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40 g of Argon is heated from 40⁰C to 100⁰C (R=2 cal/mole). Heat absorbed at constant volume is
Options
(a) 100 cal
(b) 80 cal
(c) 180 cal
(d) 120 cal
Correct Answer:
180 cal
Explanation:
Heat absorbed at constant volume = nCvdT
Now argon is monoatomic Cv = (3/2) R
Number of moles = 4/0/40 = 1
.·. Q = 1 × (3/2) × 2 × (100-40) = 3 × 60 = 180 cal.
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- Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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