Question

2 mole of an ideal gas expanded isothermally and reversibly from 1L to 10L at 300K. What is the enthalpy change??

Answer 1

Zero. It's an ideal gas undergoing a process with no temperature change.

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The enthalpy change at constant temperature is given by

`DeltaH = int_(P_1)^(P_2) ((delH)/(delP))_TdP`

But we have shown over here already that

`DeltaH = int_(P_1)^(P_2) ((delH)/(delP))_TdP`

`= int_(P_1)^(P_2) V - T((delV)/(delT))_PdP`

For ideal gases,

`PV = nRT`, so

`V = (nRT)/P`

and

`((delV)/(delT))_P = (nR)/P`

Therefore,

`color(blue)(DeltaH) = int_(P_1)^(P_2) V - T cdot (nR)/PdP`

`= int_(P_1)^(P_2) V - VdP`

`= int_(P_1)^(P_2) 0dP`

`=` `ulcolor(blue)"0 J"`