What is the difference between internal energy and enthalpy?

• fog37
In summary, the internal energy and enthalpy of a system are important concepts in thermodynamics, representing the total kinetic and potential energy of the system. Changes in these quantities can occur through heat transfer or mechanical work. Enthalpy, in particular, is useful because it allows us to focus on the effect of absorbed energy without worrying about work done by the system. In analyzing chemical reactions, only variations in internal energy and enthalpy are important, rather than knowing the specific values at a particular moment in time.
fog37
TL;DR Summary
Internal energy vs Enthalpy
Hello,

In thermodynamics, with systems being represented by gases (can liquids be also included?), the internal energy ##E_{int}## of a system represents the total kinetic energy + the total potential energy of the system: $$E_{int} = KE_{tot}+ PE_{tot}$$ The term ##KE_{tot}=Q## is also called thermal energy (random motion of all internal parts) and is proportional to the system's temperature ##T##. The term ##PE_{tot}## represents the total chemical energy stored in the chemical bonds.
Internal energy ##E_{int}## can be changed via two different mechanisms, which are heat (transfer thermal energy from the system or to the system) and/or mechanical work s done on or by the system: $$\Delta E_{int} = Q - W$$
(I guess absorbing and emitting radiation is a 3rd mechanism that is often not included). Work ##W## is nonzero when the system's shape (volume) is changed. For a gas, work is given by ##W = \int P dV##. The system (gas) has its own internal pressure ##P_{gas}## and there is also an external pressure ##P_{ext}##, generally the constant atmospheric pressure. I guess we can use either pressure in the definition of work ##W##. But the pressure ##P_{gas}## can be variable: I envision a reaction where the internal pressure change with time (ex: explosion).

If the gas is free to expand/contract and change its volume ##V##, work is nonzero and there is a change in energy.

Instead of using internal energy ##E_{int}## and its change ##\Delta E_{int}##, we can clump work with the internal energy into a new state function ##H##, enthalpy, so we don't need to track work and change in system's volume: $$[\Delta E_{int} +W] = Q$$

$$[\Delta E_{int} +W] = \Delta H$$

$$\Delta H = Q$$

which means that enthalpy changes when the system absorbs or emit thermal energy ##Q## via heat (transfer or gain of thermal energy to the environment).

Does the definition of enthalpy ##H## and its changes only apply if the internal and external pressures are both equal and constant? Why? I am missing that important point...

Thanks!

Last edited:
fog37 said:
Does the definition of enthalpy ##H## and its changes only apply if the internal and external pressures are both equal and constant?

No, the definition H := U + p·V is independent from the conditions.

fog37
Thank DrStupid.

So, ##H=U+PV##. Is the change ##\Delta H=Q## only true when ##P_{gas}=P_{ext}=P=constant##? In the scenario of a gas expanding while its internal pressure remains constant, the work done by the gas and the work done by the surrounding on the gas are equal and opposite.

$$\Delta H = H_f - H_0 = (U_f+P V_f)-(U_0+P V_0)$$
$$\Delta H = (U_f - U_0) +P (V_f - V_0)=\Delta U + P \Delta V$$

and since $$\Delta U= Q - P \Delta V$$ we get

$$\Delta H = Q$$

I am still not sure what we gain from coming up with enthalpy ##H## instead of working with ##U##...

fog37 said:
Is the change ##\Delta H=Q## only true when ##P_{gas}=P_{ext}=P=constant##?

You just need a single system. The change of enthalpy is

##\Delta H = \Delta U + p \cdot \Delta V + V \cdot \Delta p##

The thange of internal energy (without non-volumetric work) is

##\Delta U = Q - p \cdot \Delta V##

That results in

##\Delta H = Q + V \cdot \Delta p##

and with constant pressure in

##\Delta H = Q##

fog37 said:
I am still not sure what we gain from coming up with enthalpy ##H## instead of working with ##U##...

Calorimetric measurements of ##\Delta H## require constant pressure. Calorimetric measurements of ##\Delta U## require constant volume. What do you think is easier in most cases?

