Thermodynamics and Thermochemistry - Result Question 61
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61. The total energy of one mole of an ideal monatomic gas at $27^{\circ} C$ is _______ cal.
(1984,1 M)
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Answer:
Correct Answer: 61. $(900)$
Solution:
$900$ $ cal: E=\frac{3}{2} R T=\frac{3}{2} \times 2 \times 300$ $ cal$
BETA
