A) What is a is the initial thermal energy of each gas? B) What is the final the

Question

A) What is a is the initial thermal energy of each gas? B) What is the final thermal energy of each gas? C) How much heat energy is transferred, and in which direction? D) From He to O2 or O2 to He E) What is the final temperature? 2.0 g of helium at an initial temperature of 300 K interacts thermally with 8.0 g of oxygen at an initial temperature of 600. What is a is the initial thermal energy of each gas? What is the final thermal energy of each gas? How much heat energy is transferred, and in which direction? From He to O2 or O2 to He What is the final temperature?

Explanation / Answer

A)

He is monoatomic. Hence thermal energy = (3/2)*NkT

O2 is diatomic. Hence thermal energy = (5/2)*NkT

Boltzman constant k = 1.38*10^-23 J/K

Avagadro number Na = 6.022*10^23 /mol

For He, molecular weight = 4

For O2, molecular weight = 32

2 gm He contains (2/4)*6.022*10^23 = 3.011*10^23 moles

8 gm O2contains (8/32)*6.022*10^23 = 1.5055*10^23 moles

Initial thermal energy of He = (3/2)*(3.011*10^23)*(1.38*10^-23)*300 = 1869.8 J

Initial thermal energy of O2 = (5/2)*(1.5055*10^23)*(1.38*10^-23)*600 = 3116.4 J

B)

Initial total energy = Final total energy

1869.8 + 3116.4 = (3/2)*NkT + (5/2)*NkT

4986.2 = (1/2)*(1.38*10^-23)*T*[3*(3.011*10^23) + 5*(1.5055*10^23)]

Solving this, T = 436.4 K

Thus, final energy of He = (3/2)*(3.011*10^23)*(1.38*10^-23)*436.4 = 2719.9 J

Final thermal energy of O2 = (5/2)*(1.5055*10^23)*(1.38*10^-23)*436.4 = 2266.7 J

C)

Heat energy transfer from O2 = 3116.4 - 2266.7 = 849.7 J

Direction is from O2 to He.

D)

From He to O2 = -849.7 J

From O2 to He = 849.7 J

E)

Final temperature = 436.4 K (as found in part b)

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