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.

  • $\begingroup$ I figured out the answer to my question. Essentially, the approximation is not significant because it is smaller in magnitude than other approximations that are being made. I wrote up the details for my class, but since misplaced the notes. If I find them, or decide to reconstruct them, I'll post it here. $\endgroup$
    – theorist
    Mar 23, 2021 at 2:58

1 Answer 1


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. The Parr Instrument Company provides an introductory guide to bomb calorimetry (document 483M) which has the following to say:

Users interested in high precision calorimetry will understand that test results obtained in a bomb calorimeter express calorific values obtained at constant volume. This differs from the usual conditions under which fuels are burned at constant pressure rather than constant volume. If the sample contains elements which normally would increase in volume after combustion, a correction must be applied to account for the heat equivalent of this work. This value is so small as to be negligible in routine work, and it usually is disregarded.

The required pressure correction can be computed from thermal expansion and isothermal compressibility coefficients. For instance, for water one estimates that if the bomb pressure rises by $\pu{2 GPa}$ over standard pressure then the energy will increase by $\approx \pu{80 J/mol}$. Compare this correction with the heat of combustion of benzoic acid ($\approx \pu{3221 kJ/mol}$) and (assuming the coefficients for benzoic acid are of the same order of magnitude) it's clear that for condensed substances the correction is (often) negligible.

For gases, based on the assumption of ideal gas behavior, an isothermal PV change for the products will leave $\Delta U$ unchanged, so that in fact $\Delta U_r = \Delta U_r^o$.

  • 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, 2019 at 15:41
  • $\begingroup$ @theorist I looked my answer over and don't fully understand how this fails to answer your question. The comment you left promising a future answer doesn't seem to go far beyond what I have stated. $\endgroup$
    – Buck Thorn
    Jul 21, 2021 at 13:01
  • 1
    $\begingroup$ An answer requires doing a derivation that formally demonstrates why the correction is negligible on a general basis, relative to other assumptions that are being made. That's what I did for my class. $\endgroup$
    – theorist
    Jul 21, 2021 at 16:39

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