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Name
Section
TA
Chem1b Problem Set Seven Ch 1b, Problem Set 7
Due March 10, 2014
at 4 PM in the Ch1b Box
Ch1b, Problem Set Seven Due March 10 in the Ch1b Box Problems 1 and 3 are marked :-­‐) as no collaboration problems. These problems will be considered as quiz-­‐like problems. There are a total of five pages to this problem set. :-­‐) Problem 1. Rotational Spectroscopy (22 points and 2 bonus). In this problem, we will model the of CO molecule as a rigid rotor. a) Given that a photon with frequency v=1.15*1011 Hz excites 12C16O from the J=0 to J=1 state, find the equilibrium bond length of the molecule. h(J +1) h(J +1)
=
Use the formula ν =
with the given frequency, J=0, and reduced mass to 4π 2I
4 π 2 µR 2
get R = 1.13 Å. 2 points b) Find the degeneracies of the first five rotational energy levels (J=0 to 4). €
Use degeneracy = 2J+1 to get 5 points: 1 each J Degeneracy 0 1 1 3 2 5 3 7 4 9 4 points c) For J=0 to 4, calculate the frequency of light that would lead to the transition from a rotational state in the Jth energy level to a state in the (J+1)th energy level. (Note the symmetry of the rotational wavefunctions requires that light can only excite molecules from states in the Jth energy level to states in the (J+1)th energy level.) Accept answers in Hz OR wavenumber. 5 points: 1 each J Frequency (Hz) Frequency (wavenumber) 0 1.15*10^11 385 1 2.31*10^11 770 2 3.46*10^11 1160 3 4.62*10^11 1540 4 5.77*10^11 1930 d) Below is the experimental rotational absorption spectrum for a CO gas (Varberg 1992). For all eight peaks in the spectrum, determine the transition between rotational states that accompanies the absorption of light (i.e., the peak corresponds Name
Section
TA
Ch 1b, Problem Set 7
Due March 10, 2014
at 4 PM in the Ch1b Box
to a transition from state J=? to J=?). Absorption Intensity Frequency (GHz) 4 points: 0.5 each Peak 1 2 3 4 5 6 7 8 Transition 0 -­‐> 1 1 -­‐> 2 2 -­‐> 3 3 -­‐> 4 4 -­‐> 5 5 -­‐> 6 6 -­‐> 7 7-­‐> 8 e) This experimental absorption spectrum was measure at a temperature of T = 40 K. Using the Boltzmann probability, calculate the relative population in the first four excited energy levels (J=1 to 4) relative to the population in the J=0 energy level. (Hint: Make sure to account for the effect of degeneracy in calculating the population of the various energy levels.) 6 points: 1.5 each J Relative population 1 2.61 2 3.30 3 3.05 4 2.25 f) (Bonus 2 pts.) In the experimental absorption spectrum, note that the absorption intensities first increase and then decrease as a function of increasing absorption frequency. If we consider the effects of degeneracy alone, we would expect that the intensities would monotonically increase as a function of intensity. What is the Name
Section
TA
Ch 1b, Problem Set 7
Due March 10, 2014
at 4 PM in the Ch1b Box
cause of the decrease in intensity at higher frequency? (Hint: It is related to the Boltzmann probabilities.) As J increases, the energy of the state increases. According to the Boltzmann distribution, population of a state decreases as its energy increases. This effect competes with the degeneracy effect. As the energy effect begins to dominate, the population of molecules in excited rotational states begin to decrease with increasing J. Molecules may only be excited one rotational state higher than their current state, so fewer are available to be excited by higher-­‐energy photons and absorption intensity is