In most, if not all, general chemistry books, you will find that in from constant volume calorimeters, $\Delta H = \Delta U + RT \Delta n_\mathrm{g}$, which is, of course derived from $H = U + PV$.
My question is, given that the measured quantity is the change of temperature of the calorimeter, how can we assume that $T$ is constant to get to $\Delta H = \Delta U + RT\Delta n_\mathrm{g}$?
Edit:
As said in the comments, I found this link where it says that this approximation is valid for a small $\Delta T$ value. But if we have the values, why not just use them to get a more accurate number? Furthermore, for example, let's assume an experiment we measure a $\pu{10 kJ mol-1}$ heat from a $T_\mathrm{i} = \pu{298 K}$ to $T_\mathrm{f} = \pu{303 K}$, and a $n_\mathrm{i} = 0$, $n_\mathrm{f} = 2$. Then the approximation gives:
$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$
$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(2)(\pu{303 K}) - (0)(\pu 298 K)\right] = \pu{-4.96 kJ mol-1}$$
And the error is, indeed, small ($\pu{80 J}$) (side question, isn't $\Delta n$ supposed to be in moles? my dimensional analysis doesn't match if it does). However, if the reaction was from $n_\mathrm{i} = 2$ to $n_\mathrm{f} = 4$, and we keep the other values, we get:
$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$
$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(4)(\pu{303 K}) - (2)(\pu 298 K)\right] = \pu{-4.88 kJ mol-1}$$
And the error is, yes, still small but it doubled (of course). So I'm guessing the $\Delta T \approx 0$ to be valid only for small values of $\Delta T$, but also dependent on the actual values for $n_\mathrm{i}$ and $n_\mathrm{f}$. Is there a rule I'm missing?