Calculate $∆H$ for $1$ mole of real gas undergoing change of state from $\pu{10^5 Pa}$, $\pu{300 K}$ to $\pu{2e5 Pa}$, $\pu{600 K}$. The values of real gas constants are $a = 0$ and $b = \pu{50 mL/mol}$.

Given: $C_p = 20$ (in SI unit), $$ \left(\frac{∂H}{∂P}\right)_T = T\left(\frac{∂V}{∂P}\right)_T + V$$

My approach has been like this:

First, I wrote the real gas equation. Then I partially differentiated this whole equation w.r.t. $P$ keeping temperature constant so as to find $∂V/∂P$ and substituted the value of $∂V/∂P$ in the equation given in the question. My actual aim was to use the formula

$$\mathrm{d}H = nC_p\mathrm{d}t + \left(\frac{∂H}{∂p}\right)\mathrm{d}p.$$

Next, I substituted $(∂H/∂p)$ from the given equation in the the given formula, but now I am stuck here.

The given equation in question is incorrect (even dimensionally!) as stated in comment.

Real equation: $ \left(\frac{∂H}{∂P}\right)_T = V -T\left(\frac{∂V}{∂T}\right)_P $

  • $\begingroup$ Shouldn't there be a minus sign in from of the first term in your equation for partial of H with respect to P? $\endgroup$ Dec 20, 2018 at 15:23
  • 1
    $\begingroup$ No it's not. That would give it the wrong units compared to the other terms. $\endgroup$ Dec 20, 2018 at 20:08
  • 2
    $\begingroup$ I didn't guess anything. I've worked with that equation many times before, and presented its derivation in several threads on various forums. The equation is derived in every thermodynamics book that I am familiar with. $\endgroup$ Dec 20, 2018 at 20:29
  • 1
    $\begingroup$ Okay now I see where all the trouble began from. It was the question that was wrong. I worked with the corrected equation and easily reached to the solution.It was the incorrect equation that was making all the mess. Working with the incorrect equation I got terms of temperature and while I knew I had to put the limits on pressure, I was totally clueless about which value of temperature to use. I was stuck here for 2 days. Thank u so very much. I think I should delete the question now. (Or should I correct the equation and leave it?) $\endgroup$
    – Mathomania
    Dec 20, 2018 at 20:36
  • 1
    $\begingroup$ Bdw how would have we dealt if the terms of temperature didn't disappear?I mean which value of temperature should be used, 300K or 600K $\endgroup$
    – Mathomania
    Dec 20, 2018 at 20:42

1 Answer 1


You started out correctly by getting the relationship for $\mathrm{d}H$ for your van der Waals gas algebraically. The usual approach to evaluating a change like this for a real gas is to break the change into three steps (since you will typically know $C_p$ only in the limit of ideal gas behavior at low pressures):

Step 1: Calculate $\Delta H$ for the change from $\pu{10^5 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ at $\pu{300 K}$.

Step 2: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ and $\pu{600 K}$ (i.e., using the ideal gas heat capacity).

Step 3: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{600 K}$ to $\pu{2e5 Pa}$ and $\pu{600 K}$.

Based on Hess' law, add the three changes together.

For this particular problem, the 3 step procedure is not necessary because of the following:

$$\left(\frac{\partial H}{\partial P}\right)_T=b$$


$$\mathrm{d}H = C_p\mathrm{d}T + b\mathrm{d}P$$

Since $b$ is independent of temperature, $C_p$ is independent of pressure. So for this system, the equation integrates immediately to

$$\Delta H = C_p\Delta T + b\Delta P$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.