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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 81/2” 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 1.00L bulb and a 1.50L 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 mol1. 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 mol1; and c=RTB=+1.49 L2 atm mol2
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 s1 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 8.03x1013 m2 s1 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 dm3 m1 if the diffusion coefﬁcient is 5.22x1010 m2 s1. (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 2.0x104cm to 1.5x104 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 3fold.
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 m1at 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 100fold, the pressure must decrease by 100fold. Revised 9/30/10 ...
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 Fall '11
 T.Baird
 Chemistry

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