The Entropy Change of Mixing Ideal Gases
Section 3.3: The Calculation of Entropy Changes
3.13 a. Calculate∆Sfor each step of the following cycle and sum the results to obtain∆Sfor the cycle: Step 1:
1.000 mol of helium is expanded reversibly and
isothermally at 298.15 K from 10.00 L to 20.00 L.
Step 2: The gas is heated reversibly at a constant volume from 298.15 K and 20.00 L to a temperature of 473.15 K. Step 3: The gas is compressed reversibly and isothermally at 473.15 K from 20.00 L to 15.00 L.
Step 4: The gas is cooled reversibly at a constant volume of 15.00 L from 473.15 K to 373.15 K. Step 5:
The gas is compressed reversibly and isothermally at 373.15 K from a volume of 15.00 L to a volume of 10.00 L. Step 6: The gas is cooled reversibly at a constant volume of 10.00 L from 373.15 K to 298.15 K.
b. Repeat the calculation with all steps the same as in part a except that step 1 is carried out isothermally and irreversibly with a constant external pressure of 1.000 atm.
3.14 A sample of 2.000 mol of nitrogen gas (assume ideal withCV,m 5R/2) expands adiabatically and irreversibly from a volume of 8.000 L and a temperature of 500.0 K to a volume of 16.000 L against an external pressure of 1.000 atm. Find the final temperature,∆U,q,w, and∆S for this process. Find the initial and final pressures.
3.15 A sample of 1.000 mol of helium gas (assumed ideal with CV,m3R/2) expands adiabatically and irreversibly from a volume of 3.000 L and a temperature of 500. K to a volume of 10.00 L against a constant external pressure of 1.000 atm. Find the final temperature,∆U,q,w, and
∆Sfor this process. Compare each quantity with the corresponding quantity for a reversible adiabatic expansion to the same final volume.
3.16 The normal boiling temperature of ammonia is−33◦C, and its enthalpy change of vaporization is 24.65 kJ mol−1. The density of the liquid is 0.7710 g mL−1.
a. Calculateq,w,∆U,∆H,∆S, and∆Ssurrif 1.000 mol of ammonia is vaporized at−33◦C. Use the ideal gas equation of state to estimate the volume of the gas and neglect the volume of the liquid.
b. Repeat the calculations of part a, using the van der Waals equation of state to find the molar volume of the gas under these conditions (use successive approximations or other numerical procedure to solve the cubic equation) and without neglecting the volume of the liquid.
3.17 a. Find the change in entropy for the vaporization of 2.000 mol of liquid water at 100◦C and a constant pressure of 1.000 atm.
b. Find the entropy change for the heating of 2.000 mol of water vapor at a constant pressure of 1.000 atm from 100◦C to 200◦C. Use the polynomial represen- tation in Table A.6 for the heat capacity of water vapor.
3.18 a. Calculate the entropy change for the isothermal expansion of 1.000 mol of argon gas (assume ideal) from a volume of 5.000 L to a volume of 10.000 L.
b. Calculate the entropy change for the isothermal expansion of 1.000 mol of argon gas (assume ideal) from a volume of 10.000 L to a volume of 15.000 L.
c. Explain in words why your answer in part b is not the same as that of part a, although the increase in volume is the same.
3.19 a. 1.000 mol of helium is compressed reversibly and isothermally from a volume of 100.00 L and a temperature of 298.15 K to a volume of 50.00 L.
Calculate∆S,q,w, and∆Ufor the process. Calculate
∆Ssurrand∆Suniv.
b. Calculate the final temperature,∆S,q,w,∆U,∆Ssurr, and∆Sunivif the gas is compressed adiabatically and reversibly from the same initial state to a final volume of 50.00 L.
c. The gas is compressed adiabatically and irreversibly from the same initial state to the same final volume withPext 1.000 atm. What can you say about the final temperature,∆S,q,w,∆U,∆Ssurr, and∆Suniv? d. The gas is compressed isothermally and irreversibly
from the same initial state to the same final volume withPext1.000 atm. What can you say about∆S,q, w,∆U,∆Ssurr, and∆Suniv?