Thank you. So enthalpy ##H## is a state function that depends only on the initial and final state variable (##V, P,T.etc##). The pressure ##P## is the same at both states and equal to atmospheric pressure ##P_{atmos}##. However, during the transformation to get to the final state, the gas internal pressure does not have to be constant and equal to the external atmospheric pressure.

Well, constant volume requires the volume to be fixed but I guess it is hard to force the volume of reactants and products to be the same even if the reaction happens inside a fixed volume container.

So, conceptually, should enthalpy be interpreted as the thermal energy either absorbed or emitted during a chemical reaction without worrying about what mechanical ##PV## work was done or not by the system, as long as the pressure of the initial and final states is the same?

fog37 said:
So, conceptually, should enthalpy be interpreted as the thermal energy either absorbed or emitted during a chemical reaction without worrying about what mechanical ##PV## work was done or not by the system, as long as the pressure of the initial and final states is the same?

Yes, that's the basic idea behind it.

fog37
Hello DrStupid, just some remarks (feel free to provide corrections please) to end the discussion on this topic:

Say a gas absorbs thermal energy ##Q## from the environment and the gas is able to expand with the expansion happening against the external and constant atmospheric pressure ##P_{atmos}##. The provided thermal energy ##Q## will be absorbed by the gas and spent to both increase its internal energy ##U## and produce positive expansion work ##P \Delta V## with ##\Delta V= V_f - V_0 ##. The initial and final states of gas are ##(P_0, T_0, V_0)## and ##(P_f, T_f, V_f)## and their enthalpies are ##H_0 = U_0 +P_0 V_0## and ##H_f = U_f +P_f V_f##.

The change ##\Delta H= H_f - H_0##, if the external pressure the expansion fights against is constant, is ##\Delta H= H_f - H_0 = Q## meaning that the absorbed energy ##Q##, which we know produced both volume expansion and an increase in internal energy ##\Delta U##, only produces a change in ##H##...

This means that we can only focus on the effect of the absorbed energy ##Q## without worrying about any work done on or by the system when analyzing chemical reactions...Why is knowing the internal energy ##U## or enthalpy ##H## of a particular state at a particular moment not possible and not useful? Only variations in ##U## and ##H## seem to matter. Knowing how much internal energy has system has seems relevant...

Thanks!

fog37 said:
Why is knowing the internal energy ##U## or enthalpy ##H## of a particular state at a particular moment not possible and not useful?

It would be useful but it is not possible. In classical mechanics there is not even a theoretical limit for the internal energy. A system having or releasing an infinite amount of energy wouldn't violate any laws of classical physics.

This is different in relativity due to the mass-energy relation. You can use the mass of a system as a measure of its internal energy. However, this still excludes vacuum energy which remains unknown (vacuum catastrophe).

fog37
Internal energy and enthalpy are relative measures like altitude. Asking for absolute internal energy or enthalpy values is like asking how high is "high". If I say I'm 20 feet above the ground that doesn't tell you where I am relative to some zero datum like mean sea level but it does tell you I should use the stairs instead of jumping down to the ground.

In chemistry we arbitrarily define the zero datum for H to be the enthalpies of formation of pure compounds at standard state (1 atmosphere 25 C which is not the same as STP or standard temperature and pressure which is 0 C and 1 atmosphere). This allows you to do meaningful calculations like computing the heats of reactions for arbitrary chemical processes.

For internal energy you arbitrarily define what "ground state" is based on whatever is convenient for the system under study and measure changes with respect to that. The universe has no "bottom" and everything is "relative" so there is not zero milestone to measure everything from.

Temperature is about the only measurement I can think of where this is not true. There is an absolute zero temperature but this state shouldn't be confused with zero energy.

fog37
luptonma said:
Temperature is about the only measurement I can think of where this is not true. There is an absolute zero temperature but this state shouldn't be confused with zero energy.

