#69, bottom question...please help! 23 x 10 Jd. 5.16 x 104 Ja. 1.2 x 10-20 Jb. 9.14 x 10-21 Jc. 5.71 x 10-21 Jd. 1.54 x 10-20 J65. What is the average translational kinetic energy per oxygen molecule in this sample?(9) 6.022 X 10²) 5 (0.5x1.38x10 23 x 276k) = 51607KE-3 KT-3 (1.38×10 23³ x 276 ) = 5.713×10 ²166. Is this kinetic energy the same or different from the nitrogen molecules in the sample?a. Differentb. The sameb. 1.3 x 10³ m/sc. 464 m/sd. 1.11 x 10³ m/s67. What is the rms speed of the oxygen molecules in this sample? (the atomic mass number of oxygen is15.994, 1 u = 1.6605×10-27 kg)Vrms - √√31/ Ta. 835 m/s17m263(1.38X10-23) 276k5.13 x10-20kg68. Is this rms speed the same or different from the nitrogen molecules in the sample?a. Different, because the oxygen molecules have a lower rms speed because they are more massive.b. The same, because they have the same average translation KE.3un69. What is the total internal energy of 14 moles of a monatomic ideal gas at a pressure of 2.4304 atm andvolume of 0.14416 m³? (R-8.314 J/(mol-K))a. 8.52 x 104 Jb. 1.38 x 105 Jc. 1.07 x 105 Jd. 5.33 x 104 J769101Tm168.9truturnMd[258]mendeleviumMN ₂ = 4.65×10-26kgColor ClloidetalMay 10, 2023, 10:56 AM

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What is the total internal energy of 14 moles of a monatomic ideal gas at a pressure of 2.4304 atm and volume of 0.14416 m³? (R-8.314 J/(mol-K))

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter21: The Kinetic Theory Of Gases

Section: Chapter Questions

Problem 21.48AP

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Question

#69, bottom question...please help!

23 x 10 J
d. 5.16 x 104 J
a. 1.2 x 10-20 J
b. 9.14 x 10-21 J
c. 5.71 x 10-21 J
d. 1.54 x 10-20 J
65. What is the average translational kinetic energy per oxygen molecule in this sample?
(9) 6.022 X 10²) 5 (0.5x1.38x10 23 x 276k) = 51607
KE-3 KT-3 (1.38×10 23³ x 276 ) = 5.713×10 ²1
66. Is this kinetic energy the same or different from the nitrogen molecules in the sample?
a. Different
b. The same
b. 1.3 x 10³ m/s
c. 464 m/s
d. 1.11 x 10³ m/s
67. What is the rms speed of the oxygen molecules in this sample? (the atomic mass number of oxygen is
15.994, 1 u = 1.6605×10-27 kg)
Vrms - √√31/ T
a. 835 m/s
17
m
26
3(1.38X10-23) 276k
5.13 x10-20kg
68. Is this rms speed the same or different from the nitrogen molecules in the sample?
a. Different, because the oxygen molecules have a lower rms speed because they are more massive.
b. The same, because they have the same average translation KE.
3
un
69. What is the total internal energy of 14 moles of a monatomic ideal gas at a pressure of 2.4304 atm and
volume of 0.14416 m³? (R-8.314 J/(mol-K))
a. 8.52 x 104 J
b. 1.38 x 105 J
c. 1.07 x 105 J
d. 5.33 x 104 J
7
69
101
Tm
168.9
truturn
Md
[258]
mendelevium
MN ₂ = 4.65×10-26
kg
Color C
lloid
etal
May 10, 2023, 10:56 AM

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