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Work done during isothermal expansion of...

Work done during isothermal expansion of one mole of an ideal gas from 10 atm at 300 K to 1 atm (log 2 = 0.301)

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The correct Answer is:
A, C
`w=-nRT In (P_(1))/(P_(2))=(2"cal"K^(-1)"mol"^(-1))` (1 mol) (300K) In `(10/1)`
`=-1381.8` cal
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