Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Thermodynamics Problems: Equations of State, Ideal Gases, and Real Gases - Prof. Alan R. E, Exams of Physical Chemistry

Various thermodynamics problems involving equations of state for ideal and real gases, isothermal, isobaric, and isometric processes, and calculating values of constants a and b for the van der waals equation.

Typology: Exams

2010/2011

Uploaded on 10/16/2011

ames508
ames508 🇺🇸

1 document

1 / 20

Toggle sidebar

Related documents


Partial preview of the text

Download Thermodynamics Problems: Equations of State, Ideal Gases, and Real Gases - Prof. Alan R. E and more Exams Physical Chemistry in PDF only on Docsity! Page 1 of 20 Chemistry 3615 First Exam – Version A September 28, 2011 _________Answer Key__________ Name Page 2 of 20 Chemistry 3615 First Exam September 28, 2011 _____________________________ Name Answer Sheet Circle the Correct Letter. If you change your mind, completely erase the wrong answer if you are using pencil. If you are using pen completely black out the wrong answer and clearly indicate your choice for the correct answer. If you need additional scratch paper, just ask. 1. A B C D E 2. A B C D E 3. A B C D E 4. A B C D E 5. A B C D E 6. A B C D E 7. A B C D E 8. A B C D E 9. A B C D E 10. A B C D E 11. A B C D E 12. A B C D E 13. A B C D E 14. A B C D E 15. A B C D E 16. A B C D E 17. A B C D E 18. A B C D E 19. A B C D E 20. A B C D E Score = 5 x ______________ = _____________ Overall = _______________ PS Ave Raw = ________ Low PS = ______ PS Ave = _______________ Q Ave Raw = ________ Low Q = ______ Q Ave = _______________ Page 5 of 20 Other Useful Information: Table 1.5 contains experimental data. Table 1.6 contains values that are specific to the van der Waals equation of state. Page 6 of 20 1. Using the graphs on the next two pages and the information provided the data tables at the start of the exam, what is the value of the molar volume of oxygen vapor if the pressure is 1755 atm and the temperature is 310 K? (A) 0.0060 L/mol (B) 0.035 L/mol (C) 0.61 L/mol (D) 3.5 L/mol (E) 62 L/mol . . . Hence, Z = 2.4 . . ∙ ∙ ∙ . / Answer = (B) 2. Consider the following equation of state: 2 What is Tmax for this equation of state? (A) (B) (C) (D) (E) , Answer = (A) Page 7 of 20 N ot e th at s ol id li n es r ep re se n t re d u ce d t em p er at u re s. Page 10 of 20 5. What is the molar isobaric heat capacity of carbon monoxide in the limit of infinite temperature assuming ideal gas behavior? (A) (3/2)R (B) (5/2)R (C) (7/2)R (D) (9/2)R (E) (11/2)R As a diatomic molecule, carbon monoxide is linear. As such, , For an ideal gas , , , , , , 6. A 1.000 L container has 75.75 mg of an unknown gas. If the pressure in the container is 10.00 kPa and the temperature is 427.0 °C, which of the following gases could it be assuming ideal gas behavior? [from periodic table: MC = 12.011 g/mol, MH = 1.008 g/mol] (A) CH4 (B) C2H6 (C) C3H8 (D) C4H10 (E) C5H12 . . . . . . . . . . . ∙ . . . . All gases are of the form: CxHx+2 so . . . . Answer = (C) Page 11 of 20 7. Figure I contains two isotherms (T = 600K), one for a perfect gas (PV=nRT) and one for a gas governed by Equation of State 1 (b 0). Figure II contains two isotherms (T = 300K), one for a perfect gas and one for a gas governed by Equation of State 2 (a > 0). (I) Equation of State 1 p V (II) Equation of State 2 p V Which of the following statements is correct for any isothermal reversible expansion with the same initial and final molar volumes for the ideal and real gas in each case? (A) The molar work done by the ideal gas is larger for both Figure I and Figure II. (B) The molar work done by the ideal gas is larger for Figure I, but the molar work done by the real gas is larger for Figure II. (C) The molar work done by