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Question

Calculate the maximum work done in kJ expanding 16 g of oxygen at 300 K and occupying a volume of 5 dm3 isothermally until the volume becomes 25 dm3. (Ignoring the sign of value)
  1. 2
  2. 0
  3. 8
  4. 5


Solution

The correct option is A 2
Reversible work is maximum work
As the process is isothermal ,So Work
w=2.303 nRT log10(V2V1)
V1=initial volume=5 dm3
V2=final volume=25 dm3
moles of gas =1632=0.5

Putting these values in the formula above, we get
w=2.303×0.5×8.314×300 log 255=2007J

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