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What is the internal energy of 2 mol of helium at 300 K? (R = 8.31 J/mol K)
A7.5 kJ
B3.7 kJ
C12.5 kJ
D5.0 kJ
Answer & Solution
Correct answer: A. 7.5 kJ
U = n (3/2) R T = 2 × 1.5 × 8.31 × 300, about 7.5 kJ.
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