Internal Energy | Enthalpy | Thermochemistry | Change Functions

Changes in Internal Energy

Internal energy U is a function of p, V and T but since U is a state function we can relate the three values via the equation of state (e.g., pV = RT, for one mole of ideal gas).  Thus, we can always write U is a function of just two of the variables, where we use the equation of state to evaluate the third property.  We will express U as a function of V and T.

Fig. 2.21  (from Atkins) The partial derivative (∂U/∂V)T is the slope of U with respect to V with the temperature T held constant.

Fig. 2.22  The partial derivative (∂U/∂T)V is the slope of U with respect to T with the volume V held constant.

 

Fig. 2.23  An overall change in U, which is denoted dU, arises when both V and T  are allowed to change. If second-order infinitesimals are ignored, the overall change is the sum of changes for each variable separately.

Now suppose we take a system with internal energy U and volume V and adjust the volume by an infinitesimal amount.  Then the new internal energy U' will now be

Similarly, if we adjust the temperature by an infinitesimal amount, we will have

Finally, let's try changing both the volume and temperature infinitesimally, which gives us the overall infinitesimal change in internal energy.

We recognize the derivative in the first term as simply C_V (equation 3.16) and we'll rename the derivative in the second term as  .  We can now rewrite our simplified equation as

dU = pTdV + CvdT

3.38

We note that the new term p_T has units of pressure and is called the internal pressure.

Fig. 2.26  A schematic diagram of the apparatus used by Joule in an attempt to measure the change in internal energy when a gas expands isothermally. The heat absorbed by the gas is proportional to the change in temperature of the bath.

James Joule tried to measure the internal pressure of a gas in a now famous experiment.  He made an apparatus consisting of two metal bulbs joined via a stopcock and immersed in a water bath.  He expected to be able to measure a small temperature change when high pressure gas from one bulb was allowed to expand isothermally into the evacuated second bulb.  No temperature change was observed. 

  Although James Joule's apparatus was crude to the extent that even if there had been an energy change, he would not have been able to measure it, he drew some correct assumptions from his observations with regards to perfect (ideal) gas.  Since no temperature change occurred, he surmised that  dU = 0 and therefore, since w = -pΔV = 0 (no external pressure), that q = 0.  Thus, by inference p_T is also zero, This condition he attributed to a perfect gas.

  Real gases would not have a zero value for p_T but it's very small and probably Joule could not have measured it anyway.  Where does p_T  come from at the molecular level anyway?  Recall the energy curve showing the variation of energy with distance between two particles (electronic/nuclear) attractions and repulsions.

In the region where the atoms or molecules are very close together, then repulsive forces dominate and as the volume expands, the energy goes down.  Thus, for these conditions,  p_T is negative.  In the region where the atoms or molecules are close enough that attractive forces dominate, then as the volume expands, the energy goes up.  Thus, for these conditions,  p_T is positive.  For most gases at not too large pressures, the molecules don't interact very much and so there is little dependence of energy on volume so p_T is very small.  In the extreme of zero interaction,  p_T is zero.  This is the defining condition for an ideal (perfect) gas.

Joule-Thompson Effect

We can carry out a similar derivation for the change in enthalpy under infinitesimal changes in pressure and temperature to arrive at

Fig. 2.27  The apparatus used for measuring the Joule–Thomson effect. The gas expands through the porous barrier, which acts as a throttle, and the whole apparatus is thermally insulated. As explained in the text, this arrangement corresponds to an isenthalpic expansion (expansion at constant enthalpy). Whether the expansion results in a heating or a cooling of the gas depends on the conditions.

 

dH = mCpdp + CpdT

3.39

where is called the Joule-Thompson coefficient. 

Consider the energy change for a gas starting at p_i ,V_i ,T_i and expanding through a porous barrier (acts like a throttle) to p_f ,V_f ,T_f .  since the whole system is insulated, q = 0, so

 

ΔU  Uf  – Ui   Uf + pfVf  Hf

= w = piVi - pfVf = Ui + piVi = Hi.

So clearly, under these conditions, enthalpy is constant (isenthalpic)

So we can measure the Joule-Thompson coefficient under various conditions and we find that the value depends on the conditions and that in fact, it can be both negative and positive.  The sign of the isenthalp depends largely on whether the attractive or the repulsive forces dominate.

Fig. 2.31  The sign of the Joule–Thomson coefficient, µ, depends on the conditions. Inside the boundary, the shaded area, it is positive and outside it is negative. The temperature corresponding to the boundary at a given pressure is the ‘inversion temperature’ of the gas at that pressure. For a given pressure, the temperature must be below a certain value if cooling is required but, if it becomes too low, the boundary is crossed again and heating occurs. Reduction of pressure under adiabatic conditions moves the system along one of the isenthalps, or curves of constant enthalpy. The inversion temperature curve runs through the points of the isenthalps where their slope changes from negative to positive.

Fig. 2.32  The inversion temperatures for three real gases, nitrogen, hydrogen, and helium.

We calculated the heat capacity for an ideal gas previously as dH = C_pdT.  Compared to equation 3.39,
dH = mC_pdp + C_pdT, we can see that an ideal gas must have m = 0. So, the value of m is a reflection of the non-ideality of the gas, the intermolecular forces.  This also tells us that we cannot just use any gas at any set of pressures to make a refrigerator, for example.  At a given pressure, some gases may be cooling (m > 0) but others may be heating (m < 0). 

Consider a set of gases at 298K and 1 atm.

The proper choice of refrigerant will depend on both the physical properties, esp. the Joule-Thompson coefficient as well as the mechanical capacity of the equipment being used.  Thus, we cannot just exchange our ozone-depleting freon in our car's air conditioner with any other coolant unless the two gases behave similarly in the pressure - temperature ranges of the mechanical device, i.e., they must have the same sign of m at the pressures the equipment is capable of producing.  Generally, to use a more environmentally friendly coolant, we need to replace the old equipment with new equipment that will operate in the temperature range needed to make m positive.

Hence, most people are a bit reluctant to "shell out" for a whole new air conditioner for their car if the original one is working fine.

For a molecular perspective,

Consider a gas whose molecules are, on average, not interacting at all (m =0).  Then expanding the gas or compressing the gas has no effect on the potential energy and therefore also no effect on the kinetic energy.  Temperature will not change in this pressure range .  Now look at a gas under pressure that has molecules in the region where attractive forces dominate.  As we expand the gas, the molecules move further apart, against the force of attraction.  This converts the kinetic energy into potential energy (cools the gas) (m >0).  In the region where the repulsive forces dominate, as we expand the gas, the molecules actually gain kinetic energy (T rises) (m<0).