Baixe capitulo 1 peter atkins e outras Exercícios em PDF para Engenharia Química, somente na Docsity!
Thermodynamics: Examples for chapter 1.
- When 0.12 g of solid iodine was vaporized at 1670 K, the resulting gas displaced 20.2 cm^3 of dry air at 298 K and 99.9 kPa pressure. How many iodine molecules were dissociated (i.e. as atomic iodine) in the gas? The gas phase equilibrium reaction is:
I 2 (g) ⇀↽ 2I(g)
Assume that the gas mixture (of I 2 and I) behave according to the ideal gas law. Solution:
The amount of I 2 before any dissociation takes place is:
n =
m MI 2
0 .12 g 254 g mol−^1
= 4. 7 × 10 −^4 mol
After the equilibrium has been reached, the amount of I 2 is (1 − α)n and the amount of I is 2αn. Here the degree of dissociation is denoted by α. The ideal gas law is:
P V = ntotRT ⇒ ntot =
P V
RT
⇒ ntot = (1 − α)n + 2αn =
P V
RT
⇒ α =
P V
nRT
P V MI 2
mRT
(99.9 kN m−^2 ) × (20. 2 × 10 −^6 m^3 ) × (254 g mol−^1 ) (0.12 g) × (8.31 J K−^1 mol−^1 ) × (298 K)
The amount of I is 2αn = 2 × 0. 72 × (4. 7 × 10 −^4 mol) = 6. 8 × 10 −^4 mol.
- What is the molar volume of n-hexane at 660 K and 91 bar according to (a) the ideal gas law and (b) the van der Waals equation? For n- hexane, Tc = 507.7 K and Pc = 30.3 bar. If you obtain an equation that you cannot solve analytically, attempt to solve it numerically. Solution:
a)
V¯ = RT
P
(0.08314 L bar K−^1 mol−^1 )(660 K) 91 bar
= 0.603L mol−^1
b)
a =
27 R^2 T (^) c^2 64 Pc
27(0.08314 L bar K−^1 mol−^1 )^2 (507.7 K)^2 64(30.3 bar)
= 24.81 L^2 bar mol−^2
b =
RTc 8 Pc
(0.08314 L bar K−^1 mol−^1 )(507.7 K) 30 .3 bar
= 0.174 L mol−^1
P =
RT
V^ ¯ − b −^
a V^ ¯ 2 Test values for V¯ and see when you get 91 bars: V¯ = 0.39 L mol−^1. This equation can also be solved by Maxima:
a : 24.81; b : 0.174; R : 0.08314; T : 660; P : 91; r : solve(P = RT/(V - b) - a/(VV), V); float(r);
This shows three roots of which only one is real (two are complex). The last line converts the complicated algebraic form to numerical values.
- The critical temperature of gaseous ammonia is 406 K and the critical pressure is 113 bar. Use the van der Waals equation to predict its molar volume at 273 K and 1 bar. Hint: Calculate first the coefficients a and b, then write the equation for the molar volume: V¯ = (^) P +RTa/ V¯ 2 +b. Guess an initial value for the molar volume and obtain a new value from the equation. Place the obtained value in the equation again to get a new value. Repeat this procedure until the value does not change anymore. This is an example of iterative methods for solving equations. Solution:
Use the following equations derived in the lecture notes:
a =
27 R^2 T (^) c^2 64 Pc
and b =
RTc 8 Pc
This gives a = 425 L^2 kPa mol−^2 and b = 0.0373 L mol−^1. Use the above equation and iterate until you get convergence: V¯ = 22.55 L mol−^1.
- Derive the expressions for Vc , Tc and Pc in terms of the van der Waals constants a and b. Solution:
The equations in the lecture notes can be rewritten as (the definition of critical point and the equation of state):
RTc ( V¯c − b
2 a V^ ¯ (^) c^3
2 RTc ( (^) ¯ Vc − b
6 a V^ ¯ (^) c^4
Pc =
RTc V^ ¯c − b −^
a V^ ¯ (^) c^2
Division of the first equation by the second (side by side) yields V¯c = 3b. Substitution of this expression into the first equation above gives Tc =
8 a 27 Rb. Substitution of the two previous equations into the last equation above yields Pc = 27 ab 2. Note that the van der Waals constants a and b are usually calculated using the critical temperature and pressure since they are typically known more accurately than the critical volume.
- The isothermal compressibility κ of a gas is defined as:
κ = −
V
∂V
∂P
T Calculate κ for a van der Waals gas using the method of implicit differ- entiation (with respect to P ). Show that in the limit of infinite volume, it yields 1/P (the same result as the ideal gas law). Solution:
Based on the lecture notes, the van der Waals equation can be written: ( P +
a V^ ¯ 2
V¯ − b
= RT
Expand the left hand side and substitute V¯ = V /n to get:
nRT = P V − nP b +
n^2 a V
n^3 ab V 2 Implicit differentiation with respect to P at constant T gives (i.e., side by side):
0 = V + P
∂V
∂P
T
− nb −
n^2 a V 2
×
∂V
∂P
T
2 n^3 ab V 3
×
∂V
∂P
T
Solving for
(∂V
∂P
T gives: ( ∂V ∂P
T
nb − V P − n^2 a/V 2 + 2n^3 ab/V 3
a) Calculate the following partial derivatives: ( ∂P ∂V
T
and
∂P
∂T
V b) Show that the following relation holds for the mixed partial deriva- tives for this equation of state: ( ∂^2 P ∂V ∂T
∂^2 P
∂T ∂V
Solution:
a) (^) ( ∂P ∂V
T
nRT (V − nb)^2
and
∂P
∂T
V
nR V − nb
b) (^) ( ∂^2 P ∂V ∂T
nR (V − nb)^2
and
∂^2 P
∂T ∂V
nR (V − nb)^2
- Calculate the compressibility factor Z for NH 3 (g) at 400 K and 50 bar by using the attached graph and the virial equation. Solution:
By reading the value for B from the graph, using relation B′^ = B/RT , and the virial equation written in terms of P , we get:
Z = 1 + B′P + ... = 1 +
B
RT
P + ...
(− 100 × 10 −^3 L mol−^1 ) (0.0831451 bar L K−^1 mol−^1 )(400 K)
× (50 bar) = 0. 85
