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In a bomb calorimeter, the reaction occurs at a constant volume, hence we say that the heat absorbed by the water in the surroundings is equal to the change in internal energy for the reaction. But, problems in textbooks involve the reaction inside the calorimeter to be occuring at constant temperature, but for an ideal gas, internal energy depends only on temperature, so if temperature remains constant, what actually causes the internal energy change?

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    $\begingroup$ Try to pay more attention to the question formulation. While the topic may be confusing to you, the question itself, including your own prior analysis, should be clear. $\endgroup$
    – Poutnik
    Aug 29, 2023 at 17:33

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The total differential for the inner energy is:

$$ \mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V + \mu \mathrm{d} n$$

In a chemical reaction, the chemical amount $n$ for each species changes (and the chemical potential is not the same for each species). So you expect a change in inner energy (even ignoring the $T \mathrm{d} S$ term, which is discussed further here).

Another way of saying this is that you break bonds and make bonds (and they don't have the same strength, and you might have more or fewer bonds after the reaction), turning one ideal gas into another, and that changes the inner energy.

Take the simplest example:

$$\ce{2H(g) -> H2(g)}$$

The number of particles changes, you have bonds in the product and none in the reactants - surely the inner energy is allowed to change.

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  • $\begingroup$ If I for example consider the combustion of 1mol liquid oxalic acid to give carbon dioxide gas and liquid water at 27°C in a bomb calorimeter, if I subtract the values of the absolute value of the internal energy of the products from reactants, by using the formula U=f/2nRT for finding absolute value of internal energy, I find that the internal energy change for the reaction is +ve, but it is exothermic reaction so, Q is-ve, how is this consistent with the 1st law of thermodynamics? $\endgroup$ Aug 29, 2023 at 16:34
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    $\begingroup$ That formula just describes the translational energy of the gas. This is correct for mono-atomic ideal gases. Here, all your reactants and products have bonds, and most of them are not even in the gas state, so you are missing a lot of "types of energies" in your formula. $\endgroup$
    – Karsten
    Aug 29, 2023 at 16:40
  • $\begingroup$ @Kartsen , ok I understood that, can you also tell why when finding the enthalpy change for a reaction occuring in a bomb calorimeter at constant temperature, in the formula delta H=deltaU+deltaPV, in place of delta PV we use delta n(g)RT, where delta n(g) denotes difference in number of moles of gases after and before the reaction, this formula is only for ideal gases, but as in the reaction of oxalic acid is stated above, some of the products are liquid, but an ideal gas can never liquify, but still how we use the ideal gas formula for enthalpy calculations? $\endgroup$ Aug 29, 2023 at 16:57
  • $\begingroup$ The melting point of anhydrous oxalic acid is about 190 deg C. $\endgroup$
    – Poutnik
    Aug 29, 2023 at 17:30
  • $\begingroup$ @Kartsen , is the formula U=f/2nRT not valid for diatomic or triatomic gases? I thought it was valid for all ideal gases, only the values of f will change accordingly as the gas is diatonic, monoatomic or triatomic $\endgroup$ Aug 29, 2023 at 19:10

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