weaker. Problem 2. Dueling Gas Models. (15 points) At two different pressures of pure water vapor, the following temperature and molar volume data were experimentally measured (Sandler 2006). Note that molar volume refers to the volume per mole of material. T (K) v (m3/mol) at P=1.0 bar v (m3/mol) at P=40 bar 553.15 0.04581 0.000999 593.15 0.04914 0.001116 633.15 0.05238 0.0012222 673.15 0.0558 0.0013212 723.15 0.05994 0.00144 773.15 0.0657 0.0015552 823.15 0.0684 0.0016686 873.15 0.07254 0.0017784 a) Plot the temperature as a function of molar volume for P = 1.0 bar using the experimental data. On the same graph, plot the temperature that is predicted as a function of molar volume from the ideas gas equation of state. Also on the same graph, plot the temperature that is predicted as a function of molar volume from the van der Waals equation of state, using the constants a = 5.54*10-­‐6 bar m6 mol-­‐2 and b = 3.05*10-­‐5 m3 mol-­‐1 for water vapor. 4 points (3 for values, one for formatting) Name
Ch 1b, Problem Set 7
Section
Due March 10, 2014
TA
at 4 PM in the Ch1b Box
b) In a new graph, plot the corresponding results for P = 40 bar. 4 points c) Evaluate the performance if the ideal gas and van der Waals equations of state. Discuss which model is in better agreement with experimental data at high pressure and at low pressure. For the P=40 bar results, discuss why the ideal gas equation is in better agreement at high temperature. At low pressure, both models are in very good agreement with experiment, because at low pressure, most gases behave ideally. At high pressure, intermolecular interactions and particle volume become more important, so the assumptions of the ideal gas model break down and the van der Waals model, which accounts for these effects, performs better. At P=40 bar, the ideal gas model performs better as temperature increases because the particles have more kinetic energy and move faster, so intermolecular attraction is less important, and the real molar volume is closer to the ideal molar volume. 7 points: 3 for pressure discussion, 4 for temperature discussion :-­‐) Problem 3. Ideal Gas Law. (18 points) Throughout this problem, assume that gases can be described using the ideal gas law. a) If you have 175 moles of liquid oxygen at T = -­‐185 ˚C and P = 75 atm, and the density of liquid oxygen is 1.141 g/cm3, calculate the volume of the liquid. !" 𝑔
! 𝑚𝐿
! 𝐿
175 𝑚𝑜𝑙 𝑚𝑜𝑙 !.!"! 𝑔 !"""𝑚𝐿 = 4.91 L 3 points b) If the temperature of the oxygen is increased to 20 ˚C and the pressure is reduced to 1 atm, the oxygen vaporizes to a gas. Calculate the equilibrium volume of the oxygen under the new conditions. PV = nRT V = nRT/P !"# 𝑚𝑜𝑙(!.!"#!$
𝐿∗𝑎𝑡𝑚
)( !"#.!"𝐾)
𝑚𝑜𝑙∗𝐾
𝑉=
= 4.21 * 103 L 3 points ! 𝑎𝑡𝑚
c) If this sample of oxygen gas were compressed into a 10.0 L container while the temperature is kept fixed, what would be the pressure of the compressed gas? Name
Section
TA
Ch 1b, Problem Set 7
Due March 10, 2014
at 4 PM in the Ch1b Box
P = nRT/V !"# 𝑚𝑜𝑙(!.!"#!$
𝐿∗𝑎𝑡𝑚
)(!"#.!"𝐾)
𝑚𝑜𝑙∗𝐾
𝑃 = = 421 atm 3 points !".!𝐿
