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

1 Answer
Feb 11, 2018

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


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"