nsider 2.0 moles of methane (CH4 ), initially at 273 K, that is reversibly and adiabatically compressed from an initial volume of 10.00 L to a final volume of 2.00 L. 3a. Determine the final temperature, considering the gas to be described by the ideal gas equation of state. 3b. Determine the work required w, the associated heat flow q, and the internal energy change ∆U for the gas between the initial and final states.

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
Question

Physical chemistry - please refer to pics

Now, consider 2.0 moles of methane (CH4 ), initially at 273 K, that is reversibly and adiabatically compressed from an initial volume of 10.00 L to a final volume of 2.00 L.

3a. Determine the final temperature, considering the gas to be described by the ideal gas equation of state.

3b. Determine the work required w, the associated heat flow q, and the internal energy change ∆U for the gas between the initial and final states.

Integrals
r'dx
;n -1
n+1
Inx
-dx
ax -b
-In(ax - b)
=
Nomenclature – but, when in doubt, write it out!
AG Change in Gibbs free energy for an arbitrary process (energy)
AGm Change in Gibbs free energy for an arbitrary process, per mole (energy/mole)
AGn Molar Gibbs free energy change of chemical equation = (@G/Ə5)r.p (energy/mole)
AGan Standard Gibbs free change of chemical equation, per mole, between separated products and reactants, each in
the standard states for the phases specified. (energy/mole)
van der Waals equation of state
RT
an
V - nb
Virial equation of state
RT
p=
Vm
B
1+
+...
Vn V
Fundamental equations of thermodynamics
(constant composition)
Maxwell relations
(),--),
- (4),--),
(), --(#),
- (),--4),
From dU :
dU = TdS - pdv
av
as
TdS+V dp
-SdT – pdV
dA
From dH :
+
=+
se
P
dG
= -SdT +V dp
From dA:
+
as
From dG:
= +
Transcribed Image Text:Integrals r'dx ;n -1 n+1 Inx -dx ax -b -In(ax - b) = Nomenclature – but, when in doubt, write it out! AG Change in Gibbs free energy for an arbitrary process (energy) AGm Change in Gibbs free energy for an arbitrary process, per mole (energy/mole) AGn Molar Gibbs free energy change of chemical equation = (@G/Ə5)r.p (energy/mole) AGan Standard Gibbs free change of chemical equation, per mole, between separated products and reactants, each in the standard states for the phases specified. (energy/mole) van der Waals equation of state RT an V - nb Virial equation of state RT p= Vm B 1+ +... Vn V Fundamental equations of thermodynamics (constant composition) Maxwell relations (),--), - (4),--), (), --(#), - (),--4), From dU : dU = TdS - pdv av as TdS+V dp -SdT – pdV dA From dH : + =+ se P dG = -SdT +V dp From dA: + as From dG: = +
Useful info
This and the following page contain physical constants, unit conversions, and chemical reference data that will be
essential to several of the problems on the exam. You may separate them from your exam for convenience, and they
do not need to be submitted with your answers.
NA =6.022×10 mol?
g = 9.807 m/s
Pm[H,O] =1.00 g/mL
1 bar = 10° Pa
1 atm =1.013bar
