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"#