We can write the equation for the internal energy at initial or final energies as

The contribution to the internal energy from these three forms of kinetic energy are approximately 1/2kT per degree of freedom.  When we count up the degrees of freedom, we restrict ourselves to those that have energy levels that are accessible.  Generally, those are translation (3 degrees of freedom) and rotation (2 degrees if linear and 3 degrees if non-linear).  Thus, we have the following

For monatomic (Ideal) gases, the internal energy is simply

Thus, we see that the internal energy rises linearly with temperature with a slope of 3/2R.

For linear molecules, the internal energy is the sum of translational kinetic energy and rotational energy (two degrees of freedom)

Thus, we see that the internal energy rises linearly with temperature with a slope of 5/2R.

For non-linear molecules, the internal energy is the sum of translational kinetic energy and rotational energy (three degrees of freedom)

Thus, we see that the internal energy rises linearly with temperature with a slope of 2R.

Clearly, the amount of heat required to raise a system temperature depends on these different modes and non-linear molecules would require more heat than linear ones which in turn would require more heat than atomic species for a given rise in temperature.  For example, "low E" windows have argon gas injected between two (or even three) panes of glass (also coated to reduce radiative transmission of IR).  This gas has a low ability to absorb heat (no vibrations and no rotations) and hence will carry a small amount of heat between the cold pane and the warm pane, relative to air, which is composed largely of polyatomic molecules like O₂ or N₂.  Argon is also useful because it is a large heavy atom which is easier to contain than small atoms like Helium (c.f. effusion).

First Law of Thermodynamics

The First Law of Thermodynamics is stated commonly in many ways.  "Energy can neither be created nor destroyed", "the energy of the universe is constant", and perhaps the most useful one is "The energy of an isolated system is constant".  For practical reasons, it is better to restrict our measurements to small systems.

The measurments we can make, don't involve U directly.  We measure things like temperature, pressure, volume and from that we can calculate energy changes as a result of heat transfer and work.  Thus, we can write

This expression encapsulates the first law of thermodynamics in that the only two means of exchanging energy between the system and the universe are via heat transfer, q, or work, w.  Thus, the first law allows us to equate these two system processes to the change in internal energy.  The sum of q + w is obviously a state function, even though the individual system processes are not.

For example, the change in internal energy of a spring can be calculated knowing the work done when it is wound up, say 100 J, and measuring the heat lost to the surroundings, 15 J as

Keeping careful track of the sign convention, we can now determine that there is a net change to the internal energy of the spring of +85 J after it is wound up, compared to before winding.

Expansion work

When measuring work, we can have several forms.  We can measure mechanical work, |w| = f×d, from simple physics concepts, we could also measure work from electricity as the product of charge transferred and the voltage at which it was moved, say, through an electric motor.  The kind of work chemists are most often interested in is expansion work.  When a system increases its volume by pushing back the surroundings or decreases its volume when the surroundings push in on it. then work is done.  This change in volume could happen most noticeably when reactions produce gaseous products (expansion) or use up gaseous reactants (contraction). 

let's consider the infinitesimal energy change for a process done in very tiny steps.

dw = – Fdz	The minus sign tells us in this case that the system that did the work now has less energy remaining in it.
pex = F/A     or F = pex×A	We can relate the force to a change in the position of a piston of area A.  The piston contains our system inside the walls of the cylinder and any volume change involves motion of the piston.  From that motion, we get a volume change dV = Adz
dw = – pexdV

So now, we need to integrate over the total pressure and volume change to find the total work done.

IIf we can set up the conditions so that the external pressure is constant and that the volume change is done in a single "instantaneous" step then the integration becomes easy

Fig. 2.7  (from Atkins) The work done by a gas when it expands against a constant external pressure, pex, is equal to the shaded area in this example of an indicator diagram.

Since the expansion was done against a constant pressure, the integral is solved to be

Since PV is constant for an ideal gas, it is impossible to change V while keeping P constant unless either T or n changes.  So, this equation is useful in either specially controlled temperature change experiments or in cases there the volume change is caused by a chemical reaction that either creates or uses up a net amount gas molecules.

Generally, if V changes, so does P so the simple function doesn't always work.  We can explore the idea of reversible work by considering the expansion and compression of an ideal gas in a cylinder where Temperature is always constant (isothermal):

Initially, we are at state A holding the gas in a volume V_A in the cylinder with a pressure p_A.

1. We instantaneously release the pressure to p_B. The gas then expands to a new volume V_B. The pV work done (by the system) is easily measurable as the area under curve 1.

2. To reverse the process, we increase the pressure on the plunger back to p_A and the gas then compresses back to V_A. The work done (on the system) is the area under curve 2. 

Note that the work done to compress the gas was more than the work done by the gas on the surroundings.   Thus, we are loosing energy in this cyclic process.

Now consider the same process again but where we release the pressure infinitely slowly so that we are always following the exact pV isotherm.  In this case, we follow path 3 in either direction and we note that the work done on the environment by the system in the expansion phase is exactly the same as the work done by the environment on the system in the compression stage. This is reversible work. 

Note that at all times during this process, the system and the universe are at equilibrium with each other. 

The equation for this very slow process is slightly different from equation 3.8 because under the reversible conditions we describe here, the pressure of the system and the pressure of the surroundings are always equal.  Thus, we can write

If our system contains an ideal gas, whose equation of state is p=nRT/V, we get:

If the final volume is larger than the initial volume then the integral will be positive and the work will be negative.  In other words, the system did work (lost energy) by pushing back the surroundings.

If the final volume is smaller than the initial volume then the integral will be negative and so work will be positive.  The surroundings did the work on the system (which has more energy now).

Example:  what is the work done when 50 g of iron reacts with excess hydrochloric acid; a) in a closed vessel of fixed volume and b) in an open container at 25ºC.

In a chemical reaction between a Brønsted-Lowry acid and a metal, the metal is oxidized and hydrogen gas is released.  Thus, the balanced reaction is

Fe(s) + 2 H⁺(aq) → Fe²⁺(aq) + H₂(g)

a) this is simple since there is no volume change, there is no work done.  w = 0.

b) in this case, there is gas evolved, which occupies significantly more volume than the solid and the acid.  Thus, ΔV = V_f – V_i. where V_f  includes the volume of the gas, and the solution and V_i  includes the volume of the solution and the solid.  The solution volumes cancels out from the calculation (It doesn't change significantly) and the volume of the solid is small compared to the volume of the gas product so we'll ignore it.  Thus, the volume change simplifies to the volume of the gas evolved.  Since the gas is evolved at a constant pressure, we can equate the gas pressure to the external pressure, p_ex.

 ΔV = V_H2 = nRT/p_ex.

Thus, the work done, according to equation 3.9 is

w = –p_ex (nRT/p_ex) = nRT

So all we need do is determine the number of moles of gas evolved, n, and we can calculate the work done.

By integrating, we get to a macroscopic measurable
 

So, we can write the (hopefully) familiar equation for the change in internal energy of the system undergoing a process at constant volume.
 

As we hinted at earlier, if a systems energy changes via a change in temperature we might be tempted to define the amount of energy change related to the temperature change via the equation

where C is the heat capacity.  Heat capacity is an extensive property in that it depends on the amount of material.   However, heat capacities are often tabulated as intensive properties, based either on mass (specific heat) or on number of moles (molar heat capacity).  To convert the intensive property (from the table) to the extensive value we need equation 3.15, we need to multiply by mass or moles, respectively, to cancel out the units properly.  Some people actually add two extra equations to the list of things to memorize; q = mCΔT and q = nCΔT.  These latter two equations merely make explicit the conversion of an intensive property into an extensive (measurable) value.

Equation 3.15 is a straight-line equation which makes the assumption that the value for the heat capacity (the slope) is constant with temperature.  This is not always true.  A better definition of heat capacity takes this into account and allows that the heat capacity is the slope taken over an infinitesimal step in temperature.
 

Example,  What is the heat capacity of a monatomic ideal gas?

The equation for the energy of a mole of monatomic ideal gas is equation 3.3, U_m = U_m(0)+ 3/2 RT  .  If we insert this into the derivative, we get:

So clearly, a monatomic ideal gas has a heat capacity that does not depend on temperature.  The heat capacity changes as the temperature changes for polyatomic gases because the partition functions depend on the temperature (equation 2.33)  and thus, the equation of energy itself for these gases changes with temperature.

In general, the defining equation for heat capacity (eq. 3.16 at constant volume)  must be used since C_v is actually a function of T and must be considered in the integration, not as a constant.

To get to measurable numbers we need to integrate.  If we can safely assume heat capacity is constant over the temperature range we are using then we can use the simplified version

The subscript V indicates these equations apply only under constant volume conditions.

Additional example:

We have seen that if we supply heat to a system, it has different ways of handling it.  Sometimes, the system's temperature changes as the heat goes to increase the kinetic energy but other times, there seems to be little or no temperature change (phase change, for example)

Let's look at our statistical approach for some answers: 

Equation 2.27 gives our starting point.

The the instantaneous slope with respect to Temperature gives us

Remember: ν_rot is 0 for monatomic, 2 for diatomic (linear) and 3 for non-linear and we need to check the vibrational modes to see if they are active based on . 

What is C_ν,m for water at 100ºC?.  The three vibrational frequencies for water are