[Needaask]

ΔH=ΔU+PΔV at constant pressure, the work PΔV is done by the gas on the surroundings. But I don't get why ΔH=q at constant pressure. Isn't q heat gained or released? So when internal energy increases it's due to work done and heat supplied. So why is another +PΔV added and why would the enthalpy change be just the heat given out or absorbed only?

I sort of understand mathematically that ΔH=q-PΔV+PΔV so it's just equals to q. However, I'm thinking if they cancel out shouldn't there be no work done at all? But in fact in a constant pressure reaction, there is work done so I can't seem to understand this whole thing.

Thanks

[Illuminatus]

Hello Needaask,

First off, in the internal energy equation, q is heat gained by the system (if you refer to the variable without a negative sign). Work is always done by a system when there is a change in volume. Thus for a reaction which occurs at constant pressure and involves a change in volume, there is indeed work done by the system. However, enthalpy is by definition a measure of the total heat content of a system, and the change in enthalpy is thus the change in this heat content. Because you operate at constant pressure, no heat is acquired/lost through work, and thus work done does not affect the enthalpy change. As a counterexample you could imagine gas changing temperature as it does work to expand.

Keep in mind that the concept of enthalpy was designed to be a useful measurement mainly for isobaric conditions (sometimes you will see isochoric problems allude to it). Under other conditions, internal energy is usually utilized. The equation ΔH=ΔU+PΔV is an intermediate in the derivation of enthalpy; you will almost always encounter your enthalpy change in terms of heat transfer q. The +PΔV term is added for the purpose of being able to state a mechanically sound relation to internal energy, ΔU, and thus define enthalpy universally.

Enjoy

[Needaask]

Hello Needaask,

First off, in the internal energy equation, q is heat gained by the system (if you refer to the variable without a negative sign). Work is always done by a system when there is a change in volume. Thus for a reaction which occurs at constant pressure and involves a change in volume, there is indeed work done by the system. However, enthalpy is by definition a measure of the total heat content of a system, and the change in enthalpy is thus the change in this heat content. Because you operate at constant pressure, no heat is acquired/lost through work, and thus work done does not affect the enthalpy change. As a counterexample you could imagine gas changing temperature as it does work to expand.

Keep in mind that the concept of enthalpy was designed to be a useful measurement mainly for isobaric conditions (sometimes you will see isochoric problems allude to it). Under other conditions, internal energy is usually utilized. The equation ΔH=ΔU+PΔV is an intermediate in the derivation of enthalpy; you will almost always encounter your enthalpy change in terms of heat transfer q. The +PΔV term is added for the purpose of being able to state a mechanically sound relation to internal energy, ΔU, and thus define enthalpy universally.

Enjoy

Hi Illuminatus,

Ohh I think I better understand it now. The enthalpy change only tells us the heat change. But actually why must pressure be constant then? Because in most cases when we determine our enthalpy change, we use a calorimeter to find out the heat gained. So q is directly measured. So even if the pressure continuously changes won't the q remain a constant? Like when we use the first law of thermodynamics to represent this, ΔU=ΔH+PΔV so if the external P changes a lot, ΔU would still remain a constant and so would ΔH right?

Thanks for the reply