FAKULTEIT INGENIEURSWESE FACULTY OF ENGINEERING

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FAKULTEIT INGENIEURSWESE FACULTY OF ENGINEERING Ingenieurschemie 123 Engineering Chemistry 123 Toets (Eerste geleentheid) Test (First opportunity) 11 / 03 / 2010 Tydsduur / Duration 2 h Volpunte / Full marks 80 Prof EP Jacobs Eksaminator / Examiner Mede-eksaminator / Co-examiner Afdeling/ Section A: Prof KG Clarke en/ and Me MJ Janse van Vuuren Afdeling/ Section B: Prof JJ Heyns Opgestel / Set Sertifisering / Certification Gemodereer / Moderated Afdeling / Section A: Prof KG Clarke Afdeling / Section B: Prof EP Jacobs Eksaminator / Examiner Afdeling/ Section A: Me MJ Janse van Vuuren Afdeling/ Section B: Dr C Esterhuysen Mede-eksaminator / Co-examiner Instruksies 1. Skryf eerste jou naam en studentenommer op al die antwoordstukke. 2. Skryf Afdeling A en Afdeling B duidelik regs bo op aparte antwoordboeke en beantwoord Afdeling A en Afdeling B se vrae in die onderskeie antwoordboeke. 3. Beantwoord al die vrae. 4. Sakrekenaars soos voorgeskryf vir die eerste 2 jaar van die BIng mag gebruik word. 5. Slegs die Periodieke tabel met die US logo mag gebruik word. 6. Die punte word regs van die vraagnommer in vierkantige hakies aangedui. 7. Die punt word deurgaans in hierdie eksamen as desimale plekhouer gebruik. 8. Bykomende inligting en omskakelingsfaktore is agteraan die vraestel geheg. 9. Rond die molêre massas in die Periodieke Tabel af na 4 beduidende syfers. Instructions 1. First write your name and student number on all the answer books. 2. Write Section A and Section B clearly at the top right of separate answer books and answer the questions of Section A and Section B in the different answer books. 3. Answer all the questions. 4. Pocket calculators prescribed for the first 2 years of the BEng may be used. 5. Only the Periodic table with the US logo may be used. 6. The marks are indicated to the right of the question number in square brackets. 7. The period is used as decimal place-holder in this exam. 8. Additional information and conversion factors are attached at the end of the paper. 9. Round the molar masses in the Periodic Table to 4 significant figures. Toetsweek 2010: Eerste geleentheid Afdeling A // Section A 1 6

Afdeling A/ Section A [Volpunte = 40/ Full Marks = 40] Skryf asseblief Afdeling A op die buiteblad van die toetsboek. Please write Section A on the front cover of the test book. Vraag A1 / Question A1 [5 Punte / 5 Marks] i. Die drywingsgetal (N p ) is n dimensielose getal wat algemeen gebruik word en verband hou met die drywing uitset van roerders. Dit word gedefineer as ii. met P: drywing uitset van die dryfrat ρ: digtheid van die fluïed n: rotasie spoed van die roerder d: diameter van die roerder Bereken die drywings uitset van n dryfrat (in SI eenhede) indien die drywingsgetal n eenheidsgrootte het (N p =1) vir die volgende kondisies. ρ = 0.0543 lb m / vt 3 n = 600 min -1 (rotasies per minuut) d = 2 duim (3) Skryf die SI-eenheid vir drywing in basiese SI-eenhede en gebruik dit om te wys hoekom die drywingsgetal dimensieloos is. (2) i. The power number (N p ) is a commonly-used dimensionless number relating to the power dissipated by stirrers and is defined as ii. where P: power dissipated by impeller ρ: density of the fluid n: rotational speed of the stirrer d: diameter of stirrer Calculate the power dissipated by an impeller (in SI units) if the power number is unity (i.e. N P =1) for the following conditions. ρ = 0.0543 lb m / ft 3 n = 600 min -1 (rotations per minute) d = 2 inch (3) Write the SI unit of power in basic SI units and use this to demonstrate why the power number is dimensionless. (2) Toetsweek 2010: Eerste geleentheid Afdeling A // Section A 2 6

Vraag A2 / Question A2 [17 Punte / 17 Marks] n Houer met n volume van 400 dm 3 bevat 800 g van n gas mengsel. Die gas mengsel bevat 87 massa % argon en 13 massa % water. Die houer bevat ook 1 L silika gel om van die water te absorbeer (waterdamp word deur die silika gel geabsorbeer en kan dus so van die mengsel verwyder word). Na n gedeelte van die water verwyder is, daal die druk met 0.07 atm. Die temperatuur bly konstant by 218 o C. Aanvaar dat die ideale gas wet geld en bereken: i. die hoeveelheid mol van elk argon en water en die totale hoeveelheid mol gas in die houer voor enige water geabsorbeer word (5) ii. die totale durk voor enige water geabsorbeer word (3) iii. die parsiële druk van argon en water voor enige water geabsorbeer word (4) iv. die totale druk na die water geabsorbeer is (2) v. die parsiële druk van argon en water na die water geabsorbeer is (3) A container with a volume of 400 dm 3 contains 800 g of a gas mixture. The gas mixture contains 87 mass % argon and 13 mass % water. The container also contains 1 L silica gel to absorb some of the water (water vapour absorbs on the silica gel and is thus removed from the gas). After a part of the water is removed, the pressure has decreased by 0.07 atm. The temperature remains constant at 218 o C. Assume the ideal gas law is applicable and calculate: i. the number of moles of each argon and water and the total number of moles of gas in the container before any water is absorbed (5) ii. the total pressure before any water is absorbed (3) iii. the partial pressure of argon and water before any water is absorbed (4) iv. the total pressure after the water is absorbed (2) v. the partial pressure of the argon and the water after the water is absorbed (3) Vraag A3 / Question A3 [11 Punte / 11 Marks] i. Koolstofdioksied word in n 0.5 m 3 tenk in n 20 m 3 stoorkamer by seevlak gehou. Ongelukkig was die tenk beskadig en het die koolstofdioksied in die stoorkamer in gelek totdat die drukmeter op die tenk n druk van 0 Pa (meterdruk) gegee het. As die aanvanklike druk in die tenk 3.01325x10 5 Pa (absoluut) was en die temperatuur konstant by 25 o C gebly het, bereken die mol koolstofdioksied wat in die stoorkamer gelek het. Aanvaar ideale gas gedrag. (5) ii. Indien aanvaar word dat die koolstofdioksied wat in die stoorkamer gelek het n gelyke hoeveelheid mol lug in die stoorkamer verplaas het, bereken of die koolstofdioksied die dodelike perk van 8 massa% oorskry. (6) Aanvaar: - die ideale gas wet geld - die molekulêre massa (gewig) van lug = 29 g/mol Toetsweek 2010: Eerste geleentheid Afdeling A // Section A 3 6

