CHEM* Physical Chemistry FALL Assignment #1: SOLUTIONS. Gases. Thermodynamics: heat and work

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1 CHEM* hysical Chemistry FALL 2005 Assignment #1: SOLUTIONS Gases. Thermodynamics: heat and work 1. 1 mol o nitrogen and mol o hydrogen are injected into a container o volume 10 dm at 298K. What are the partial pressures and the total pressure? (Ans. N 2, 2.45 atm; H 2, 7.5 atm; total, 9.80 atm). 1 mol N 2 + mol H 2 in a 10 dm container (10 dm = 10L) at 298K Assuming ideal gases ==> EOS: =nrt artial pressures o nitrogen and hydrogen in the mixture: (N 2 ) = n(n 2 ) [RT/] and (H 2 )= n(h 2 ) [RT/] Data: n(n 2 ) = 1 mol n(h 2 ) = mol R= atm L/K mol T= 298K = 10L (N 2 ) = 2.45 atm ; (H 2 ) = 7.4 atm ; total = (N 2 ) + (H 2 ) = 9.79 atm 2. The composition o dry air at sea level in mass per cent is approximately; N 2, 75.52; O 2, 2.15; Ar, 1.28; CO What is the partial pressure o each component when the total pressure is 1 atm? T = 1 atm = total pressure rom Tables: Mr (N 2 ) = 28.2 g/mol; Mr(O 2 ) = 2 g/mol ; Mr(Ar) = 9.95 g/mol; Mr(CO 2 ) = g/mol In 100 g o air: g N 2 ; 2.15 g O 2 ; 1.28 g Ar; g CO 2 Number o moles o each gas in the mixture ==> ni = mi / Mr. For instance or N 2 : n(n 2 ) = g / 28.2 g.mol -1 = mol Following the same proceeding or O 2, Ar, and CO 2 :

2 n(o 2 ) =0.72 mol ; n(ar) = 0.02 mol; n(co 2 ) = mol total number o moles = n T = n(n 2 )+n(o 2 )+n(ar)+n(co 2 )=.445 mol x(n 2 )= molar raction o N 2 in the mixture = n(n 2 )/n T = 2.678/.445 = x(o 2 ) = x(ar) = x(co 2 ) =.05x10-4 artial pressure o N 2 in the mixture = (N 2 ) = x(n 2 ) T = atm = atm artial pressure o O 2 = atm artial pressure o Ar = atm artial pressure o CO 2 =.05x10-4 atm. The synthesis o ammonia (NH ) rom H 2 and N 2 is an important technological process. A vessel o volume 22.4 dm contains 2 mol o H 2 and 1 mol o N 2 at K. What is the mole raction o each component, their partial pressures, and the total pressure? A vessel o 22.4 dm = 22.4 L contains 2 mol o H 2 and 1 mol o N 2 at K Total number o moles = n T = Assuming ideal behaviour ==> EOS = n RT ==> = mol ( atm L/K mol) K / 22.4 L = atm Molar ractions: x(h 2 ) = 2/ x(n 2 ) = 1/ artial pressures: (H 2 ) = x(h 2 ) = (2/) atm = 2 atm (N 2 ) = x(n 2 ) = (1/) atm = 1 atm

3 4. In a respiration experiment, a bacterial consumed 22.7 cm o oxygen gas, measured at a pressure o 79 cm (o Hg) and at a temperature o 2.8 o C. Calculate the number o moles o oxygen used (Ans x 10-4 mol) =22.7 cm o oxygen gas equivalent to dm =79 cm (o Hg) = 790 mm(hg) = 1.09 atm (1 atm = 760 mmhg) t=2.8 o C ==> T= K Ideal gases: =nrt ==> n = / RT n (O 2 ) = 1.09 atm L / ( atm L K -1 mol K) = mol 5. An evacuated bulb weighs 82. g when empty and 89.6 g when illed with an unnamed gas at a pressure o one atmosphere and a temperature o 25 o C. When the bulb is illed with O 2, it weighs 84.7 g at the same temperature and pressure. What is the molecular weight o the unnamed gas? (Ans. 97.) w (unknown) = weigh unknown gas = 89.6 g 82. g = 7. g w (O 2 ) = weigh O 2 = 84.7 g 82. g = 2.4 g both ideal gases ==> = nrt (n = w/mr) == > = (w/mr) RT,, and T are the same or both gases ==> (w (unknow)/mr (unknown) = /RT = (w(o 2 )/Mr(O 2 )) Mr(unknown gas) = w(unknow) Mr(O 2 ) / w(o 2 ) Mr (unknown gas) = (7. g / 2.4 g) 2 g/mol = 97. g/mol 6. An ideal gas is compressed isothermally by a orce o 1.0 N acting through 0.5 m. Calculate the heat change. (Ans. 0.5 J), ideal gas compression ( T cte) ( T,, ) isothermal = ideal gas( T,, ) i i F= 1.0 N and L = 0.5 m From lecture:

