calculate the maximum work done in expanding 16 g of oxygen at 300K and occupying a volume of 5 dm^3 isothermally until the volume becomes 25 dm^3.
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Answer: -2.01 x 10(3)
Explanation: Maximum work is obtained in isothermal reversible process.
The expression for the maximum work is as follows:
w=−2.303 NRT logV1V2
The number of moles of oxygen is NO2=Molecular weight Weight=3216=0.5
The temperature is T =300 K
The initial volume is V1 =5 dm3
The final volume is V2 =25 dm3
Substituting values in the above reaction, we get
w=−2.303×3216×8.314×300×log525=−2.01×103 J
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