5.4: First Law of Thermodynamics
- Page ID
- 158435
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- Determine how heat and work are related to the overall internal energy of a system
- Calculate the work done by/on a system at constant pressure (PV work)
- Calculate changes in internal energy and enthalpy using the First Law of Thermodynamics
- Define a state function and determine which variables are state functions
The First Law of Thermodynamics - Law of Conservation of Energy
The first law of thermodynamics states that energy can not be created or destroyed, that is, the energy of the universe is constant. This allows us to define a thermodynamic system, which in the case of chemistry is a reaction, with the surroundings being everything else.
Lets consider the combustion of hydrogen
\[\ce{2H2 + O2 -> 2H2O}\]
there is bond potential energy in the hydrogen and oxygen bonds along with kinetic energy as they translate, rotate and vibrate. This is a combustion reaction and from common experience we know that energy is released. That is, the bond potential energy of the product is lower than the reactant, and so energy is released to the surroundings.
Internal Energy, U is the symbol we use to describe the internal energy of a system, although many textbook use E.
Convention for Energy Flow
So what is positive energy? The convention thermodynamics use is that positive energy is when the system gains energy, which is when either heat is transferred to, or work is done on the system. This requires differentiating the system from the surroundings. In chemical thermodynamics, the system represents the internal energy of the atoms of the reactants as they become products.
Figure\(\PageIndex{1}\): Diagram of internal energy
As the energy of the universe is constant, any energy lost by the system must be gained by the surroundings, and vice-versa, any energy gained by the system must be lost by the surroundings.
Internal Energy is given the symbol U, and represents the energy of the system
\[\Delta E_{Universe} = \Delta E_{System} + \Delta E_{Surrounding} = 0\]
therefore,
\[ \Delta E_{System} = - \Delta E_{Surrounding}\]
Noting that: ESystem = U, we can write this as
\[\Delta U = - \Delta E_{Surrounding}\]
That is, the heat lost by the system is gained by the surroundings, and remember, that in chemistry, the reaction is the system, so.
\[\Delta E_{Reaction} = -\Delta E_{Surrounding}\]
- EReaction < 0, the system loses energy, (and the surrounding gains energy).
- EReaction > 0, the system gains energy. (and the surrounding losses energy).
Must Heat Flow From a System of High Energy to a System of Low Energy? No, although it can. It depends on the temperature differential.
Direction of Heat Transfer: Heat energy flows from a hot object to a cold object, even if the cold object has more energy. To understand the direction of heat flow we need to invoke the second law of thermodynamics and the concept of entropy, which will be covered in chapter 18, where we will learn that the direction of a spontaneous process is the direction that maximizes the entropy of the universe. If you drop a mL of boiling water into the North Atlantic Ocean, heat will transfer from the boiling water to the cold ocean until they are at the same temperature, even though the ocean obviously has more energy then a mL of water. There are two observations we can make about heat energy transfer:
- Heat Flows from a substance at a higher temperature to the substance at a lower temperature, with the hot substance cooling and the cold substance warming.
- Once the temperature of the two substances equal, the system is at thermodynamic equilibrium. This is actually a dynamic equilibrium, in that energy is flowing between the substances, but the rate of flow in both directions is the same, so there is no net transfer of energy between the substances.
Pressure-volume work*
The kind of work most frequently associated with chemical change occurs when the volume of the system changes owing to the disappearance or formation of gaseous substances. This is sometimes called expansion work or PV-work, and it can most easily be understood by reference to the simplest form of matter we can deal with, the hypothetical ideal gas.
The figure shows a quantity of gas confined in a cylinder by means of a moveable piston. Weights placed on top of the piston exert a force f over the cross-section area A, producing a pressure P = f / A which is exactly countered by the pressure of the gas, so that the piston remains stationary. Now suppose that we heat the gas slightly; according to Charles’ law, this will cause the gas to expand, so the piston will be forced upward by a distance Δx. Since this motion is opposed by the force f, a quantity of work f Δx will be done by the gas on the piston. By convention, work done by the system (in this case, the gas) on the surroundings is negative, so the work is given by
\[w = – f Δx \label{3-1}\]
When dealing with a gas, it is convenient to think in terms of the more relevant quantities pressure and volume rather than force and distance. We can accomplish this by multiplying the second term by A/A which of course leaves it unchanged:
\[ w = -f Δx \dfrac{A}{A} \label{3-2}\]
By grouping the terms differently, but still not changing anything, we obtain
\[ w = -\dfrac{f}{A} Δx A \label{3-3}\]
Since pressure is force per unit area and the product of the length A and the area has the dimensions of volume, this expression becomes
\[w = –P ΔV \label{3-4}\]
Enthalpy
Enthalpy (\(H\)) comes from the Greek enthalpein, meaning (“to warm”). which is the measure of the heat of a system. From above we know the change in internal energy (U) of a system is equal to the heat and work done on the system.
\[ΔU=Q+W \label{1st}\]
We know the work of expansion is \(PV\) work, which for a system at constant pressure is
\[W=-PΔV \label{work}\]
For a system at constant pressure, the change in enthalpy \(ΔH\) is \(Q_p\), so
\[Q_p = ΔH \label{heat}\]
substituting the definitions of work (Equation \ref{work}) and heat (Equation \ref{heat}) into Equation \ref{1st} at constant \(P\) gives,
\[ΔU= ΔH -PΔV\]
rearranging to solve for enthalpy gives:
\[ΔH = ΔU + PΔV\]
Since PΔV is the negative of the work of expansion, enthalpy is the internal energy of the system minus the work of expansion. Note this equation is under conditions of constant pressure, which are the only conditions we will use in this class, and a more general equation for enthalpy is H=U+PV
We often call enthalpy the heat of the system, and when a reaction gives off energy, we often call it the enthalpy of reaction, and tabulate enthalpies of reaction to determine the energy changes that occur as reaction proceed from reactants to products.
State Functions
Enthalpy and Internal Energy are state functions. A state function is a function whose value is defined by the state, its pressure, temperature, composition, and amount of substance, and not the process of how it was obtained. So the difference between two state functions is path independent. Thus any solution of NaCl at 25°C and 1 atm that contains a mixture of 1 mol NaCl and 100 mol H2O has the same internal energy and the same enthalpy as any other solution with the same specifications. It does not matter whether the solution was prepared by simply dissolving NaCl(s) in H2O, by reacting NaOH(aq) with HCl(aq), or by some other method. Therefore, no matter how you go from reactant to product, 
and 
will be the same, and are state functions.
Is work a state function? No, as work is path dependent. If you look at Figure\(\PageIndex{2}\) you have a nice analogy for understanding the state function. The gravitational potential energy difference between an object at a height of the summit of Mt. Kilimanjaro and it's base is the same, no matter how you made it to the summit. The green and yellow path "X" and"Y" represent different paths, and the work required does depend on the path.
Figure\(\PageIndex{1}\): Diagram showing difference between path dependent function and a state function. The work required to reach the summit is not a state function, and depends on the path, but once you are on the top, the view is the same, no matter how you got there.
Example \(\PageIndex{1}\)
A sample of an ideal gas in the cylinder of an engine is compressed from 500. mL to 40.0 mL during the compression stroke against a constant pressure of 8.08 x 105 Pa. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy
of the gas in joules?
Solution
Contributors:
- Robert Belford (UA of Little Rock)
- Ronia Kattoum (UA of Little Rock)