Exam_I_F10_solutions

Exam_I_F10_solutions - Chem 300 ! ! Exam I! Fall 2010 !...

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Unformatted text preview: Chem 300 ! ! Exam I! Fall 2010 ! Name ! Solutions READ EACH QUESTION CAREFULLY. This is a closed book exam. You may use one 8-1/2” x 11” sheet of equations and information during the exam. You may use a calculator but you may not share the calculator with other students. To receive full credit you must show all you work clearly. Having the correct answer without the correct work will not receive full credit. Remember to give units and use significant figures when needed. Make your answers clear so I can understand what you are doing. Forgot something on your equations sheet? I will give some information for points taken from your exam score. If something is not clear, ASK!!! Good luck ☺ Question 1 ! ! ! ! out of 20 Question 2 ! ! ! ! out of 20 Question 3 ! ! ! ! out of 30 Question 4 ! ! ! ! out of 30 ! ! out of 100 ! ! ! Total! ! ! Revised 9/30/10 1. Ideal gases a. (6 pts) Explain how the experiments of Boyle, Charles, and Avogadro led to the formulation of the ideal (perfect) gas equation of state. (Discussion question from Chapter 1) Boyleʼs law: V∝1/P Charlesʼ law: V∝T Avogadroʼs law: V∝n Together they lead to V ∝ nT nRT , which in turn lead to V = P P b. (7 pts) A 1.00-L bulb and a 1.50-L bulb, connected by a stopcock, are filled with argon at 0.75 atm and helium at 1.20 atm respectively, at the same temperature. Calculate the total pressure and the partial pressure of each gas after the stopcock has been opened. (Problem 7 from HW #1 and in class exercise) Weʼll let 1 be the conditions before the stopcock is opened and 2 for after. Since weʼre at constant T and n we can use the relationship P1V1 = P2V2 to determine the partial pressures of each individual gas. V2 is the total volume (1L + 1.5 L = 2.5L) ! ! c. (7 pts) The pressure inside a meteorological balloon increased to 51 Torr after rising from sea level at 20°C to its maximum altitude where the temperature is −25°C. By how much will the balloon expand or contract as it rises? (Variation of problem 14 back of chapter 1 and in class exercise) PiVi Pf V f V f PiT f (1atm ) ( 248 K ) = ; = = = 12.6 Ti Tf Vi Pf Ti ( 51torr ) (1atm / 760torr ) ( 293K ) Therefore the balloon will expand by 12.6 times. 2. Real gases a. (3 pts) Write down the van der Waals equation of state. ⎛ an 2 ⎞ RT a P + 2 ⎟ (V − nb ) = nRT or P = −2 ⎜ V⎠ Vm − b Vm ⎝ Revised 9/30/10 b. (4 pts) Explain the physical significance of the two coefficients. The “a” coefficient is part of a correction factor that corrects for the pressure drop that results from attractive interactions in a real gas and strongly correlates with boiling point. The “b” coefficient is part of a correction for the fact the volume of the gas molecules is not always negligible in a real gas. c. (4 pts) The van der Waals coefficients are related to the critical constants. What is the physical significance of the critical temperature? (discussion question 6) The critical temperature is the temperature above which a gas cannot be compressed into a liquid, no matter the pressure or volume. d. (9 pts) Calculate the molar volume of CO2 at 300 K and 50 atm, given that the second virial coefficient (B) of CO2 is -0.0605 L mol-1. For the calculation of the molar volume use the explicit solution of the virial equation of state and choose the most physically reasonable solution. (in