4 - Compression work + - Ideal gas and isothermal process ==>)U = 0 - First law: )U = q + w w = 1.0 N 0.5 m = 0.5 N m = 0.5 J )U = 0 = q + w ==> q = -w = -0.5 J (exothermic process to keep at T cte) The work done on the system is completely returned to the surroundings in the orm o heat. 7. Chapter 2 (Tinoco et al., 4 th edition). ROBLEMS: 2,, 4, 6, 7, 9, 10 Library Reserve desk 8. Calculate the work done when 1 mole o gas (assumed ideal) expands a) reversibly and isothermally rom 400 kn.m -2 to 150 kn.m -2 at 298K b) irreversibly and isothermally rom 400 kn.m -2 to 150 kn.m -2, in one stage, by a sudden direct reduction o pressure to 150 kn.m -2 at 298K. Show the work done in each case, using a sketched diagram. How much heat is absorbed in each case and what changes occur in the internal energy? [Ans. a) kj, )U=0, q= 2.42 kj b) )U=0, w= 1549 J, q=-1549 J] For both (a) and (b) Taking into account that 1 a = 1 N.m -2 and 1 atm = ka ==> 1 = 400 kn.m -2 2 = 150 kn.m -2 = 400 ka =.948 atm = 150 ka = atm gas ideal ==> = nrt 1 = 1mol atm L K -1 mol K /.948 atm= 6.19 L 1 = 1mol atm L K -1 mol K / atm= L

5 (a) A w = -I (nrt/) d= - nrt ln( 2 / 1 ) i B Reversible path Ininitely small steps A (400 kn.m -2, 6.19 L, 298K) B (150 kn.m -2, L, 298K) Isothermal expansion (ideal gas) ==> )U = 0 ; )U = q + w => q = - w w = B d reversible gas w = nrt remember A B A d x2 dx x x = ln x 2 x1 1 ( T = cte) ( 8.14J. K mol ) 298K ln B d B w = nrt = nrt ln = mol A A w = J (expansion the sign is correct) q = w = J (endothermic process to keep the temperature constant) b) A (400 kn.m -2, 6.19 L, 298K) B (150 kn.m -2, L, 298K)

6 A w =-) w B Irreversible - sudden reduction o pressure to 150 kn.m -2 at 298K Isothermal expansion (ideal gas) ==> )U = 0 ; )U = q + w => q = - w w = B A d irreversible = 150 kn. m 2 w = = 50kN. m 2 (16.52L 6.19L) 2 m Units! 1 L = 1 dm = 1x10 m w = 50kN. m 10. dm 1x10 = 549.4N. m = J dm w = J (expansion the sign is correct) q = w = J (endothermic process to keep the temperature constant) The amount o heat and work are dierent or each path! Work done by the system in a reversible path gives the maximum eiciency! 9. A quantity o mol o an ideal gas initially at a pressure o 15.0 atm and 00K is allowed to expand isothermally until its inal pressure is 1.00 atm. Calculate the work done i the expansion is carried out, a) against vacuum b) against a constant ernal pressure o 1.00 atm c) reversibly [Ans: b) J and c) J]

7 (a) Expansion against vacuum ==> = 0 w = 0 (b) Expansion against a constant pressure, = 1.00 atm mol o an ideal gas initially at a pressure o 15.0 atm and 00K i = nrt/ i = nrt/ t w = i d ( cons tan t pressure) = 1 atm = w = ( ) i = nrt ( nrt ) i J 1.00 w = nrt 1 = 0.85 mol K 1 = 978.7J =.979kJ i Kmol 15.0 (c) Reversible expansion w = d reversible gas w = nrt remember i i d x2 dx x x = ln x 2 x1 1 ( T = cte) w = nrt i d = nrt ln i = 0.85mol ( 8.14J. K mol ) 00 K nrt ln nrt i w = 0.85mol i 15.0 ( 8.14J. K mol ) 00 K ln = 0.85 mol ( 8.14J. K mol ) 00 K ln = J 1.00 w = 5741 J (expansion the sign is correct) Reversible expansion gives the largest amount o work.