3.20 2.000 mol of helium is expanded adiabatically and irreversibly at a constant external pressure of 1.000 atm from a volume of 5.000 L and a temperature of 273.15 K to a volume of 25.000 L. Calculate∆S,∆Ssurr, and∆Suniv. State any approximations or assumptions.
3.21 a. Calculate the entropy change for the following reversible process: 2.000 mol of neon (assume ideal withCV,m3R/2) is expanded isothermally at 298.15 K from 2.000 atm pressure to 1.000 atm pressure and is then heated from 298.15 K to 398.15 K at a constant pressure of 1.000 atm. Integrate on the path representing the actual process.
b. Calculate the entropy change for the reversible process with the same initial and final states as in part a, but in which the gas is first heated at constant pressure and then expanded isothermally. Again, integrate on the path representing the actual process. Compare your result with that of part a.
c. Calculate the entropy change of the surroundings in each of the parts a and b.
d. Calculate the entropy changes of the system and the surroundings if the initial and final states are the same as in parts a and b, but if the gas is expanded
irreversibly and isothermally against an external pressure of 1.000 atm and then heated irreversibly with the surroundings remaining essentially at equilibrium at 400 K.
3.22 Find∆H,∆S, andqfor the reversible heating of 0.500 mol of benzene from 25◦C to 100◦C at a constant pressure of 1.000 atm. The normal boiling temperature of benzene is 80.1◦C and the enthalpy change of vaporization is 30.8 kJ mol−1.
3.23 a. A sample of 2.000 mol of neon is expanded reversibly and adiabatically from a volume of 10.00 L and a temperature of 500.0 K to a volume of 25.00 L. Find the final temperature,q,w,∆U,∆S, and∆Sunivfor the process. State any assumptions or approximations.
b. The same sample is restored to its original state and is first expanded adiabatically and irreversibly at a constant external pressure of 1.000 atm to a volume of 25.00 L, then cooled reversibly to the same final temperature as in part a at a constant volume of 25.00 L. Find the final temperature for the irreversible step, and findq,w,∆U, and∆Sfor this entire
two-step process. What can you say about∆Sunivfor each step of this two-step process?
3.24 1.000 mol of carbon tetrafluoride is vaporized at the normal boiling point of−128◦C and 1.000 atm (bothPandT held constant). The molar volume of the liquid is 44.9 cm mol−1. The enthalpy change of vaporization is 12.62 kJ mol−1. Findq,w,∆S, and∆Hfor this process.
3.25 A sample of 2.000 mol of a monatomic ideal gas is expanded and heated. Its initial temperature is 300.0 K and its final temperature is 400.0 K. Its initial volume is 20.00 L and its final volume is 40.00 L. Calculate∆S.
Does the choice of path between the initial and final states affect the result?
3.26 The molar heat capacity of water vapor is represented by CP,m30.54 J K−1mol−1+(0.01029 J K−2mol−1)T whereTis the absolute temperature.
a. Find∆Hfor heating 2.000 mol of water vapor at a constant pressure of 1.000 atm from 100◦C to 500◦C.
b. Findq,w, and∆Ufor the process.
c. Find∆Sfor the process.
3.4 Statistical Entropy
Since molecules can occupy various states without changing the macroscopic state of the system of which they are a part, it is apparent that many microstates of a macroscopic system correspond to one macroscopic state. We denote the number of microstates corresponding to a given macrostate byΩ. The quantityΩis sometimes called the thermodynamic probabilityof the macrostate. The thermodynamic probability is a measure of lack of information about the microstate of the system for a particular macrostate. A large value corresponds to a small amount of information, and a value of unity corresponds to knowledge that the system is in a specific microstate.
For a macrostate specified by values ofU,V, andn, Boltzmann defined astatistical entropy:
SstkBln(Ω)+S0 (definition of statistical entropy) (3.4-1) wherekBis Boltzmann’s constant, equal toR/NAv. The constantS0is a constant that we usually set equal to zero. The task of this section is to show that the statistical entropy can be equivalent to the thermodynamic entropy for a model system that is called a lattice gas. This system contains a numberNof noninteracting point-mass molecules confined in a container with volumeV. We assume that the particles obey classical mechanics. The lattice gas is in an equilibrium macrostate specified by values ofU,V andN. The microstate of the lattice gas is specified by the positions and velocities of all of the particles.
