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.
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 

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