R = 8.314 J/K-mol
= 8.206x10 L-atm/K-mol
= 8.206x10° m-atm/K-mol
= 760 Torr
1 mbar = 10 bar
= 62.36 L-Torm/K-mol
1J = 10² L-bar = 9.870×10° L-atm= 0.2390 cal
1m = 10° cm = 10° mm = 10° µum = 10° nm = 1010 A
1 L-atm = 101.3 J
1 m = 10° L = 10° cm
1 kg = 1000 g
1 Watt (W) = 1 J/s
1 Volt (V) = 1 J/Coulomb
1 Coulomb (C) = 6.24 x 1018 elementary charges
F = 9.645 x 10 C/mol
At 298 K (only):
RT = 2.478 kJ/mol (= 24.46 L-atm/mol)
RT/F = 0.0257 V (= 25.7 mV)
Physical Properties of Selected Substances
mol. wt.
AH
Cp.m
AH
kJ/mol
melting temp AHap
kJ/mol
boiling temp
3,m
"fus
g/mol
44.0
kJ/mol
J/(mol-K) J/(mol-K)
K
K
CO,(g)
CH,(g)
CH,OH(I)
CH,OH(g)
n-C,H4()
H,(g)
H,O(1)
H,O(g)
Hg(1)
1,(8)
L(g)
HI(g)
O,(g)
Properties of pure phases are for standard conditions of 298 K and I atm.
Phase transition enthalpies are valid at the indicated phase transition temperature, at 1 atm.
-393.5
37.1
16.04
-74.8
35.3
0.94
91
8.2
112
32.04
-238.66
81.6
3.16
175.2
35.29
337.2
32.04
-200.66
43.9
86.2
-198.7
204.3
13.1
178
28.9
342
2.0
116.1
28.8
18.0
-285.8
69.9
75.3
6.0
273.15
40.6
373.15
18.0
-241.8
188.8
33.6
200.6
76.0
2.3
234.3
59.3
629.7
253.8
116.14
54.4
15.5
386.8
41.8
458.4
253.8
+62.44
260.69
36.9
127.91
+26.48
206.6
29.2
32.0
205.1
29.4
Transcribed Image Text:Useful info This and the following page contain physical constants, unit conversions, and chemical reference data that will be essential to several of the problems on the exam. You may separate them from your exam for convenience, and they do not need to be submitted with your answers. NA =6.022×10 mol? g = 9.807 m/s Pm[H,O] =1.00 g/mL 1 bar = 10° Pa 1 atm =1.013bar R = 8.314 J/K-mol = 8.206x10 L-atm/K-mol = 8.206x10° m-atm/K-mol = 760 Torr 1 mbar = 10 bar = 62.36 L-Torm/K-mol 1J = 10² L-bar = 9.870×10° L-atm= 0.2390 cal 1m = 10° cm = 10° mm = 10° µum = 10° nm = 1010 A 1 L-atm = 101.3 J 1 m = 10° L = 10° cm 1 kg = 1000 g 1 Watt (W) = 1 J/s 1 Volt (V) = 1 J/Coulomb 1 Coulomb (C) = 6.24 x 1018 elementary charges F = 9.645 x 10 C/mol At 298 K (only): RT = 2.478 kJ/mol (= 24.46 L-atm/mol) RT/F = 0.0257 V (= 25.7 mV) Physical Properties of Selected Substances mol. wt. AH Cp.m AH kJ/mol melting temp AHap kJ/mol boiling temp 3,m "fus g/mol 44.0 kJ/mol J/(mol-K) J/(mol-K) K K CO,(g) CH,(g) CH,OH(I) CH,OH(g) n-C,H4() H,(g) H,O(1) H,O(g) Hg(1) 1,(8) L(g) HI(g) O,(g) Properties of pure phases are for standard conditions of 298 K and I atm. Phase transition enthalpies are valid at the indicated phase transition temperature, at 1 atm. -393.5 37.1 16.04 -74.8 35.3 0.94 91 8.2 112 32.04 -238.66 81.6 3.16 175.2 35.29 337.2 32.04 -200.66 43.9 86.2 -198.7 204.3 13.1 178 28.9 342 2.0 116.1 28.8 18.0 -285.8 69.9 75.3 6.0 273.15 40.6 373.15 18.0 -241.8 188.8 33.6 200.6 76.0 2.3 234.3 59.3 629.7 253.8 116.14 54.4 15.5 386.8 41.8 458.4 253.8 +62.44 260.69 36.9 127.91 +26.48 206.6 29.2 32.0 205.1 29.4
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

Blurred answer
Knowledge Booster
Thermodynamics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY