# Calculating enthalpy change for a real gas

I know this question would be closed citing it to be a homework question. I posted it earlier also. But believe me I have tried it for myself before and that too many times but then couldn't arrive at a solution maybe due to lack of my conceptual understanding. Any help would greatly be appreciated.

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 $$b = \pu{50 mL/mol}$$.

Besides, we have been given $$C_p = 20$$ (in SI unit) and the relation

$$\frac{∂H}{∂P} = T\frac{∂V}{∂P} + V$$

(the partial derivative has been taken keeping the temperature constant on both sides of equality.)

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.

• Shouldn't there be a minus sign in from of the first term in your equation for partial of H with respect to P? – Chet Miller Dec 20 '18 at 15:23
• The equation should read $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$ – Chet Miller Dec 20 '18 at 16:16
• It is ∂V/∂P and not ∂V/∂T – Mathomania Dec 20 '18 at 19:58
• 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. – Chet Miller Dec 20 '18 at 20:29
• 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?) – Mathomania Dec 20 '18 at 20:36

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:

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

So,

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

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

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