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LECTURE 8

Page history last edited by fwhitfie@gmu.edu 14 years, 5 months ago

Dr Cooper's comments: Number fundamental equations (Eq. 1,  Eq. 2 etc)

 

Calculating Entropy Change in Specific Processes

 

For adiabatic changes, q=0 and the entropy change is 0.  As long as the initial  and final states of a system are connected by any reversible adiabatic path, the entropy change is going to be zero.

 

Entropy Change for an Ideal Gase (Isotherm):

Formula

 

Formula

 

Formula **  This is the equation for entropy change in a system whether the path is reversible or not because entropy is a state function.

 

 

Entropy Change for Ideal Gas (Isobar):

Formula

 

Formula

 

 

Entropy Change for Ideal Gas (Isochore):

 

Formula

 

Formula

 

Entropy Change for Reversible Phase Changes:

 

     Melting                                                                                                         Boiling

Formula                                                                           Formula

 

Formula (constant pressure)                                                                   Formula (constant pressure)

 

Formula                                                                                          Formula

what do the subscripts mean? 

 

Entropy Change for Reversible Phase Changes:

 

When matter changes phases, the order in which molecules are held together as well as the localization of energy changes. This results in an entropy change. The entropy of matter increases as it changes from solid to <  liquid to< gas.  At the temperature where the transition is taking place, the heat transferred from the system to the surroundings is reversible because the two phases are at equilibrium.

 

Table 3.1 gives us the standard entropies (and temperatures) of phase transitions, and table 3.2 gives us the standard entropies of vaporization of liquids.

 

 

 

Trauton's Rule: (Trouton's)

 

Trauton's rule specifically applies to vaporization. According to Trauton's rule, a change in volume takes place when a liquid evaporates to become a gas, and all liquids can be expected to have similar standard entropies of vaporization.  However, there are exceptions to this rule. Water, due to its hydrogen bonding that holds the molecules together, shows an entropy of vaporization higher than the majority of liquids.

 

Problem # 1:

Calculate the entropy change when Argon at 25 C, and 1.00 bar in a container of volume 0.500 dm^3  is allowed to expand to 1.00 dm^3 and is simultaneusly heated to 100 C.

 

We can solve this problem in two steps:

 Step 1: Change in volume (Isothermal expansion from 0.500 to 1.000 dm^3)

 

Formula          Formula            1 bar = 10 ^5 Pa ,   1 Pa m^3 = 1 J

 

 

  Formula 

 

Formula

 

Step 2: Change in temperature (Heating at constant volume)

 

Formula

 

 

Why the gap? Steps missing? 

 

 

 

 

Step 2.

Formula

 

 

 

 

Lastly, we add the change in entropy of both steps together to obtain the total entropy  change: 

ΔSTotal = ΔSstep 1 + ΔSstep 2                                                                ΔSstep 1 = 0.06 J K-1  

ΔSTotal= 0.06 J K-1  + 0.12  J K-1                                                         ΔSstep 2 = 0.12  J K-1

ΔSTotal = 0.18 J K-1

 

 

 

 

The Standard Molar Entropy ΔS0

This is found by calculating the entropy change of a substance when it warmed from absolute zero to the standard temperature (the experimenter in interested in).

However this is impossible to calculate definitively.

 

The third law of thermodynamics takes this into account with the Nernst Heat Theorem:

Which states that as Temperature goes to Zero Kelvin (units?) The change in entropy will also go to zero.

However, this only is valid for a perfectly crystalline substance because when the temperature is at zero Kelvins (units?)  heat energy does not exist in the system and the structure is perfectly ordered leaving no other way to disperse the energy because there is no disorder in the system.

The calculation for the change in entropy is describe by the integral:

Formula equation 1

 

Where Cp is the heat capacity at constant pressure and T is temperature.

However, heat capacity is difficult to calculate at Absolute Zero zero.

In this difficult area The Debye extrapolation is used. Which states that that Cp is proportional to T3

This is used to approximate heat capacities at temperatures below 10 Kelvin. The equation is as follows:

Formulaequation 2            Where a is just a proportionality constant.

 


Figure 1 below in the top graph shows how the Debye approximation might be used to estimate the entropy of a

substance approaching 0 K.


                 (Figure 1)

 

 

Figure 1 above in the first graph shows the Debye approximation with this it becomes possible to calculate the entropy by using

the basic template of equation 1 in the second graph the changes in entropy are shown at the transition points.


The entropy of a molecule at a particular temperature is found by adding up all the changes in entropy from T=0 to the temperature

in quesion plus the changes of entropy of the state changes.

