ExamplE 2.12 (continued)
The last term is the conversion factor between joules and liter#atmospheres.
This is the ideal gas law.
The process is adiabatic, so q 5 0.
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2.8 | More on Heat Capacities 57
Using the properties of logarithms and evaluating each integral at its limits, we get 2R ln Vf
Vi 5CV ln Tf
Ti (2.44)
for an adiabatic, reversible change in an ideal gas. Again using properties of loga- rithms, we can get rid of the negative sign by taking the reciprocal of the expression inside the logarithm:
R ln Vi
Vf5CV ln Tf Ti
Recognizing that Cp5CV1R, we rearrange it as Cp2CV5R and substitute:
(Cp2CV ) ln Vi
Vf5CV ln Tf Ti Dividing through by CV:
(Cp2CV) CV ln Vi
Vf 5ln Tf
Ti The expression Cp/CV is usually defined as g:
g; Cp
CV (2.45)
We can rearrange the equation relating volumes and temperatures above to get aVi
Vfb
g21
5 Tf
Ti (2.46)
It can be shown that g2 1 equals 23 for a monatomic ideal gas. Thus, aVi
Vfb2/35 Tf
Ti (2.47)
for an adiabatic, reversible change of a monatomic ideal gas. If we did this in terms of pressure instead of volume, we would find that
apf
pib2/55 Tf
Ti (2.48)
If equations 2.47 and 2.48 were combined algebraically, one would derive p1V15/3 5p2V25/3 (2.49) for an adiabatic, reversible change of a monatomic ideal gas.
ExamplE 2.13
For an adiabatic, reversible change in 1 mole of an inert monatomic gas, the pres- sure changes from 2.44 atm to 0.338 atm. If the initial temperature is 339 K, what is the final temperature?
Solution Using equation 2.48,
a0.338 atm
2.44 atmb2/55 Tf
339 K Solving:
Tf5154 K
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58 Chapter 2 | The First Law of Thermodynamics
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2.9 Phase Changes
So far, we have discussed only physical changes in our system, like changes in pressure or temperature. Now we consider other changes, beginning with phase changes. Consider a substance such as H2O at 0 K and 1 atm. As heat is added to the H2O, its temperature will rise at a rate governed by its heat capacity. Figure 2.12 shows what happens in a qualitative way. At 0 K, the water is in the form of ice and has zero thermal energy. As we heat it up, it will continue to gain energy until it melts at 273 K. At this point, the addition of more heat does not increase its temperature because all of the energy goes into melting the ice until it is completely melted. This is a phase change. Once all the ice has melted, the temperature continues to rise until we reach the boiling point at 1 atm, which is 373 K. At the boiling point, we encounter the next phase change, from liquid ExamplE 2.14
For any gas, equation 2.48 can be written as apf
pib(
g21)/g
5 Tf Ti
For an adiabatic, reversible change in 1 mole of CO2, the pressure changes from 2.44 atm to 0.338 atm. If the initial temperature is 339 K, what is the final tempera- ture? Ignore any contributions from vibrational energy.
Solution
The major difference between this example and the previous one is that the value of g will be different. For a linear molecule, CV has contributions from translational ener- gies (CV (trans) 5 3/2R) and rotational energies (CV (rot) 5 R). Therefore, we have
CV5 5/2R Because Cp5 CV1 R, we also have
Cp5 7/2R Gamma, therefore, is
g 5 Cp CV5
7 2R
5 2R
5 7 5 The exponent in the expression is thus
g 21 g 5
7 521
7 5
5 2 7 Substituting:
a0.338 atm
2.44 atmb2/75 Tf 339 K Solving:
Tf 5 193 K
Note how different this answer is from the previous exercise’s answer.
5/2R is 20.78 J/mol?K.
7/2R is 29.10 J/mol?K.
You can verify this answer by using the numerical values for 5/2R and 7/2R from above.
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2.9 | Phase Changes 59 to vapor. Again, the temperature will remain constant during the phase change. (One
additional phase change, from solid to gas, does not normally occur with H2O under these conditions.)
In most cases, changes in phase (solid L liquid, liquid L gas, solid L gas) occur under experimental conditions of constant pressure, so that the heat involved, q, is also equal to DH.† For example, for the melting of ice at its normal melting point of 0°C
H2O (s, 0°C) h H2O (,, 0°C)
a certain amount of heat is required per gram or per mole in order to change the phase. However, during the phase change, the temperature does not change; phase changes are isothermal.
