derive the relation between change in internal energy and change in enthalpy for a system in which the reactants and products are gases
Dear student
Please find the solution to the asked query:
From First law of Thermodynamics we have:
ΔU = q +w
ΔU is the change in internal energy, q is the heat transfered, w is the work done
If a change is brought about at constant pressure , in that case volume changes . Let volume change from VA to VB at constant pressure P. The work done by the system is given by:
w = -P (VB- VA)
Substituting this value in the above equation:
ΔU = q - P (VB- VA)
or UB -UA = q - P (VB- VA)
(UB +PVB) - (U A +PVA) =q
The quantity , U +PV is called the enthalpy of the system and is denoted by H
Thus [ HB -HA = ΔH = q] and
H = U +PV
Hope this information will clear your doubts regarding the topic. If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible.
Regards
Please find the solution to the asked query:
From First law of Thermodynamics we have:
ΔU = q +w
ΔU is the change in internal energy, q is the heat transfered, w is the work done
If a change is brought about at constant pressure , in that case volume changes . Let volume change from VA to VB at constant pressure P. The work done by the system is given by:
w = -P (VB- VA)
Substituting this value in the above equation:
ΔU = q - P (VB- VA)
or UB -UA = q - P (VB- VA)
(UB +PVB) - (U A +PVA) =q
The quantity , U +PV is called the enthalpy of the system and is denoted by H
Thus [ HB -HA = ΔH = q] and
H = U +PV
Hope this information will clear your doubts regarding the topic. If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible.
Regards