Question

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.​

Answer 1

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 logV1​V2​​

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