# How Safe Is The Constant Pressure Assumption For Enthalpy?

1. May 6, 2015

### DocZaius

I am curious about a statement that is usually tacked onto the end of a derivation showing enthalpy with constant pressure being equal to the heat into the system.

First Law of Thermodynamics
$\Delta U = Q - W_{by}$

Define Enthalpy and look at its change:
$H=U+PV \\ \Delta H = \Delta U + \Delta (PV)$

Plug in First Law:
$\Delta H = Q - W_{by} + \Delta (PV)$

Under constant pressure:
$\Delta H = Q - P\Delta V + P\Delta V \\ \Delta H = Q$

And here is the part that I've heard and read more than once (Khan uses it in his enthalpy video for example). It is said that considering enthalpy at a constant pressure is useful for us, since that is the condition under which most of our chemical experiments occur (e.g. those done inside test tubes, etc.). But I always wonder why we can say that. My intuition would be that when a chemical reaction occurs, particularly a violent one, the pressure conditions at the center of the reaction would not necessarily adhere to those of its environment. I understand that an open test tube would be connected to the atmospheric pressure of the room, but surely for a short period of time and under very local conditions (e.g. at the center of the reaction) there must be a significant pressure gradient around the reaction? I would think that force per area in surfaces considered around the reaction would be relatively high. And is it not during that short period of time and at that specific location that we are considering quantities such as enthalpy?

My intuition appears to be wrong in this case, since clearly people have been using enthalpy under constant pressure as a useful quantity for a long time. But I am curious why it is that before any equilibrium is reached, we can safely assume constant pressures in energetic chemical reactions.

Last edited: May 6, 2015
2. May 6, 2015

### dipole

You've not understood the basic assumption of all thermodynamics, which the consideration of systems under equilibrium. You need to go back to chapter 1 page 1 of your thermo book of choice and re-read this.

3. May 6, 2015

### DocZaius

Shouldn't it be called thermostatics then? :P Anyhow, fair enough but if one got from point A (pre-reaction) to point B (post-reaction) under conditions that are clearly not quasi-static, I don't understand why we get to usefully apply these quantities that were derived assuming constant equilibrium.

In other words, I'm OK with my book announcing that the basic assumption will be that all considered systems will be under equilibrium. But I don't understand why it proceeds to tackle systems that clearly don't stay under equilibrium.

P.S. And does that mean then that this enthalpy-at-constant-pressure quantity would then only be useful for slow reactions that approximate equilibrium as best as possible?

Last edited: May 6, 2015
4. May 7, 2015

### Staff: Mentor

Don't forget that the work done by the system on the surroundings for a closed system between two thermodynamic equilibrium states is always given by $W=\int P_{Int} dV$, where PInt is the pressure at the interface with the surroundings (atmospheric pressure in your example). (See my recent Physics Forums Insight article https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/). This equation applies irrespective of whether the process is reversible or irreversible, and/or how non-uniform the conditions are within the reaction vessel. Only in the case of a reversible process is Pext = P, where P is the pressure that one would calculate from the equation of state for the gas.

However, for an irreversible process, if the final equilibrium state of a system has been attained via spontaneous reaction at constant interface pressure (PInt), the final equilibrium pressure of the system Pfinal will be equal to PInt (which is the same as in the initial equilibrium pressure of the system). So, $W=P_{final}ΔV=P_{Int}ΔV=PΔV$.

Chet