Bomb Calorimeter, enthalpy

Main Question or Discussion Point

I am going to define these things as I understand them. Please correct/refine them as necessary.

A bomb calorimeter is a small volume capsule within a tank of water. A reaction is carried out within this small capsule and the heat evolved is transferred to the water and the heat change of the water is measured. But what about the pressure change within the small capsule? Isn't that a representation of heat as well? Is enthalpy the heat change only? Or does it somehow take into account the pressure as well?

Borek
Mentor
Bomb calorimeter is a constant volume device, so pdV is zero.

Do you know difference between enthalpy (H) and internal energy (U)?

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Just because the volume is constant in that tiny chamber doesn't mean the pressure will stay the same. Right? Essentially, an explosion is happening in that chamber, and I am assuming that the pressure is going to increase. Is that pressure change accounted for in the enthalpy? It seems like it isn't, at least in the measurements done by a bomb calorimeter.

I do not know what the difference between internal energy and enthalpy. I guess the equation I am seeing says that Enthalpy = U + pV. Well in my bomb calorimeter, the p is increasing (but not being measured) the V is staying the same, and the U is the heat evolved?

Borek
Mentor
Yes the pressure changes, no there is no work done as volume is constant. H = U + PV - there is a reason why bomb calorimeter keeps the volume constant. (Hint: if volume is constant, is there any difference between U and H? The equation you wrote tells all you need to know at this moment.)

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There is another angle possible on this issue:
in the definition H=U+pV, some textbooks state p=external pressure, some state p=internal pressure, but most of the state p=pressure and leave it to the reader or context or assume internal pressure=external pressure (reversible thermodynamics).
If you follow O Levenspiel (1996) you take p=p(external) and all the above questions are answered.

Just because the volume is constant in that tiny chamber doesn't mean the pressure will stay the same. Right? Essentially, an explosion is happening in that chamber, and I am assuming that the pressure is going to increase. Is that pressure change accounted for in the enthalpy? It seems like it isn't, at least in the measurements done by a bomb calorimeter.

I do not know what the difference between internal energy and enthalpy. I guess the equation I am seeing says that Enthalpy = U + pV. Well in my bomb calorimeter, the p is increasing (but not being measured) the V is staying the same, and the U is the heat evolved?

I see where you are confused. Here's the first law of thermodynamics...

∆U = Q - W
where W = P∆V​
There are two forms of energy, heat (Q) and work (W). Also, energy cannot be created nor destroyed, only transferred from one system to another.

First, a bomb calorimeter is a sealed tiny chamber usually made of steel (∆V=0, so W=0, therefore ∆U=Q) submerged in a fluid filled container, which is open to the environment (constant pressure so Q = ∆H).

This is probably the point of confusion...
The definition of enthalpy (∆H) is that it is equal to heat transferred (Q) into or out of the system when the system is under constant pressure.
Within the sealed container, pressure is NOT constant. Yes, pressure does increase during a combustion. But this is irrelevant since Q of the combustion reaction is measured by the change in temperature of the surroundings, such as the fluid and the steel, which are at constant pressure.

Putting it all together...
Energy cannot be created nor destroyed, only transferred. The energy released by the reaction in the sealed chamber is in the form of heat (Q) (since there is no change in volume, W=0), which is transferred to its surroundings, in this case the steel and water. The steel and water are open to the environment (constant pressure), therefore Q = ∆H.