the real gas is larger for Figure I, but the molar work done by the ideal gas is larger for Figure II. (D) The molar work is done by the real gas is larger for both Figure I and Figure II. (E) There is insufficient information to answer this problem. Work = area under the curve. For Equation of State 1, the upper curve is the real gas as the denominator is smaller for a real gas than an ideal gas undergoing the same compression step. As a consequence, more work is done by the system expandng for the real gas. For Equation of State 2, the lower curve is the real gas as the first term is an ideal gas, and something is subtracted for all molar volumes. Hence, more work is done by the system expanding for the ideal gas. Answer = (C) Page 12 of 20 8. Consider the following unbalanced reaction. 1 SiCl4 (g) + 4 H2O (g) → 1 Si(OH)4 (g) + 4 HCl(g) If 8.49 grams of SiCl4 and 5.40 grams of H2O react until the limiting reagent is consumed inside a 0.500 m3 reaction vessel, what is the partial pressure exerted by the residual reactants if all gases behave ideally and isothermally at 450 °C? [from periodic table: MH = 1.008 g/mol, MO = 16.00 g/mol, MCl = 35.45 g/mol, MSi = 28.086 g/mol] (A) 601 Pa (B) 1.20x103 Pa (C) 2.40x103 Pa (D) 3.01x103 Pa (E) 4.20x103 Pa Moles SiCl4 = (8.49 g)/(169.886 g/mol) = 0.0500 mol Moles H2O = (5.40 g)/(18.016 g/mol) = 0.300 mol As one requires 4 mol H2O/mol SiCl4 for the reaction SiCl4 is limiting (none left) Moles H2O consumed = 0.0500 mol SiCl4 x (4 mol H2O/1 mol SiCl4) = 0.200 mol Moles H2O remaining = 0.300 mol – 0.200 mol = 0.100 mol Moles of Si(OH)4 formed = 0.0500 mol SiCl4 x (1 mol Si(OH)4/1 mol SiCl4) = 0.0500 mol Moles of HCl formed = 0.0500 mol SiCl4 x (4 mol HCl/1 mol SiCl4) = 0.200 mol As all gases are ideal. . . . . Answer = (B) 9. Assuming one can write the G as an exact differential of two variables, T and p, yields Making use of the the fact this is an exact differential, indicate which of the following relationships is correct or that none of the above are correct. (A) (B) (C) (D) (E) None of the above are correct Answer = (D) Page 15 of 20 – , ∆ , , ∆ , ∆ , , , ∆ ∆ n = 0.5000 mol U Q W AB (isometric) (TB – TA) UCA 0 BC (isobaric) (TC – TB) UBC – WBC -pB(VC – VB) CA (isothermal) 0 -WAB -nRTln(VB/VA) Cycle UAB + UBC = 0 QAB + QBC + QCA (=-Wcycle as check) WBC + WCA n = 0.5000 mol U Q W AB (isometric) 18.71 kJ 18.71 kJ 0 BC (isobaric) -18.71 kJ --31.18 kJ 12.47 kJ CA (isothermal) 0 6.850 kJ -6.850 kJ Cycle 0 -5.62 kJ 5.62 kJ Page 16 of 20 15. The following equation of state applies at low pressures: If Z = 0.9980 at T = 200.0 K and p = 7.250 x 103 Pa with a Boyle temperature of 346.8 K for the gas, calculate the values of a and b. a /Pa•m6•K•mol-2 b /m3•mol-1 (A) 367.8 1.070 x 10-4 (B) 228.6 2.286 x 10-4 (C) 36.78 1.070 x 10-5 (D) 22.86 2.286 x 10-5 (E) 3.678 1.070 x 10-6 At TB, Furthermore, Hence, . . . . . . . and . . . . ∙ ∙ ∙ Answer = (B) Page 17 of 20 16. For the phase diagram below for a one component system, English letters (A – S) indicate points, Greek letters () indicate single phases or coexisting phases, and Tx indicate isotherms. The isotherms are represented as dashed lines except where they coincide with the phase boundaries (solid lines). Indicate which statement is false. (A)  →  at point K. True –  is infinite at the critical point. (B) M is the critical point. False - F = 2 + C – P = 2 + 1 – 3 = 0 M is a triple point (C) The system has 2 degree of freedom at point E. True - F = 2 + C – P = 2 + 1 – 1 = 2 (D) , , and  could be gas, liquid and solid, respectively. True, as T goes to infinity one has a gas (). As T decreases liquid () would form before solid (). (E) The system has 1 degree of freedom at point H. True - F = 2 + C – P = 2 + 1 – 2 = 1
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved