In a bomb calorimeter, why is dH not equal to dU
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In a bomb calorimeter, why is dH not equal to dU

[From: ] [author: ] [Date: 11-12-07] [Hit: ]
At constant pressure the heat change q is equal to the change in enthalpy and at constant volume q, the heat change, is equal to the change in internal energy.In a bomb calorimeter dV = 0 but dP does not.Therefore, dH does not equal dU.......
Since dH = dU + PdV, and bomb calorimeters maintain a constant volume (dV=0), why is dH not equal to dU? When does the equation dH = dU + dnRT apply?

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I remember mixing this up. At constant pressure the heat change q is equal to the change in enthalpy and at constant volume q, the heat change, is equal to the change in internal energy.

By definition H = U + PV
therefore dH = dU + PdV + VdP
In a bomb calorimeter dV = 0 but dP does not. Therefore, dH does not equal dU. It equals
dU + VdP

If you assume ideal behavior, PV=nRT so it seems to me that dH = dU + d(nRT) is always applicable for a gaseous system where ideal behavior can be assumed. However, the only time I've ever used it is at constant temperature when it reduces to
dH = dU + RTdn and dn is the change in number of moles of gas during the reaction.

(Off-topic, but I think deltas are more appropriate than derivatives for thermochemistry since it involves overall changes rather than instantaneous changes. I know,nobody cares about that stuff.)

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