2
$\begingroup$

Suppose we carry out a reaction in a bomb calorimeter whose starting temperature is $298.15\ \mathrm K$. Here we assume $\Delta V$ is close enough to zero that we consider the process to be at constant volume.

Based on $\Delta T$ and the calorimeter constant, we determine the internal energy change for the reaction at constant $T$ and $V$, which I'll call $\Delta U_\mathrm r$. (We can assume constant $T$ because we know how much heat flow would be required to return the calorimeter to its starting temperature.)

A typical pchem textbook will then say you can calculate what $\Delta H$ would be if the reaction were instead carried out at the same temperature, but at a constant external pressure of $1\ \mathrm{bar}$, namely $\Delta H_\mathrm r^⦵$, from the $\Delta U$ we determined in the calorimeter, as follows:

$$\Delta H_\mathrm r^⦵=\Delta U_\mathrm r^⦵+\Delta pV=\Delta U_\mathrm r^⦵+p\,\Delta V\approx\Delta U_\mathrm r^⦵+\Delta n_\text{gas}RT$$

But: we didn't determine $\Delta U$ at standard state ($1\ \mathrm{bar}$). We determined it at constant volume. Thus, strictly speaking, when we do the following calculation using the $\Delta U$ from the calorimeter

$$\Delta H_\mathrm r^⦵\approx\Delta U_\mathrm r^⦵+\Delta nRT$$

aren't we additionally making the approximation that $\Delta U_\mathrm r=\Delta U_\mathrm r^⦵$? I.e., aren't we ignoring the volume-dependence of $U$? Typically, this is a very good approximation since, for most substances, $U$ is only weakly dependent upon $V$. But it seems it's an approximation nonetheless. If so, it is one that is not (but should be) explicitly acknowledged in these textbooks.

There are other approximations, of course, such as the one shown above, that $p\Delta V\approx\Delta n_\text{gas}RT$ (ignore volume change in liquids and solids, and use ideal gas law to calculate volume change due to change in number of moles of gas), but I'm specifically wondering about this one.

$\endgroup$
  • 1
    $\begingroup$ Please note that, since 1982, the standard state pressure is $p^\circ=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ (i.e. no longer $1\ \mathrm{atm}$). $\endgroup$ – Loong Jan 26 at 10:55
0
$\begingroup$

To quote Sime's textbook on Physical Chemistry, experiment 3 on the measurement of heats of combustion of hydrocarbons using bomb calorimetry: "we will assume that $\Delta U = \Delta U_r^o$; that is, we will assume that $\Delta U$ is independent of pressure". There is no further justification of the statement.

However, based on the assumption of ideal gas behavior, then an isothermal PV change for the products will leave $\Delta U$ unchanged, so that in fact $\Delta U_r = \Delta U_r^o$. To illustrate graphically:

enter image description here

So, if the ideal gas behavior is justified (usually this just means that the pressure and density of gaseous products is not too high), then it is safe to assume that $\Delta U_r = \Delta U_r^o$.

$\endgroup$
  • 1
    $\begingroup$ I'm afraid that doesn't address my question. I'm not asking if the assumption that $\text{ΔU}_r=\text{ΔU}_r^⦵$ can be made, I'm asking if it is being made. In the case of ideal gases, U is always independent of V, so that assumption can always be made. $\endgroup$ – theorist Jan 25 at 15:41
  • $\begingroup$ @theorist I can't speak for every bomb calorimetry experiment ever made, but typical pchem books generally stick to simpler systems in which gases behave ideally, so I would hazard to guess that, yes, it is being made. $\endgroup$ – Try Hard Jan 25 at 15:56

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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