5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 6
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16Problem set #4 solutions for 5.112, Fall 2011
Page 6 of 16
(
)
(
)
−21
−4
-1
-1
1.15341 × 10
2.4 × 10
mol
0.08206 L atm mol
K
K
−14
P =
= 2.273 × 10
atm
−12
1.0 × 10
L
-14
The atoms exert a pressure of 2.3 x 10
atm.
Question 9 (out of 2 points)
Equal numbers of atoms of Helium-4 and Xenon-129 exert the same pressure at constant
volume and temperature, despite the large difference in the masses. Based on kinetic theory,
explain what this tells you about the velocities of the atoms.
Since pressure is defined as the force exerted on a unit area of the walls of a container by atoms of
the gas, an average helium atom must exert the same force as an average xenon atom. According
to the definition of an ideal gas, these collisions must be elastic. Since force exerted in an elastic
collision by an impact is proportional to the momentum of the atom, helium atoms must move
much faster than xenon atoms to exert the same force.
Question 10 (out of 6 points)
The “escape velocity” necessary for objects to leave the gravitational field of the earth is 11.2
km/s. Calculate the ratio of the escape velocity to the root-mean-square speed of helium,
argon, and xenon atoms at 2200 K. Does your result help explain the low abundance of the
light gas helium in the atmosphere? Explain.
To calculate the rms speeds of helium, argon, and xenon atoms at 2000 K (1 significant figure!),
-1
we substitute the molar masses in kg mol
of the three gases successively into the expression
-1
-1
, taking T = 2000 K and R = 8.315 J mol
K
.
For He
(
)
(
)
-1
-1
3 8.315 J mol
K
2200 K
=
= 3.703 x 10
3
-1
v
m s
(
)
−3
rms
-1
4.0026 x 10
kg mol
= 3.7 x 10
3
-1
v
m s
(correct sig figs)
rms
3
-1
escape velocity
3.703 x 10
m s
=
Ratio of
3
-1
root-mean square velocity
11.2 x 10
m s
= 0.33061
= 0.33
For Ar
(
)
(
)
-1
-1
3 8.315 J mol
K
2200 K
=
= 1.1172 x 10
3
-1
v
m s
(
)
rms
−3
-1
39.948 x 10
kg mol
= 1.1 x 10
3
-1
v
m s
(correct sig figs)
rms
3
-1
escape velocity
1.12 x 10
m s
(
)
=
Ratio of
100
3
-1
root-mean square velocity
11.2 x 10
m s
= 0.10465
=0.10 (correct sig figs)
For Xe
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