- #1
Steveku
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May you professors please check part b of my work because my answer does not match the answer in the back of the book. Thank you.
Question:
(a) What is the total random kinetic energy of all the molecules
in 1 mol of hydrogen at a temperature of 300 K? (b) With what
speed would a mole of hydrogen have to move so that the kinetic
energy of the mass as a whole would be equal to the total random
kinetic energy of its molecules?
Answer to part (a):
Since KE (total) = 1.5nRT, the total random kinetic energy is (1.5)(1)(8.314)(300) =
3738.42 J/mol
Answer to part (b):
The atomic weight of hydrogen is 1.01 grams per mole
So, the mass of one mole is 0.001 kg.
We have to find the velocity
By the kinetic energy formula,
KE = (.5)(m)(v)^2
(.5)(m)(v)^2 = 3738.42 J/mol
So, v = 2734.3811 m/s
However, the answer page in my booklet tells me that the correct answer for part b is 1.93 km/s. Please advise me and tell me what I did wrong. Thanks again.
Steve
Question:
(a) What is the total random kinetic energy of all the molecules
in 1 mol of hydrogen at a temperature of 300 K? (b) With what
speed would a mole of hydrogen have to move so that the kinetic
energy of the mass as a whole would be equal to the total random
kinetic energy of its molecules?
Answer to part (a):
Since KE (total) = 1.5nRT, the total random kinetic energy is (1.5)(1)(8.314)(300) =
3738.42 J/mol
Answer to part (b):
The atomic weight of hydrogen is 1.01 grams per mole
So, the mass of one mole is 0.001 kg.
We have to find the velocity
By the kinetic energy formula,
KE = (.5)(m)(v)^2
(.5)(m)(v)^2 = 3738.42 J/mol
So, v = 2734.3811 m/s
However, the answer page in my booklet tells me that the correct answer for part b is 1.93 km/s. Please advise me and tell me what I did wrong. Thanks again.
Steve