Here, ΔH = Enthalpy change, ΔU = Change in internal energy, P = pressure and V = volume.
I know that ΔH(Enthalpy) is heat given/taken to/from system at constant pressure. But I have been practicing questions recently on this topic and came to this confusion.
Let me first put out the question I solved:
"One mole of an ideal gas (Cv = 3/2*R) is heated at constant pressure reversibly at 1 atmosphere from 25°C to 100°C. Calculate ΔU and ΔH, Take R =2 cal/K.mol ".
I proceeded to solve the question, Since
ΔH = nCpΔT = 1 * 5/2 * R * 75 = 375 cal.(Correct answer) and
ΔU = nCvΔT = 1 * 3/2 * R * 75 = 225 cal.(Correct answer)
Now I found out the work done in the process = -PΔV = -nR(T2 - T1) = -150 cal.
Calculation for work done is as follows:
W = -Pext(V2-V1) = -P[(nRT2/P)-(nRT1/P)] = -nR(T2-T1) = -12(373-298) = -150 cal.
But if I try to find out the ΔU using the equation ΔH = ΔU +Δ(PV), ΔU = 375 - (-150) = 525 cal.(Wrong)
But if I use first law of thermodynamics, ΔU = q + w, ΔU = 375 + (-150) = 225 cal, which is the correct answer.
So, my question is why should I use first law of thermodynamics here even though the process is taking place at constant pressure??