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"