KEY POINTS FOR CHAPTER SEVEN
8.2 CONSTANT PRESSURE PROCESSES
For any closed system process in which the system changes from State 1 to State 2,
H = H2 – H1 = (U2 + P2V2) – (U1 + P1V1) = (U2 – U1) + (P2V2 – P1V1) = U + (PV) H = U + (PV) . . . [5] But U = Q + W; therefore H = Q + W + (PV)
Suppose the pressure of the system is kept constant and is the same as the pressure of the sur- roundings; furthermore, if the only type of work done is PV-work, then,
W = –Psurr Vsyst = –Psyst Vsyst
and H = Q + W + P V
= Q – P V + P V = Q
Thus, for a constant pressure process in which Psyst = Psurr:
HP = QP Constant PressureProcess, Only PV-Work . . . [6]
QP is the heat for the process; we use the subscript P to remind ourselves that the process is a con- stant pressure process. Therefore, for a constant pressure process where Psyst = Psurr and only PV- work is done against the surroundings, the change in enthalpy for the process is just the heat! If the constant pressure process is just a heating or cooling process (no chemical reaction or phase changes in the system), then we know that the heat is just given by the familiar QP = CP T, and
Const P, only PV-Work, Simple
Heating or Cooling Process HP = QP = CP T = ncP T = mcP T . . . [7]
For constant pressure systems in which only solids or liquids are involved, usually (PV) = P V is very small because such systems are more or less incompressible ( V 0), so that
H = U + (PV) U
Therefore, for processes involving small volume changes,
H U Solid/Liquid Systems,Constant Pressure . . . [8]
On the other hand, for reactions in which gases are involved, V may be very large. For example, the volume at STPof one mole of liquid water is only about 18 cm3, while the volume at STPof one
mole of water vapor is about 22 400 cm3. Since the former volume is negligible when compared
with the latter, we can say that 22 400 + 18 22 400 without introducing any significant error. Therefore, for reactions involving gases, it is usually accurate to say
Vreaction Vgases . . . [9]
For a real gas, H = CP T only when the gas is heated or cooled at constant pressure. However,
when an ideal gas is heated or cooled, we still use H = CP T, even though the pressure may not be constant during the process. This is a consequence of the fact that for an ideal gas the enthalpy
is determined only by the temperature. 2 Similarly, for the same reason, for the heating or cooling
of an ideal gas U = CV T even though the volume may not be constant during the process. For
most real gases as well, to a good approximation, when we heat or cool a real gas we also can say that H CP T and U CV T, even though the pressure and volume may change during the
process.
Example 8-1
One mole of nitrogen gas is compressed from an initial state of 1.00 bar and 25°C to 10.0 bar and 100°C. Assuming the gas behaves as an ideal gas, what is H for this process? For N2 gas,
cV = 20.81 J mol–1 K–1.
Solution
For one mole of an ideal gas, H = cP T and cP = cV + R;
therefore: H = (cV + R) T
= (20.81 + 8.314)(100 – 25) = 2184 J = 2.184 kJ.
Ans: 2.18 kJ
This problem also can be solved as follows:
H = U + (PV)
= U + P2V2 – P1V1
= U + nRT2 – nRT1
= U + nR T
Note how, for an ideal gas, H depends only on temperature and not on pressure! For one mole of an ideal gas, U = cV T;
so that U = (20.81)(100 – 25) = 1560.8 J
Therefore: H = U + nR T
= 1560.8 + (1)(8.314)(100 – 25) = 1560.8 + 623.55
= 2184 J = 2.184 kJ (Same answer as before.)3
Example 8-2
One mole of gaseous A is injected at 25.0°C into a rigid reactor having a total constant volume of
V = 1.00 L, and initially containing 2.00 moles of liquid B. The reactor is sealed and the following
chemical reaction proceeds to completion, with the evolution of 285.0 kJ of heat:
A(g) + 2B(liq) 3C(s) + 2D(g)
At the completion of the reaction the temperature in the reactor is 500 K. The molar volumes of liq- uid B at 25.0°C and of solid C at 500 K are 24.0 mL and 14.2 mL, respectively. What is H for this process? The vapor pressures of liquid B and solid C may be assumed to be negligible.
Solution
The volume VB of liquid B initially present is
VB= nBvB = (2 mol)(24.0 mL mol–1) = 48.0 mL
Therefore the initial volume occupied by gas A is
VA = V – VB = 1000 – 48.0 = 952 mL = 952 × 10–6 m3
and the initial pressure in the reactor is
P1= PA = n RT V A 1 A = ( .1 00 8 314 298 15)( . )( . ) 952 10 6 × = 2 604 10 6 . × Pa
Since the reaction proceeds to completion, from the reaction stoichiometry 3.00 moles of solid C and 2.00 moles of gaseous D will be produced.
The final volume of gaseous D in the reactor will be
VD = V – VC = 1000 – 3 × 14.2 = 957.4 mL = 957.4 × 10–6 m3
and the final pressure P2 will be
P2 = PD = n RT V D 2 D = ( . )( . )( ) . 2 00 8 314 500 957 4 10× 6 = 8 684 10 6 . × Pa
Since the volume of the system is constant, V = 0 and therefore W = WPV = 0. So, U = Q + W = Q = –285.0 kJ and H = U + (PV) = U + P2V2 – P1V1 But V2 = V1 = V = constant, therefore, H = U + (P2 – P1)V = –285 000 + (8.684 × 106 – 2.604 × 106)(0.00100) = –285 000 + 6084 = –278 916 J = – 278.9 kJ Ans: – 278.9 kJ Exercise 8-1
One mole of a real gas initially at a pressure of 1.98 atm and contained in a flask having a volume of 8.27 L is suddenly expanded by opening a valve connecting the flask to a second flask, the latter having a volume of 7.64 L and containing a vacuum. During the process, 0.71 J of heat is ab- sorbed by the gas from the surroundings. The final pressure of the gas is 1.033 atm. (a) What is U for this process? (b) What is H?
8.3 THERMOCHEMISTRY
Thermochemistry focuses on the heats associated with chemical reactions. Using as a basis the formulas U = Q + W, H = U + (PV), UV= QV, and HP= QP, it is possible to find H for any process, including chemical reactions.
For now, we shall deal with the enthalpy changes that occur during phase changes taking place at constant pressure. Commonly encountered phase changes include:
liquid (liq) vapor (V) solid (S) liquid (liq)
solid (S) vapor (V) solid A solid B 4
4 A typical example of solid A solid B would be the transition at some temperature of one crystalline structure to