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O.75 mole isothermally at 300 k from 15 l to 25 l the maximum work is

O.75 mole isothermally at 300 k from 15 l to 25 l the maximum work is

Grade:11

1 Answers

Arun
25750 Points
6 years ago
Dear student
 
w =  –n R T ln (V2/V1)
 
w = – 2.303 * nRT log (V2/V1)
 
w = –2.303 * 0.75 * 8.314 * 300 log (25/15)
 
on solving
w = –955.5 Joules
 
Regards
Arun (askIITians forum expert)

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