CHEM 331

October 3, 2000

Dr. Davis

 

Adiabatic Expansion/Compression

 

This paper illustrates calculations of energy and entropy changes for adiabatic processes in a gas.  By definition, the starting point for dealing with any adiabatic process is the fact that q=0.  However the energy and entropy changes that accompany an adiabatic process depend on (a) what sort of adiabatic process and (b) what sort of gas.  Here we will consider the following four examples:

  1. reversible, adiabatic compression of an ideal gas
  2. irreversible, constant pressure compression of an ideal gas
  3. reversible, adiabatic compression of a van der Waals gas
  4. irreversible, constant pressure compression of a van der Waals gas

 

 

 


1. Reversible, adiabatic compression of an idea gas.  Consider one mole of  an ideal gas initially at 298K and 10.0 atm pressure.  The gas is compressed adiabatically and reversibly until the volume has been reduced by half.  What is the energy change DU and what is the entropy change DS?  Assume that the heat capacity of the gas is CV,m= .

 

In an adiabatic process, no heat is gained or lost by the system.  Therefore our starting point is dq = 0.  From the First Law, this implies that dU = dw.  Now we consider the particular case of a reversible process in an ideal gas.  If the process is reversible, the work is given by dw = -pdV.  If the process involves an ideal gas, then dU = nCV,mdT.  Therefore, for this case we have

 

where we have also substituted the ideal gas law for the pressure.  Integration of both sides of this equation will not work, since the temperature on the right hand side is not constant, but varies with V according to some relation that is not given.  However, integration becomes possible if we simply divide both sides of the above by the temperature.  This gives

 

 

Integration of both sides then gives

 

 

 

This can be simplified to give

 

Eq. 1

 

Where c = CV.m/R ,  where Ti and Vi describe the initial state (before expansion or compression) and Tf and Vf describe the final state.  Note that this relationship proves that the final state has only one degree of freedom if it was reached via an adiabatic process.  In other words, either the final volume or the final temperature suffices to define the final state, since one determines the other – so long as the initial state was also specified. 

 

 

Applying Eq. 1 to the problem at hand, we have c = 3/2, Ti = 298K, pi = 10.0 atm and Vi = 2.45L and Vf = ˝Vi = 1.22L.     Therefore Tf =  473K. 

 

Note that the temperature is not constant in this adiabatic compression.  Adiabatic processes are not isothermal!  When a gas is compressed its energy goes up due to the work done on the system.  If no heat can escape (as in an adiabatic process), then this increased energy will inevitably lead to a temperature increase.

 

Now we can obtain the energy increase DU.  For an ideal gas process, we have DU = nCV.mDT.  Since DT has been found to be 473 – 298 = 175, we have DU = 2180 J.

 

We can also obtain the entropy change DS.  For a reversible process, dS = dq/T.  However the reversible process in this problem is also adiabatic, so dq = 0.  Therefore we have DS = 0 as well.

 

 

2. Irreversible, adiabatic compression of an ideal gas.  Consider one mole of  an ideal gas initially at 298K and 10.0 atm pressure.  The gas is compressed adiabatically and irreversibly at a constant pressure until the volume has been reduced by half.  What is the energy change DU and what is the entropy change DS?  Assume that the heat capacity of the gas is CV,m= .

 

This case differs from the previous one in being irreversible.  Whereas in the first case the pressure was increased gradually from 1 atm to its final value (so that the external and internal pressures were always essentially equal), here we increase the pressure instantly to its final value and maintain it there throughout the compression.  In other words we are forcing the compression by using a greater pressure than necessary.

 

The initial volume may be calculated from the ideal gas law to be 2.45 L.  Therefore the final volume is half of this, or 1.22 L.

 

The starting point is just as before.  Since dq = 0, we have dU = dw.  However we no longer have dw = - pdV, where p is the pressure of the system.  Rather we have dw = - pf dV, where pf is the constant, external pressure that is producing the work.  Since we are still assuming an ideal gas in this case, we again have dU = nCV,mdT.  Equating these expressions for dw and dU, we obtain

 

Integration of both sides then gives

 

 

Thus, just as in the previous case, we obtain a relationship between the initial and final states.  Also as before, either the final temperature or the final volume is sufficient to determine the final state, as long as the initial state is known.  

 

Solving for the final temperature in the above equation gives

 

Eq. 2

 

Substituting the data given in this problem gives Tf = 3Ti =  894K.  It follows that the final pressure is 60.1 atm.

 

Note that the final temperature is much higher than for the reversible case, even though both compressions were carried out adiabatically to the same final volume.  Why is the temperature higher in the irreversible compression?  It is because the applied, external pressure was greater resulting in more work done on the gas.  More work means a larger energy increase, which (since no heat can escape) means that the temperature increase must be greater.

 

Now we can obtain the energy increase DU.  As before, for an ideal gas process, we have DU = nCV.mDT.  Since DT is now known to be 894 – 298 = 596, we have DU = 7430 J.

 

Obtaining the entropy change DS is more difficult than before, since the process is now irreversible.  We know that dS = dq/T only for a reversible process; the relation can’t be applied directly to the compression in this problem.  The trick is to construct an alternate, reversible path between the same initial and final states.  The initial state is Ti = 298K and Vi = 2.45 L.  The final state is Tf = 894K and Vf = 1.22 L.  There are many possible reversible paths between these two states.  We will use the following two-step path: (1) compress the gas reversibly and isothermally until it reaches its final volume of 1.22 L, then (2) warm the gas reversibly at constant volume until it reaches its final temperature of 894K.  We can calculate DS for each step and then add the two contributions together.  For the first step we have

 

 

where we have used the fact that dU = 0 for an isothermal process in an ideal gas.  Integrating both sides of the above, we obtain

 

 

For the second step we have

 

Integrating both sides of the above we obtain

 

 

Therefore the total entropy change is given by

 

Eq. 3

 

Substituting the data for this problem gives DS1 = -5.80 J/K , DS2 = +13.70 J/K , and a total DS of +7.90 J/K.  Note that the entropy change is different than for the reversible case, where it was zero.  This does not contradict the fact that S is a function of state, since the final states are different for the reversible and irreversible processes.  The irreversible path ends with a temperature of 894K while the reversible process between the same two temperatures had a final temperature of only 473K.

 

Note too that the entropy change is positive.  This illustrates the Second Law, which states that any adiabatic, irreversible (spontaneous) process will have DS > 0.

 

 

3. Reversible, adiabatic compression of a van der Waals gas.  Consider one mole of  a van der Waals gas initially at 298K and 10.0 atm pressure.  The gas is compressed adiabatically and reversibly until the the volume has been reduced by half.  What is the energy change DU and what is the entropy change DS?  Assume that the heat capacity of the gas is CV,m = 3R/2.  Assume that the van der Waals parameters are those for CO2, a = 3.640 atm L2/mol2 and b = 0.04267 L/mol.

 

This problem is the same as the first case, except that we must use the van der Waals equation of state instead of the ideal gas law.  This has a number of implications.  First of all, we can no longer simply assume dU = nCV,mdT.  Instead we must use the general expression

 

 

where pT is given by the so-called thermodynamic equation of state

 

Eq. 4

 

Using the van der Waals equation of state to evaluate the partial derivative in Eq. 4, we obtain for the van der Waals gas

 

Eq. 5

 

Using this substitution and equating dU to dw, we have

 

 

Next we substitute the van der Waals equation for p in the right hand side of the above equation.  This gives

 

 

Dividing both sides by T then gives

 

 

Finally, integrating both sides, we obtain the van der Waals analog of Eq. 1

 

Eq. 6

 

We can solve Eq. 6 for the final temperature if we first obtain the initial volume.  This can be calculated from the van der Waals equation using the method of successive approximations.  “Solving” the van der Waals equation for V, we obtain

 

Eq. 7

 

Starting with an initial guess for the solution V (which may be conveniently obtained from the ideal gas law), one substitutes the initial guess into the right hand side of Eq. 7 and then solves for an improved estimate of the volume on the left hand side.  Then the improved value is substituted again into the right hand side and the process repeated.  Using the values given in this problem, we start with an initial guess of Vi = 2.45 L (from the ideal gas law).  Substituting this into the right hand side, along with the CO2 van der Waals parameters then gives a new value of Vi = 2.35 L.  Continuing this process successively, one finds that the volume converges to two decimal places at 2.34 L.  It follows that Vf =˝Vi = 1.17 L.

 

Substituting this final volume and the initial temperature and volume into Eq. 6 gives a final temperature of Tf = 479 K.  Note that this is just 6 degrees higher than the final temperature in the ideal case.  Except at rather high pressures, the van der Waals equation does not differ greatly from the ideal.

 

Now we can obtain the energy increase DU.  For a non-ideal gas process, we must use the general expression dU = nCV.mdT + pTdV.  Therefore we have

 

 

Integrating both sides we get for the internal energy change in a van der Waals gas,

 

Eq. 8

 

The first term in Eq. 8 is 2257 J, slightly larger than in the ideal case due to the higher final temperature for the van der Waals gas.   The second term, which does not appear at all for the ideal gas case, is –168 J, much smaller but still significant.  Thus the DU for the van der Waals case is 2090 J, not much less than in the ideal gas case.  

 

The entropy change DS is the same as for the ideal case.  For a reversible process, dS = dq/T.  However the reversible process in this problem is also adiabatic, so dq = 0.  Therefore we have DS = 0 as well.  The fact that we are dealing with a van der Waals gas now does not matter.

 

 

4. Irreversible, adiabatic compression of a van der Waals gas.  Consider one mole of  a van der Waals gas initially at 298K and 10.0 atm pressure.  The gas is compressed adiabatically and irreversibly until the volume has been cut in half.  What is the energy change DU and what is the entropy change DS?  Assume that the heat capacity of the gas is CV,m= .  Assume that the van der Waals parameters are those for CO2, a = 3.640 atm L2/mol2 and b = 0.04267 L/mol.

 

The starting point is similar to the previous example, except that the work is now irreversible,  and constant pressure.  Equating dU and dw as before, we obtain

 

 

but with p now constant at its final value.  Substituting the van der Waals expression for pT and integrating both sides, we obtain

 

 

Replacing pf with its van der Waals equivalent and rearranging gives

 

Eq. 9

 

This expression relates the final temperature to Ti, Vi, and Vf.  The initial and final volumes are the same as in the previous example, namely 2.34 L and 1.17 L, respectively.  Solving for the final temperature thus give Tf = 1008 K.  As for the ideal gas, the irreversible compression results in a much larger temperature increase than the reversible compression.   In fact, for the van der Waals case, the irreversible compression to the same final volume results in a temperature a full 114 degrees higher than in the reversible case.  The reversible/irreversible difference in the ideal case was only 6 degrees. 

 

Eq. 8, derived in the previous example, applies to the irreversible case as well, although the final temperature will be different than before.  Using Tf = 1008 K in Eq. 8 we obtain DU = 8700 J, 1270 J higher than for the ideal case.

 

Now we must obtain an expression for the entropy change for the irreversible, van der Waals case.  As for the ideal case, we devise a two step, reversible path between the initial state (Ti = 298K, Vi = 2.34 L) and the final state (Tf = 1008K, Vf = 1.17 L), this time consisting of (1) an isothermal compression to the final volume Vf, followed by (2) a constant volume expansion to the final temperature Tf. 

 

For the first step the entropy change is given by

 

 

Integrating both sides we obtain the result

 

 

For the second step, just as in the ideal case, we have

 

 

Thus the total entropy change is given by

 

Eq. 10

 

This is very similar to the ideal case in Eq. 3, and reduces to it in the case where b = 0.  Substitution of the values for the final and intial temperatures and volumes gives DS1 = -5.92 J/K, DS2 = +15.20 J/K , and a total DS of +9.28 J/K.  As before the total DS is positive, as it must be for an adiabatic, reversible change.

 

 

 

 


The results for these four sample calculations are summarized in the Table below.

 

 

ideal, reversible

ideal, irreversible

vdW, reversible

vdW, irreversible

Tf

473 K

894 K

479 K

1008 K

DU

2180 J

7430 J

2090 J

8696 J

DS

0

+7.90 J/K

0

+9.28 J/K

 

 

 

 

 


Exercises

 

 

  1. Repeat the ideal, irreversible calculation, but let the compression this time be not to half the initial volume but instead to the same temperature (473 K) as in the reversible case.  Calculate the final volume and pressure as well as DU and DS.

  2. Repeat the van der Waals, irreversible calculation, but let the compression this time be not to half the initial volume but instead to the same temperature (479 K) as in the reversible case.  Calculate the final volume and pressure as well as DU and DS.