class exercise and problem 8, homework #1). P= RT Vm ⎛ B⎞ RT RTB 2 ⎜ 1 + V ⎟ ; P = V + V 2 ; PVm − RTVm − RTB = 0 ⎝ m⎠ m m −b ± b 2 − 4 ac ; where a=P=50atm, b=-RT=-24.62 L atm mol-1; and c=-RTB=+1.49 L2 atm mol-2 2a Substituting we get: RT ± ( RT )2 − 4 ( P ) ( RTB ) 2P −1 Vm = 24.62 L ⋅ atm ⋅ mol ± ( 24.62 L ⋅ atm ⋅ mol ) −1 2 ( − 4 ( 50 atm ) 1.49 L ⋅ atm ⋅ mol 2 −2 ) ; 100 atm 24.62 L ⋅ atm ⋅ mol −1 ± 24.37 L ⋅ atm ⋅ mol −1 Vm = = 0.422 L ⋅ mol −1 100 atm The other answer (0.0707) is too far from the ideal gas prediction to be physically possible. Vm = RT 24.62 L ⋅ atm ⋅ mol 1 = = 0.493L ⋅ mol −1 P 50 atm Revised 9/30/10 3. Kinetic model of gases a. (8 pts) What is the relative rate of diffusion of carbon monoxide (CO) and oxygen (O2) at a given temperature and pressure? Which is faster? M O2 rCO = rO2 M CO 32 amu = 1.07 28 amu = CO will diffuse 1.07 times faster than oxygen. b. (8 pts) When we are studying the photochemical processes that can occur in the upper atmosphere, we need to know how often atoms and molecules collide. At an altitude of 20 km the temperature is 217 K and the pressure is 0.050 atm. What is the mean free path of N2 molecules? Take σ = 0.43 nm2. (problem 24, chapter 1) σ = ( 0.43x10 −18 m 2 ) (1L / 1000 m 3 ) = 4.30 x10 −22 L ⋅ m −1 RT = 2σ PN a λ= 17.807 L ⋅ atm ⋅ mol −1 = 0.973µ m 2 4.30 x10 −22 L ⋅ m −1 ( 0.05 atm ) 6.022 x10 23 mol −1 ( ) ( ) c. (8 pts) How many collisions per second do the N2 molecules undergo? (Problem 25, chapter 1) 3RT = M c= Z= 5.412 x10 3 kg ⋅ m 2 ⋅ s −2 ⋅ mol −1 = 425 m / s 30 g 10 −3 kg / g ⋅ mol −1 ( ) c 425 m / s = = 4.37 x10 8 s −1 λ 0.973m d. (6 pts) The mean speed of a gas is calculated by multiplying each speed by the fraction of molecules that have that speed, and then adding all the products together. With this in mind, show how you would calculate the mean speed of molecules in a gas in the interval from 300 to 600 m s-1 using the Maxwell distribution function (f(s)). DO NOT do any actual calculations! (in class exercise) s= ∫ 600 300 sf ( s ) ds Revised 9/30/10 4.The Liquid State a. (7 pts) How far will a protein with a diffusion coefficient of 8.03x10-13 m2 s-1 diffuse in 1s? (problem 14, Chapter 11) ( ) x 2 = 2 Dt = 2 8.03x10 −13 m 2 s −1 (1s ) = 1.267 x10 −6 m / 1s b. (7 pts) Use Fickʼs first law to calculate the flux of nutrient molecules down a concentration gradient of 0.1 mol dm-3 m-1 if the diffusion coefficient is 5.22x10-10 m2 s-1. (problem 13, chapter 11) ⎛ ∂c ⎞ J = − D ⎜ ⎟ = − 5.22 x10 −10 m 2 s −1 0.1mol ⋅ L−1 ⋅ m −1 = −5.22 x10 −11 mol ⋅ L−1 ⋅ m ⋅ s −1 1L ⋅ 1000 m −3 = −5.22 x10 −8 mol ⋅ m −2 s −1 ⎝ ∂x ⎠ t ( )( ) ( ) c. (8 pts) We said in class that a decrease in capillary radius from 2.0x10-4cm to 1.5x10-4 cm by cholesterol deposits will increase the resistance to blood flow. By how much must blood pressure increase to maintain blood flow with such a narrowing? Q= 4 πΔP1 R14 πΔP2 R2 = 8η L 8η L ΔP2 R14 24 = 4= = 3.16 . Blood pressure must increase by over 3-fold. ΔP1 R2 1.5 4 d. (8 pts) Use the Laplace equation to calculate the difference in pressure on either side of a curved surface of water (γ=72 nN m-1at 298K) in a tube with a radius of 0.10mm. How does the pressure difference change in a 10 mm tube? (problem 15, chapter 16) ( ) −9 −1 2γ 2 72 x10 N ⋅ m ΔP = = = 1.44 x10 −3 N ⋅ m −2 −3 r 0.1x10 m Since the radius increases by 100-fold, the pressure must decrease by 100-fold. Revised 9/30/10 ...
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