- #1

wintermute++

- 30

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delta H = delta U + P(delta V)I've been able to reduce the two equations so that delta U = q given a constant-pressure and volume container. Of course, delta H is also equal to q. So from that, the internal energy and enthalpy change would be equal. Or so I would assume.

What is throwing me off is the constant-temperature. If the value of q is changing but the temperature is constant and so is the ability to work... I just don't understand how the temperature cannot change given the other conditions if a change in heat occurs.

Any help?