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"#