d) Using parts (a) and (b), calculate the expansion ratio of oxygen (volume of gas divided by volume of liquid) and comment in 1-­‐2 sentences about why liquid gases can be very dangerous if not handled properly. Expansion ratio = 4.21 * 103 L/4.91 L = 857 (1 point) Since liquid gases expand to many times their volume upon returning to gaseous state, they can put immense pressure on their containers or even rupture the container. This can lead to property damage, serious injury, and death. (See the following Wikipedia article: http://en.wikipedia.org/wiki/Boiling_liquid_expanding_vapor_explosion.) Note, however, that the overpressurization and expanding gases will not necessarily involve a fire. 2 points for reasonable answers e) A sample of 199 L of air at 20 ˚C and 1.0 atm is condensed to liquid, and the primary components of the air (O2 and N2) are purified and weighed. It is found that the sample of air consists of 55.68 g O2 and 180.6 g N2. Using this information, determine the partial pressures of O2 and N2 in the original sample of air. We can first determine the total number of moles in the sample of air. n = PV/RT (!.!𝑎𝑡𝑚)(!""𝐿)
𝑛=
= 8.27 moles total 𝐿∗𝑎𝑡𝑚
(!.!"#!$
𝑚𝑜𝑙∗𝐾
)(!"#.!"𝐾)
Then, calculate the moles of O2 and N2. nO2 = 55.68 g (1 mol/32g) = 1.74 moles O2 nN2 = 180.6 g (1mol/28g) = 6.45 moles N2 We can determine the mole fractions of O2 and N2 and use those to calculate 𝑃
𝑛
𝑉
partial pressures, since for an ideal gas 𝑃𝑖 = 𝑛𝑖 = 𝑉𝑖 xO2 = 1.74 mol / 8.27 mol = 0.21*(1 atm) = 0.21 atm (O2) xN2 = 6.45 mol / 8.27 mol = 0.78*(1 atm) = 0.78 atm (N2) 1 point for each correct answer, 2 points for calculations f) Neglecting any possible error due to the use of the ideal gas law, explain why these two values in part (e) do not exactly sum up to the original pressure of the sample of air (1 atm). These values do not sum up to 1 atm because there are more components of air than just O2 and N2. (Note: The remaining 1% consists of argon, carbon dioxide, and small amounts of other gases.) 2 points Name
Ch 1b, Problem Set 7
Section
Due March 10, 2014
TA
at 4 PM in the Ch1b Box
Problem 4. Van der Waals Equation of State. (17 points) Suppose that sulfur hexafluoride is formed from vaporized molecular sulfur and fluorine via the following exothermic reaction. S2(g) + 6F2(g)  2SF6(g) a) If you react 5 moles of S2 with 6 moles of F2 at 20 ˚C in a 20 L container, what is the pressure before and after the reaction? The reaction is performed in a container of fixed volume. Treat all gases as ideal, and assume the reaction goes to completion. However, note that not all of the reactant may be consumed. The reaction will consume all 6 moles of F2 and only 1 mole of S2. Therefore, the product will be 2 moles of SF6, and 4 moles S2 will remain. In effect, we have 11 moles total  6 moles total and we can use these totals to calculate pressure. Before: P1 = n1RT/V P1 = (11 mol)(0.08206 L*atm/mol*K)(293.15K)/20L = 13.2 atm After: P2 = n2RT/V P2 = (6 mol)(0.08206 L*atm/mol*K)(293.15K)/20L = 7.22 atm 1 point for identifying moles on each side 2 points for calculations 1 point for each correct pressure b) Recognizing that SF6 is not an ideal gas, use the following pressures and volumes for 3 moles of pure SF6 at 20 ˚C to calculate the van der Waals constants (a and b) of SF6. Hint: Do not look up these values from a reference text, and be careful not to round until the end of your calculations; rounding errors will accumulate significantly. Pressure Volume 6.79 atm 10 L 15.1 atm 4.1 L Using the van der Waals equation of state (OGC page 389), 𝑛!
𝑃 + 𝑎 ! 𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇 𝑉
we can set up the following set of equations: 6.79 𝑎𝑡𝑚 + 𝑎
15.1 𝑎𝑡𝑚 + 𝑎
(!𝑚𝑜𝑙)!
𝐿∗𝑎𝑡𝑚
10𝐿 − 3𝑚𝑜𝑙 ∗ 𝑏 = 3𝑚𝑜𝑙 0.08206 𝑚𝑜𝑙∗𝐾
!"𝐿 !
(!𝑚𝑜𝑙)!
293.15𝐾 𝐿∗𝑎𝑡𝑚
4.1𝐿 − 3𝑚𝑜𝑙 ∗ 𝑏 = 3𝑚𝑜𝑙(0.08206 𝑚𝑜𝑙∗𝐾)(293.15𝐾) !.!𝐿 !
and use them to solve for a and b. 𝑎 = 6.59
𝑎𝑡𝑚∗𝐿!
𝑚𝑜𝑙!
