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??

  • $\begingroup$ PDelta V=+150, not -150 $\endgroup$ Commented May 8, 2021 at 3:05
  • 1
    $\begingroup$ You have made a mistake in sign of work as $P\Delta V=150$. Also, first law of thermodynamics is valid everywhere, not just at constant pressure. $\endgroup$
    – V.G
    Commented May 8, 2021 at 3:12
  • $\begingroup$ @LightYagami, Can you please refer the post again, I have added the calculation for work done, to my knowledge I have learnt the formula for work done as -PextΔV. Do correct me if I am wrong. $\endgroup$
    – Srini
    Commented May 8, 2021 at 3:43
  • $\begingroup$ @Srini First you have written $-P\Delta V=-150$ which is fine. But when you did the calculation for $\Delta U$, you are subtracting -150, where 150 had to be subtracted. $\endgroup$
    – V.G
    Commented May 8, 2021 at 4:30
  • $\begingroup$ @LightYagami, Oh got it thank you so much!! $\endgroup$
    – Srini
    Commented May 8, 2021 at 5:40

1 Answer 1


Your trouble comes from the two opposed concepts used in the world to define the work $\pu{p\Delta V}$.

For theoretical scientists, the work is positive when work is done on the system, when the gas is compressed by an external force. And in such a compression, the volume decreases, so $\pu{\Delta V}$ is negative. So in order for the work to be positive, the work must be : $\pu{w = - p\Delta V}$. So that the total change of internal energy is $\pu{\Delta U = q + w}$.

For practical scientists, and specially for engineers, the work is considered positive when the gaz expands, because the gaseous system is working like a machine that must produce work when heat is given to it. In this case, the work is : $\pu{w = + p\Delta V}$. As a consequence, the internal energy is what remains in the system when heat has been given to it and some work has been produced. So here : $\pu{\Delta U = q - w}$.

Finally both concepts leads to the same final result : $\pu{\Delta U = q - p\Delta V}$

  • $\begingroup$ Thank you so much, got a clearer vision about this . $\endgroup$
    – Srini
    Commented May 10, 2021 at 9:38

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