2.5 g of helium at an initial temperature of 350 K interacts thermally with 7.6
ID: 986596 • Letter: 2
Question
2.5 g of helium at an initial temperature of 350 K interacts thermally with 7.6 g of oxygen at an initial temperature of 600 K
a) What is the initial thermal energy of each gas?
b) What is the final thermal energy of each gas?
c, d, e, in picture below.
Part c How much heat energy s transtered, and in which diection?n How much heat energy is transferred, and in which direction? Etrans Submit My Answers Give Up Part D From He to O2 O From 02 to He From 02 to He Submit My Answers Give Up Part E What is the final temperature? Express your answer using three significant figures. 2Explanation / Answer
2.5 g of helium at an initial temperature of 350 K interacts thermally with 7.6 g of oxygen at an initial temperature of 600 K
a)
moles of He = 2.5 / 4 = 0.625
moles of O2 = 7.6 / 32 = 0.2375
initial thermala energy of He U = n Cv dT
= 0.625 x (3/2) R x T
= 0.625 x 1.5 x 8.314 x 350
initial thermala energy of He U = 2728 J
initial thermala energy of O2 U = 0.2375 x (5/2) x 8.314 x 600
initial thermala energy of O2 U = 2962 J
b) final thermal energy of each gas
U total = 2962 + 2728 = 5690 J
0.625 x (3/2) R x T + 0.2375 x x (5/2) R x T = 5690
T = 447 K
the new enrgies are
final thermal energy of He = 0.625 x (3/2) R x T = 0.625 x (3/2) x 8.314 x 447 = 3484 J
final thermal energy of He = 0.2375 x (5/2) R x 447 = 2206.6 J
c)
heat energy transformed = Q = delta U He = 3484 - 2728
= 756 J
756 J heat transfered from the Oxygen to helium
d) from O2 to He
e) 447 K
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