Problem 20.52 Part A 2.2 g of helium at an initial temperature of 350 K interact
ID: 1655841 • Letter: P
Question
Problem 20.52 Part A 2.2 g of helium at an initial temperature of 350 K interacts thermally with 8.4 g of oxygen at an initial temperature of 610 K What is the initial thermal energy of each gas? Express your answer using two significant figures. Enter your answers numerically separated by a comma. E, (He)E, (02)- Submit My Answers Give Up Part B What is the final thermal energy of each gas? Express your answer using two significant figures. Enter your answers numerically separated by a comma. Submit My Answers Give Up Part C How much heat energy is transferred, and in which direction? trans =Explanation / Answer
(A)
weight of helium (He) = 2.2g
Number of Moles n = (2.2 / 4) * 6.022*10^23 = 3.312*10^23 moles
Weight of oxygen(O2) = 8.4 g
Number of moles n = (8.4 / 32) * 6.022*10^23 = 1.58*10^23 moles
lnitial thermal energy for He (monoatomic),
E(He) = (3/2)nkT = (3/2) * (3.312*10^23) * (1.38*10^(-23)) * 350
E(He) = 2399.61 J
lnitial thermal energy for O2 (diatomic),
E(O2) = (5/2)nkT = (5/2) * (1.58*10^23) * (1.38*10^(-23)) * 610
E(O2) = 3325.11 J
(B)
Initial total energy = Final total energy
2399.61 + 3325.11 = (3/2) * nkT + (5/2) * nkT
5724.72 = 1.38*10^(-23) * T * [3/2 * (3.312*10^23) + 5/2 * (1.58*10^23)]
Final temperature T = 465.15 K
Final energy of He,
Ef = (3/2) * (3.312*10^23) * (1.38*10^(-23)) * 465.15
Ef (He) = 3189.14 J
Final energy of O2,
Ef = (5/2) * (1.58*10^23) * (1.38*10^(-23)) * 465.15
Ef (O2) = 2535.57 J
(C)
Direction of heat transfer is from O2 to He.
Heat transfer,
Q = 3325.11 - 2535.57
Q = 789.53 J
(D)
From He to O2 = -789.53J
From O2 to He = 789.53 J
(E)
Final temperature,
T = 465.15 K
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