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5 mole of an ideal gas change it's volume from 10 L to 20 Lit. reversible isothermally at 300 K then work done by gas is.......KCal. [Given R = 2 Cal] |
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Answer» Correct answer is (2) For isothermal reversible process W = -2.303 nRT log (v2/v1) = -2.303 x 5 x 2 x 300 log (20/10) = -2072.7 cal = -20727 x 103 cal = -2 K Cal = |W| = 2 K cal |
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