
)
2
2
πr
A
-3
π 0.15 x 10
m
= 0.1415 Pa
= 0.14 Pa (correct sig figs)
So the pressure exerted by the light on the surface is 0.21 Pa.
Question 8 (out of 6 points)
In an optical atomic trap, beams of tightly focused laser light replace the physical walls of
conventional containers. They briefly (for 0.500 s) exert enough pressure to confine 695
–15
3
sodium atoms in a volume of 1.0 X 10
m
. The temperature of this gas is 0.00024 K, the
lowest temperature ever reached for a gas.
a) Compute the root-mean-square speed of the atoms in this confinement.
3RT
=
The root-mean-square speed of the molecules of a gas is v
rms
M
-1
-1
-1
where M is the molar mass of the gas. We substitute M = 0.02299 kg mol
, R = 8.315 J mol
K
,
and T = 0.00024 K to compute
(
)
(
)
−4
-1
-1
3 8.315 J mol
K
2.4 x 10
K
3RT
=
=
v
rms
-1
M
0.02299 kg mol
= 0.5103 m s
-1
= 0.51 m s
-1
-1
v
= 0.51 m s
.
rms
-1
-1
Common errors here are to use M in g mol
instead of kg mol
or to use R in the wrong units.
b) Assume ideal gas behavior to compute the pressure exerted on the “walls” of the optical
bottle in this experiment.
Since we are making the assumption that the Na atoms are acting as an ideal gas, we can use the
equation PV = nRT. We have to first determine V. The volume of the trap is expressed in liters (1
-3
3
-15
3
L = 10
m
) and we can convert the volume of 1.0 x 10
m
to L;
The amount of its contents is expressed in moles
695 atoms
= 1.1541 x 10
-21
n=
mol
23
-1
6.022 x 10
atoms mol
Then, the problem becomes a substitution into PV = nRT;
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