8 10. a) Calculate the work done when the volume o a gas at 0 o C increases by 1 ml against a pressure o 1 atm. (a) b) What is the work done when the volume increase is 1L against a pressure o 1 atm? (Ans J, J) w = d ( cons tan t pressure) = 1 atm w = ( ) = atm 1x10 L = L atm i Units! i 1 L = 10 m 1L atm = atm = x10 N m 2 5 m x10 N m 2 5 N m w = 0.001L atm 1.012x10 x10 = L atm (b) J w = J x 1000 = J 11. A lead bullet with an initial temperature o 25 o C and a velocity o 4.00 x 10 2 m s -1 is stopped by impact with a brick wall. Assuming no loss o heat rom the bullet to the wall, what will the temperature o the bullet be ater impact? The speciic heat o lead is approximately. x10-2 cal g -1 o C -1 (Ans. 604 o C) lead bullet (at t=25 o C) v = 4.00x10 2 m.s -1 E K = ½ m v 2 = ½ m (4.00x10 2 ) 2 m 2.s -2 On impact kinetics energy, E k, is converted to heat (q ) q v = m C v )T= m.x10-2 cal g -1 o C -1 )T

9 )U = E K = q v ½ m ((4.00x10 2 ) 2 m 2 s -2 )= m (.x10-2 cal g -1 o C -1 ) (4.184 J.cal -1 ) )T )T = 579 o C ==> 579 o C = T 25 o C ==> T = 604 o C 12. During expansion against a constant ernal pressure o 1 atm, a gas absorbs 200 J o heat while its volume increases rom 1.00 dm to dm. Calculate U or this change. (Ans x 10 2 J) )U = q + w q = 200 J w = - 1 atm ( ) L = - 9 L atm => w = J g o crystalline urea [(NH 2 ) 2 CO] is burned in a bomb calorimeter or which the net heat capacity is 5.0 kj K -1. A temperature rise rom 21.4 to o C is observed. The products o the reaction are CO 2 (g), H 2 O (l), and N 2 (g). What is U or the process? What is U per mole. (Ans kj, kj mol -1 ) Bomb calorimeter: = cte )=0 => w = 0 (no work done) Combustion reaction (NH 2 ) 2 CO (s) + /2 O 2 (g) ==> N 2 (g) + CO 2 (g) + 2 H 2 O (l) )U = q v = 0 = )U reaction + C )T (C = net heat capacity = 5.0 kj. K -1 ) )U reaction = - C )T = kj. K -1 ( ) K = kj per 1.72 g o crystalline urea exothermic reaction ==> O.K. Mr (urea) = 60.0 g.mol -1 U = (17.9 kj /1.72g) 60.0 g. mol = 78.2kJ mol

10 14. I the average 60 kg person generates about x 10 4 kj o heat per day rom ood consumed, calculate the temperature rise assuming the body is adiabatic and the heat capacity equal to 4.18 J K -1 g -1. adiabatic ==> no transer o heat into or out o the system q= 0 q converted = q generated ==> m C )T = 1.046x10 4 kj.day -1 (1 day) = 1.046x10 7 J g (4.18 J.K -1 g -1 ) )T = x10 7 J )T = 41.7 o C Good Luck!!!!

Useful Information to be provided on the exam: 1 atm = 760 mm Hg = 760 torr = lb/in 2 = 101,325 Pa = kpa. q = m C T. w = -P V.

Useful Information to be provided on the exam: 1 atm = 760 mm Hg = 760 torr = lb/in 2 = 101,325 Pa = kpa. q = m C T. w = -P V. Chem 101A Study Questions, Chapters 5 & 6 Name: Review Tues 10/25/16 Due 10/27/16 (Exam 3 date) This is a homework assignment. Please show your work for full credit. If you do work on separate paper, attach