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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 $

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  • $\begingroup$ 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
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    $\begingroup$ No it's not. That would give it the wrong units compared to the other terms. $\endgroup$ Dec 20, 2018 at 20:08
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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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

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

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