Page 211 - Theory and Problems of BEGINNING CHEMISTRY

200 KINETIC MOLECULAR THEORY [CHAP. 13

13.15. (a) Calculate the “average” velocity of O 2 molecules at 1.00 atm and 300 K. (b) Does the pressure matter? (c) Does
the identity of the gas matter?
2
Ans. (a) 1 mv = 6.21 × 10 −21 J (from the prior problem)
2
1g 1kg

= 32.0 amu = 5.32 × 10 −26 kg
m O 2 23
6.02 × 10 amu 1000 g
5
2
v = 2.33 × 10 (m/s) 2
v rms = 483 m/s (about 0.30 miles/s)
The square root of the average of the square of the velocity, v rms , is not the average velocity, but a
quantity called the root mean square velocity.
(b) The pressure of the gas does not matter.
(c) The identity of the gas is important, because the mass of the molecule is included in the calculation.
(Contrast this conclusion with that of the prior problem.)
13.16. Explain why neon atoms obey the gas laws the same as nitrogen molecules.

Ans. The gas laws work for unbonded atoms as well as for multiatom molecules.
13.17. (a) Calculate the square of each of the following numbers: 1, 2, 3, 4, and 5. (b) Calculate the average of the numbers.
(c) Calculate the average of the squares. (d) Is the square root of the average of the squares equal to the average of the
numbers? (e) Explain why quotation marks are used with “average” velocity in the text for the velocity of molecules
with average kinetic energies.

Ans. (a)
Number Square

1 1
2 4
3 9
4 16
5 25
(b)3
(c)11
√
(d) The square root of the average of the squares ( 11) is not equal to the average of the numbers (3).
(e) The velocity equal to the square root of the quotient of twice the average kinetic energy divided by the
2
molecular mass, is not really the average velocity. That is, the square root of v does not give v, but the
root mean square velocity.

13.18. Contrast the motions of the molecules of a sample of gas at rest to those in a hurricane wind.
Ans. At rest, as many molecules are traveling on average in any given direction as in the opposite direction, and
at the same average speeds. In a hurricane, the molecules on average are traveling somewhat faster in the
direction of the wind than in the opposite direction.

13.19. Oxygen gas and sulfur dioxide gas are at the same temperature. What is the ratio of the “average” velocities of their
molecules?
Ans. Since the temperatures are the same, so are the average kinetic energies of their molecules. From Graham’s
law,

v 1 MM 2 64.06
= = = 1.415
v 2 MM 1 32.00
The oxygen molecules are moving, on average, 1.415 times as fast as the SO 2 molecules.