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When 5 moles of He gas expand isothermally and reversibly at 300   K from 10 litre to 20 litre, the magnitude of the maximum work obtained is J . [nearest integer ] (Given : R = 8 . 3   J   K - 1   mol - 1 and log 2 = 0 . 3010 )

Correct answer

0

Step-by-step solution

Work done in isothermal reversible process is calculated by the equation W = - 2 . 303 nRTlog V 2 V 1 n = Number   of   moles T = Temperature = 300 K V 2 = Final   volume = 20   litres V 1 = Initial   volume = 10   litres W = - 2 . 303 × 5 × 8 . 3 × 300 log 20 10 = - 8630 . 38   J

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