i. Carbon dioxide is kept in a 0.5 m 3 tank inside a 20 m 3 storeroom at sea level.. Unfortunately the tank had become damaged and the carbon dioxide had leaked into the storeroom until the pressure gauge on the tank registered 0 Pa (gauge). If the initial pressure in the tank was 3.01325 x 10 5 Pa (absolute) and the temperature was constant at 25 o C throughout, calculate the moles of carbon dioxide that leaked into the storeroom. Assume ideal gas behaviour. (5) ii Assuming that the carbon dioxide that leaked into the storeroom displaced an equal number of moles of air in the storeroom, calculate whether the carbon dioxide exceeds the lethal limit of 8 mass %. (6) Assume: the ideal gas law applies the molecular mass (weight) air =29 g/mol Vraag A4 / Question A4 [7 Punte / 7 Marks] n 100 ml Oplossing wat 40 massa % etanol in water bevat word tot 54 massa % gekonsentreer deur die oplossing in n fles te kook, die resulterende etanol en water dampe te kondenseer en in n bottel te versamel. Die kookproses is gestaak nadat 48 ml van die gekonsentreerde oplossing in die bottel versamel is. Bereken die ooreenstemmende massa % etanol wat in die fles oorbly. Aanvaar alle digthede = 1 g/ml. A 100ml solution containing 40 mass % ethanol in water is to be concentrated to 54 mass % by boiling the solution in a flask, condensing the resultant ethanol and water vapours and collecting these in a bottle. Boiling was stopped when 48mls of the concentrated solution had been collected in the bottle. Calculate the corresponding mass % ethanol remaining in the flask. Assume all densities =1g/ml. Toetsweek 2010: Eerste geleentheid Afdeling A // Section A 4 6

Bykomende inligting vir Afdeling A / Additional information for Section A Die volgende vergelykings kan sonder bewys gebruik word / The following equations can be used without proof: 1. Ideale gas vergelyking / Ideal gas equation: PV = n R o T Met / with R o = universele gaskonstante / universal gas constant = 8.314 J/ (mol K) = 8 314 J/ (kmol K) T = absolute temperatuur / absolute temperature 2. Van der Waals vergelyking / Van der Waals equation: met / with 2 n a P + ( V nb) nr T 2 V = o R o = universele gaskonstante / universal gas constant = 8.314 J/ (mol K) = 8 314 J/ (kmol K) T = absolute temperatuur / absolute temperature 3. Parsiële druk = mol fraksie x totale druk Partial pressure = mole fraction x total pressure Toetsweek 2010: Eerste geleentheid Afdeling A // Section A 5 6

NUTTIGE OMSETTINGSFAKTORE / CONVERSION FACTORS Grootheid / quantity Massa / mass Ekwivalente waardes / Equivalent values 1 kg = 1 000 g = 0.001 metrieke ton / metric tonne = 2.20462 lb m = 35.27392 oz 1lb m = 16 oz = 453.593 g = 0.453593 kg Lengte / length 1 m = 100 cm = 1 000 mm = 0.001km = 39.37 duim / inch = 3.2808 vt / ft = 1.0936 yd 1 vt / ft = 12 duim / inch = 0.3048 m = 304.8 mm Volume Krag / force Druk / pressure Energie / energy Drywing / power Temperatuur/temperature 1 m 3 = 1 000 liter / litre = 10 6 cm 3 = 35.3145 vt 3 / ft 3 = 220.83 imp gall = 264.17 US gall 1 vt 3 / ft 3 = 1 728 duim 3 / inch 3 = 28.317 liter / litre = 0.028317 m 3 1 N = 1 kg m/s 2 = 10 5 dynes = 0.22481 lb f 1 atm = 1.01325 x 10 5 N/m 2 (Pa) = 101.325 kpa = 1.01325 bar = 760 mmhg by 0 C (torr) = 14.696 psi 1 J = 1 N m = 10 7 erg = 2.778 x 10 7 kw h = 0.23901 cal = 0.7376 ft lb f = 9.486 x 10 4 Btu 1 W = 1 J/s = 0.23901 cal/s = 1.341 x 10 3 hp = 9.486 x 10 4 Btu/s T( F) = 1.8 C + 32 T ( C) = (5/9)( F 32) T (K) = C + 273.15 T( R) = F + 459.6 Toetsweek 2010: Eerste geleentheid Afdeling A // Section A 6 6