 

Formula equation 3

                                        |--1st integral---|     |--fusion---|        |--2nd integral--|    |-vaporization-|   |--final integral-|

 

S(0) is the entropy at 0k which is 0. The first integral  calculates the entropy change from 0K up to the temperature of fusion,the equation for the entropy of fusion is added to this. The second integral calculates the entropy from fusion to vaporization, the equation for the entropy of vaporization is added to this.  The final integral calculates the entropy up to the temperature that is desired. 


 

Problem Using the Debye Approximation:

 

A certain solid has a Cp,m at 4.2 K of 0.43 J K-1mol-1. What is the molar entropy at this temperature?

 

According to the debye approximation Cp becomes aT3 

                                               1 

Formula

 

 

The T's cancels and we get:

Formula

Cp,m  is equal to aT3so:

Formula

 

Formula


Standard Reaction Entropy

 

The change in molar entropy caused by reactants in their standard states changeing to products in their standard states.

The equation:

 

Formulaequation 4

 

Which outlines that the standard reaction entropy is the sum of the molar entropies of the products subtracted from the sum of the molar entropies of the reactants. The v is the stoichemetric equivalents of the products and reactants.

 

For a reaction where:  Formula

 

Formula

                 |-------Products---------|      |------Reactants---------|

 

 

----as a note gas can better distribute energy than a solid so in general the standard entropy if there is a net formation of gas

        in a reaction will be positive where as if gas is consumed the standard entropy will be negative.

 

 The tablebelow basically shows that the more complex a molecule is the higher its molar entropy will be:

 

           (Figure 2)

 

           


Problem:

 

Calculate the standard reaction entropy of the formation of water:

Formula  at 25º C.

 

place in the stochiometic amounts of the products and reactants:

Formula

 

Calculate inserting the standard molar entropies of products and reactants found in literature:

Formula

 

Formula

 


Spontaneity of a Reaction

 

The Purpose of calculating entropies is to be able to find out if a reaction is spontaneous but the standard molar entropy changes of a reaction does not indicate whether a reaction is spontaneous.

 

In order to find out if a reaction is spontaneous it is necessary to know the change in entropy of the surroundings as well as the system

 

Then, if the sum of the surroundings and the system is positive then the reaction will be spontaneous.

 

 


 

Clausius Inequality

 

Reversible work nets the maximum work possible:

 

Formula(equation 5)

 

by rearranging the above equation:

 

Formula

 

Since internal energy is a state function the sum of nonreversible work should equal the sum of reversible work.

It does not matter if the work is reversible because internal energy depends on the initial and final states only.

With this reasoning:

 

Formula

 

then rearrange the equation to get:

Formula

 

dividing by Temperature (T) yields:

Formula                                               Where Formula is equal to the change in entropy.

 

Rearranging the equation further yields:

Formula

 

finally, it can be stated that:

Formula(equation 6) The Clausius Inequality

 

using this inequality (equation 6) to considering a system of constant volume in thermal equilibrium with its surroundings at temperature T. We find that:

Formula   where in a system at constant volume Formula

(equation 7)

 

from this we can conclude:

 

Formulaat constant volume

(equation 8)

 

From this equation we can determine that for a spontaneous reaction:

 

---The internal energy in a constant volume system will be less than or equal to zero because the only way to

     reduce internal energy will be to pump heat out into the surroundings increasing the entropy of the surroundings. 

 

At constant internal energy and volume Formula the systems change in entropy should be greater than or equal to zero and

at constant entropy and volume the change in internal energy of a system will be Formula

 

From this equation 8 can be rearranged to:

Formula

 

Formula

 (equation 9)

at constant temperature and volume Formula

 

Formula where A= Helmholtz energy

 (equation 10)             

 

From equation 9 Formula it can be determined that:

 

For a spontaneous reaction at constant Temperature and Volume dA<0

generally either dU is decreased or TdS is increased to satisfy this condition.

But it should be remembered that entropy is the only critical factor in a spontaneous change.

 

 


Maxium work out of a system:

 

Can be found by using equation 7 where:

Formula and from equation 8:Formula

 

since Formula it can be found that:

 

Formularearranging this equation for dw yields: Formula and according to this

the maximum work is:

 

Formula

(equation 11)

 

Which states the maximum work is equal to the Helmholtz energy.

 

So the Helmholtz energy determines if a reaction is spontaneous (at constant volume and temperature)

and give the maximum amount of work of a system.


 

Sections in black done by szarco@gmu.edu 

 

Sections in reddish brown done by fwhitfie@gmu.edu 

 

 

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