How about mass (due to the positive energy theorem) or entropy (due to the third law of thermodynamics)? There are a lot of other absolute scales.

DrStupid said:
It would be useful but it is not possible. In classical mechanics there is not even a theoretical limit for the internal energy. A system having or releasing an infinite amount of energy wouldn't violate any laws of classical physics.

This is different in relativity due to the mass-energy relation. You can use the mass of a system as a measure of its internal energy. However, this still excludes vacuum energy which remains unknown (vacuum catastrophe).

Ok, I can see how knowing the overall total energy ##E_{tot}## that a system has (the sum of all forms of energy) may not be feasible. But even in circumstances and mechanics problems where we limit ourselves to analyze the mechanical energy of the system ##E_{mech}=(PE +KE)##, which one specific type of energy a physical system may have, it seems that what matters is the change ##\Delta E_{mech}## instead of the value ##E_{mech}##...

Energy, not matter its form (chemical, kinetic, thermal, potential), is a relative quantity that depends on the reference frame. However, in the same reference frame, we could compare two different systems based on their energies. I guess both those two energies are relative in value so once again the energy difference is what really matters...

luptonma said:
Internal energy and enthalpy are relative measures like altitude. Asking for absolute internal energy or enthalpy values is like asking how high is "high". If I say I'm 20 feet above the ground that doesn't tell you where I am relative to some zero datum like mean sea level but it does tell you I should use the stairs instead of jumping down to the ground.

In chemistry we arbitrarily define the zero datum for H to be the enthalpies of formation of pure compounds at standard state (1 atmosphere 25 C which is not the same as STP or standard temperature and pressure which is 0 C and 1 atmosphere). This allows you to do meaningful calculations like computing the heats of reactions for arbitrary chemical processes.

For internal energy you arbitrarily define what "ground state" is based on whatever is convenient for the system under study and measure changes with respect to that. The universe has no "bottom" and everything is "relative" so there is not zero milestone to measure everything from.

Temperature is about the only measurement I can think of where this is not true. There is an absolute zero temperature but this state shouldn't be confused with zero energy.

Temperature is proportional to the system's thermal energy which is the overall kinetic energy of the system and kinetic energy is relative so temperature would appear to be relative too...

Temperature is a state function...I wonder if all state functions, like volume, pressure, etc. are relative...

Apollinecathenna said:
The above equation about Internal energy and Enthalpy is very useful to me.
I think Enthalpy is the heat energy and Internal energy is sum of potential energey and kinetic energy

Thanks Apollinecathenna. To be more specific, what does ##\Delta H## tell you that ## \Delta U## does not?

1. What is the difference between internal energy and enthalpy?

Internal energy is the total energy contained within a system, including the kinetic and potential energies of its particles. Enthalpy, on the other hand, is the measure of the total heat content of a system at a constant pressure. It includes the internal energy of the system as well as the work done by or on the system.

2. How are internal energy and enthalpy related?

Enthalpy (H) is equal to the internal energy (U) plus the product of the pressure (P) and volume (V) of the system. This can be expressed as H = U + PV. Therefore, internal energy and enthalpy are directly related, but enthalpy also takes into account the work done by or on the system.

3. Can internal energy and enthalpy change during a physical or chemical process?

Yes, both internal energy and enthalpy can change during a physical or chemical process. Internal energy can change due to changes in temperature, pressure, or volume of the system. Enthalpy, on the other hand, can change due to changes in temperature, pressure, or the amount of heat added to or removed from the system.

4. How do we measure internal energy and enthalpy?

Internal energy and enthalpy are measured in units of energy, such as joules (J) or calories (cal). They can be measured using various instruments such as calorimeters, which measure changes in temperature, and bomb calorimeters, which measure the heat released or absorbed during a chemical reaction.

5. Are internal energy and enthalpy conserved in a closed system?

Yes, according to the first law of thermodynamics, energy cannot be created or destroyed in a closed system. Therefore, the total internal energy and enthalpy of a closed system will remain constant, although they may change in form or distribution within the system.

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