We now specify the positions of the particles in a new way. We mentally divide the volume of the box into a numberMof rectangular cells of equal size stacked in a three-dimensional array (lattice) as shown schematically in Figure 3.11. Since the cell boundaries are imaginary the molecules pass freely through them. Instead of giving the values of three coordinates to specify the location of one particle we specify which cell it occupies. This“coarse-grained” descriptiongives less precise information about the location of the particles than specifying exact coordinate values, but we can increase the precision by making the cells smaller. We continue to specify the velocity of each particle by specifying the values of three velocity components.
Figure 3.11 The Lattice Gas.
We assume that the probability that a randomly chosen molecule has a particular velocity is independent of the probability that it has a particular position. It is a fact of probability theory that the number of ways of accomplishing two independent events is the product of the number of ways of accomplishing each event. The thermodynamic probability is therefore the product of two factors, one for the coordinates and one for the velocities:
ΩΩcoordΩvel (3.4-2)
We now seek a formula to representΩcoordfor our lattice gas. The number of possible coordinate states for a single molecule is equal to the number of cells,M. Since the molecules are mass points, the presence of one molecule in a cell does not keep other molecules from occupying the same cell. Any state of a second molecule can occur with any state of the first molecule, so the number of possible coordinate states for two molecules isM2. Any state of a third molecule can occur with any state of the first pair of molecules, so the number of possible states for three particles isM3. For a system ofNmolecules,
ΩcoordMN (3.4-3)
E X A M P L E 3.13
Calculate the value ofΩcoordand ln(Ωcoord) for 1.000 mol of an ideal gas in a volume of 24.4 L if cells of 0.500 nm on a side are taken.
Solution
Vcell(0.500×10−9m)31.25×10−28m3 M V
Vcell 0.0244 m3
1.25×10−28m3 1.95×1026 ΩcoordMN(1.95×1026)6.02×1023101.58×1025 This is such a large number that we are not able to give any significant digits.
ln(Ωcoord)6.02×1023ln(1.95×1026)3.65×1025
The value ofΩcoordin Example 3.10 is calculated on the assumption that the particles are distinguishable from each other. When quantum mechanics is studied, it is found that identical particles must be treated as inherently indistinguishable from each other.
The number of ways of rearrangingNdistinguishable objects isN! (Nfactorial), which
is defined as the product of all of the integers starting withNand ranging down to unity:
N!N(N−1)(N−2)(N−3) … (2)(1) (definition) (3.4-4) If the particles are actually indistinguishable, we have overcounted the value ofΩby this factor. We should replace the expression in Eq. (3.4-3) by
Ωcoor MN
N! (3.4-5)
so that
ln(Ωcoor)Nln(M)−ln(N!) (3.4-6)
For very large values ofN the use of the exact expression forN! is inconvenient.
For fairly large values ofNwe can applyStirling’s approximation:
N!≈(2πN)1/2NNe−N (3.4-7)
ln(N!)≈1
2ln(2πN)+Nln(N)−N (3.4-8)
For very large values ofNwe can neglect the first term on the right-hand side of this equation:
ln(N!)≈Nln(N)−N (3.4-9)
With this approximation
ln(Ω)≈Nln(M)−Nln(N)+N (3.4-10)
E X A M P L E 3.14
Find the value of ln(Ωcoord) in Example 3.13 using Eq. (3.4-8).
Solution
ln(Ωcoord)3.64×1025−(6.022×1023) ln(6.022×1023)+6.022×1023 4.03×1024
Ωcoorde4.03×1024101.75×1024
Exercise 3.14
a. List the 36 possible states of two dice and give the probability for each sum of the two numbers showing in the upper faces of the dice.
b. Determine how many possible states occur for four dice.
c. Determine how many possible states occur for two “indistinguishable” dice, which means that there is no difference between a four on the first die and a five on the second die, or between a five on the first die and a four on the second, etc. Explain why the correct answer is not equal to 18.