𝐿
𝑏 = 0.0747 𝑚𝑜𝑙 2 points for calculations 2 points for each correct answer (Note to Grader: Please be lenient when grading the exact numbers for this problem. It is up to you how lenient to be depending on how the students solve the problem, but the students may be off for the constants. Depending on what equation solver they used, they could be off by about 1-­‐5% even with the correct calculations. It is suggested that a leniency of about 5-­‐10% for each value be considered on this problem.) Name
Section
TA
Ch 1b, Problem Set 7
Due March 10, 2014
at 4 PM in the Ch1b Box
c) Using the van der Waals constant for the excluded volume (b) that was obtained in part (b), calculate the effective radius of a molecule of SF6 in Angstroms. Assume that the molecule is spherical. First, we must convert b (which is in L/mol) to Å3 per atom. 𝐿 0.001𝑚!
1 𝑚𝑜𝑙
10!" Å!
Å!
0.0747 =
124
𝑚𝑜𝑙
1𝐿
6.022 ∗ 10!" 𝑎𝑡𝑜𝑚𝑠
𝑚!
𝑎𝑡𝑜𝑚
We then plug this into the equation for volume of a sphere, V = 4/3πr3. However, since we saw in class that each molecule only contributes half to the excluded, we must use V = 2/3πr3. 2
124Å! = 𝜋𝑟! 3
r = 3.90Å 2 points for conversion, 2 points for calculations, 2 points for correct answer Problem 5. Ideal Gases and the First Law. (28 points) Several processes performed on one mole of an ideal gas are illustrated in the diagram below: D P1,V1,T1 P1,V2,T3 E P A P2,V2,T1 B C P2,V2,T2 V a) Process A corresponds to a reversible, isothermal expansion. Calculate the change in internal energy during the course of this process. Also calculate the amount of heat absorbed and work performed by the system. Express you answers in terms of the macroscopic variables that describe the initial and final states (i.e., P1, V1, T1, P2, V2). In this problem we use the sign convention ∆U = q – w. Take off only half a point for all sign errors. ∆U = 0 because U is a function only of temperature for an ideal gas. dw = PdV = (R*T1)/V Name
Section
TA
Ch 1b, Problem Set 7
Due March 10, 2014
at 4 PM in the Ch1b Box
"V %
"V %
dv. Integrate to get w = RT1 ln$ 2 ' and q = w = RT1 ln$ 2 '
# V1 &
# V1 & 4 points
b) Process B corresponds to a reversible, adiabatic expansion. It is followed by Process C, in which € the system is heated € at constant volume. For both Processes B and C, calculate change in internal energy, as well as the heat absorbed and work performed by the system. Express you answers in terms of the macroscropic variables that describe the initial and final states. Process B: q = 0 because the process is adiabatic. Thus ∆U = -­‐w = CvdT = (3/2)Rdt for an ideal gas. ∆U = (3/2)R * (T2-­‐T1) w = -­‐(3/2)R * (T2-­‐T1) Process C: No pressure-­‐volume work is done, so w = 0 Thus, ∆U = (3/2)R * (T1-­‐T2) and q = (3/2)R * (T1-­‐T2) 6 points
c) Comparing the results of parts (a) and (b), show that the change in internal energy for part A is the same as the combined change for Processes B and C. Why is this the case? Add the ∆U values to get 0 for Process B combined with Process C. This is the case because internal energy is a state function—only initial and final conditions matter, not the path between states. 3 points d) Suppose that P1 = 2.0 atm, V1 = 5.0 L, and V2 = 10 L. Calculate T1 and T3 (the initial and final temperatures for Process D). 4 points (2 for each temp.) T1 = P1*V1/R = 121 K T3 = P1*V2/R = 244 K e) Use your results from part (d) to calculate the change in internal energy associated with Process D, as well as the amount of heat absorbed and work performed by the system. ∆U = (3/2)R * (T3-­‐T1) = 1.53 kJ w = P1*(V2-­‐V1) = 1.01 kJ (watch the units. Must convert atm to Pa and L to cubic meters). Name
Section
TA
Ch 1b, Problem Set 7
Due March 10, 2014
at 4 PM in the Ch1b Box
q = 2.54 kJ 5 points f) Using the values obtained above, calculate change in internal energy associated with Process E, as well as the amount of heat absorbed and work performed by the system. E = A -­‐ D, so ∆U = -­‐1.53 kJ w = 0 because volume doesn’t change, so q = -­‐1.53 kJ 4 points g) Use your results from parts (e) and (f) to obtain a numerical value for the change in internal energy associated with Process A. ∆U = 0 (still). 2 points