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))
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))
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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#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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3bf8d9f-79bd-47af-b660-4bc5b007f58d%2Fc0a5fbbe-2e2c-488c-97e3-6bf0bb67c6c2%2Ftamukf_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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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