H2O can exist at 0°C as either a solid or a liquid. Because there is no DT, equation 2.9 does not apply. Instead, the amount of heat involved is proportional to the amount of material. The proportionality constant is called the heat of fusion, DfusH, so that we have a simpler equation:
q5m #DfusH (2.50)
The word fusion is a synonym for “melting.” If amount m is given in units of grams, DfusH has units of J/g. If the amount is given in units of moles, equation 2.50 is more properly written as
q5n#DfusH (2.51)
and DfusH is a molar quantity with units of J/mol. Because freezing and melting are simply opposite processes, equations 2.50 and 2.51 are valid for both processes. The process itself dictates whether the label exothermic or endothermic is appropriate.
For melting, heat must be put into the system, so the process is endothermic and the value of DH for the process is positive. For freezing, heat must be removed from the system, so freezing is exothermic and the value for DH is negative.
Figure 2.12 Qualitative behavior of the thermal energy and the heat capacity for water from 0 K to over 373 K. Both graphs start at a y value of zero.
DfusH Thermal
Energy
0 273 373
Temperature (K) Energy or Heat Capacity
CapacityHeat
DvapH
†Changes in pressure can also cause phase changes. We will consider this in Chapter 6.
ExamplE 2.15
The heat of fusion DfusH for water is 334 J/g.
a. How much heat is required to melt 59.5 g of ice (about one large ice cube)?
b. What is the value of DH for this process if it occurs at constant pressure?
Solution
a. According to equation 2.50,
q 5 (59.5 g)(334 J/g) q 5 1.99 3 104 J
b. Because heat must be put into the system in order to go from solid to liquid, the DH for this process should reflect the fact that the process is endothermic.
Therefore, DH 5 11.99 3 104 J.
DH 5 q at constant pressure.
Changes in volume when going from solid to liquid, or from liquid to solid, are usually negligible, so that DH<DU. (Water is an obvious exception. It expands approximately 10% when freezing.) On the other hand, the change in volume in going from a liquid to a gas (or a solid to a gas) is considerable:
H2O (,, 100°C) hH2O (g, 100°C)
In going from a liquid to a gas, a process called vaporization, again the temperature stays constant while the phase change occurs, and the amount of heat necessary is
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60 Chapter 2 | The First Law of Thermodynamics
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again proportional to the amount. This time, the proportionality constant is called the heat of vaporization, DvapH, but the form of the equation for calculating the heat involved is similar to equation 2.50:
q5m#DvapH for amounts in grams (2.52) or equation 2.51:
q5n#DvapH for amounts in moles (2.53) The heat involved in the reverse process, condensation, can also be calculated with equations 2.52 and 2.53 with the understanding that once again we will have to keep track of which direction heat is going. When determining work for a vaporization or sublimation, it is common to neglect the volume of the condensed phase, which is usually negligible. The following example illustrates.
ExamplE 2.16
Calculate q, w, DH, and DU for the vaporization of 1 g of H2O at 100°C and 1.00 atm pressure. The DvapH of H2O is 2260 J/g. Assume ideal gas behavior. The density of H2O at 100°C is 0.9588 g/cm3.
Solution
Using equation 2.52, the heat and DH for the process are straightforward:
q 5 (1 g)(2260 J/g) 5 2260 J into the system q 5 DH 5 12260 J
In order to calculate the work, we need the volume change for the vaporization. For the process H2O (,) SH2O (g), the change in volume is
DV 5 Vgas2 Vliq
Using the ideal gas law, we can calculate the volume of the water vapor at 100°C 5 373 K:
Vgas5
(0.0555 mol)a0.08205 L#atm
mol#Kb(373 K) 1 atm
Vgas 5 1.70 L
The volume of liquid H2O at 100°C is 1.043 cm3, or 0.001043 L. Therefore, DV 5 Vgas 2 Vliq 5 1.70 L 2 0.001043 L < 1.70 L 5 Vgas
In this step, we show that the volume of the liquid is negligible with respect to the volume of the gas, so to a very good approximation DV 5 Vgas. To calculate the work of the vaporization:
w 5 2pext DV
w 5 2(1.00 atm)(1.70 L)a101.32 J 1 L#atmb w 52172 J
Because DU 5 q 1 w,
DU 5 2260 J 2 172 J DU 5 2088 J