Ryan Lenet MISE - Physical Basis of Chemistry Third Set of Problems - Due December 11, 2005

Transcription

1 Ryan Lenet MISE - Physical Basis of Chemistry Third Set of Problems - Due December 11, 2005 Submit electronically (digital drop box) by Sunday, December 11 by 6 pm. Note: When submitting to digital Drop Box label your files with your name first and then the label - HmwkThree. Please put you name in the Header along with the already-inserted page #. The following information may be useful when solving the problems: Ideal gas law: PV = nrt ; where R = L atm mole -1 K mole = x things. The atomic weight of an element listed in the periodic table is the mass of one mole of atoms in grams. This is termed the molar mass (i.e., mass per mole) of the element. The molecular weight - determined from molecular formula as the sum of the atomic weights of all of the constituent atoms - is the mass of one mole of molecules in grams. This is termed the molar mass (i.e., mass per mole) of the molecule. 1 atmosphere (atm) = 760 mm Hg = 76 cm Hg = kilopascals cm 3 = 1 L ; K = ºC ; 1 calorie (cal) = joules (J). 1. Given: 3 identical containers (same volume), at 1 atm pressure and 0 ºC. Each flask is filled with a different gas. Flask A: CO Flask B: NO 2 Flask C: He Answer each of the following and briefly explain your answer using the concepts covered in class: (a) Which flask contains the greatest number of moles of gas? Avagadro theorized that equal volumes of gases at the same pressure and temperature contain the same number of molecules. This implies that equal volumes of gases at the same pressure and temperature contain equal number of moles.

2 Therefore, all three flasks would contain the same number of moles of gas since they are all the same volume, pressure and temperature. (b) Which flask contains the largest total mass of gas? CO = 28 amu NO2 = 46 amu He = 4 amu mass because NO2 has the largest total mass. Flask B contains the largest total (c) energy? In which flask will the gas molecules have the greatest average kinetic CO E = mv E = (.028 kg)( m/s ) ^2 = Joules NO2 E = (.046 kg) ( m/s) ^2 = Joules He E= (.004 kg)(1,304.7 m/s) ^2 = Joules All three gases will have the same average kinetic energy. (d) In which flask will the molecules have the greatest average velocity? Vav = ( 3RT/Mkg)^ CO =.028 kg/ mol Vav = [ ( 3) (8.314 J/ mol K )( 283 K) / kg/mol) ^1/2 ] = m/s NO2 =.046 kg/ mol Vav = [ ( 3) (8.314 J/ mol K )( 283 K) / kg/mol) ^1/2 ] = m/s He = kg/mol Vav= Vav = [ ( 3) (8.314 J/ mol K )( 283 K) / kg/mol) ^1/2 ]= 1,304.7 m/s He will have the greatest average velocity since it is the lightest gas. The lighter the gas the quicker the speed. The heavier the gas the slower the gas will move.

3 2. A mixture of gases effusing A chamber is filled with a gaseous mixture consisting of 1.00 mole of He gas and 4.00 moles of SO 2 gas at a certain temperature. [The atomic weight of He is 4.00 g/mole and the molecular weight of SO 2 is 64.0 g/mole] The mixture is allowed to escape into a series of two evacuated chambers ( 1 & 2 ) connected by pinholes. Please refer to the diagram below. 1 2 Initial Chamber You may assume that all three chambers have the same volume and are rigid. You may also assume that both pinholes are exactly the same area. Finally, temperature is constant throughout the entire process. [Recall the kinetic theory lecture slides (especially #15) depicting the process and relevant relationship to determine how the composition of a mixture of gases changes when the mixture effuses (leaks) through a pinhole into an evacuated chamber.] A useful proportion when working in moles is the mole percent. If we have a mixture of substances, A and B, the mole % is defined as: mole % A = moles A moles A + moles B 100 % = moles A total moles 100 % mole % B = moles B moles A + moles B 100 % = moles B total moles 100 % Of course, mole % A + mole % B = 100 %. Now we are ready to go. Please show your work. (a) Determine the mole percent of each gas in the initial chamber before effusion occurs. Mole % of A 1 mole of He / 1 mole of He + 4 moles of SO2 = 1/5 Mole % of B 4 moles of SO2/ 1 mole of He + 4 moles of SO2 = 4/5 Mole % of A He = 1/5 (100%) = 20% of He Mole % of B SO2 = 4/5 ( 100%) = 80% of SO2 (b) Determine the ratio of moles He : moles SO 2 in chamber 1 a few moments after the gaseous mixture begins to effuse through the pinhole into evacuated chamber 1. Now, determine the mole percent of each gas in this effused mixture. VHe/VSO2 = (M SO2/ M He)^1/2 = (64g of SO2/ 4g of He)^1/2 = 4

4 Helium will travel 4 times faster into chamber one than SO2. There is 80 % of He in chamber 1 and 20% of SO2 in chamber 1. (c) Determine the ratio of moles He : moles SO 2 in chamber 2 a few moments after the gaseous mixture begins to effuse through the pinhole into evacuated chamber 2. Now, determine the mole percent of each gas in this effused mixture. The same calculations could be made when the gases go into chamber 2. If that is true then the ratio will be 1:16, 96 % of He and 4% of SO2. (d) In class, we said that the effusion process creates a gaseous mixture that is enriched in the gas of lower molar mass. Thus, if more than one effusion occurs, each newly effused mixture is even more enriched in the gas of lower molar mass. Compare your results to (a), (b), and (c) and briefly comment on the validity of these assertions. This was evident in the results above. As the two gases moved away from the first chamber, the gas with a lower molar mass, which was the Helium, moved quicker into the other two chambers. Since helium is a lighter gas it traveled faster than the SO2. The effusion rate of helium was quicker than SO2. From the results above, there was more helium in the following chambers as the gases moved away from the first chamber. This was present in the Manhattan Project. 3. Real gases - relaxing some of the postulates from kinetic theory. Recall the postulates of the kinetic theory of gases (see kinetic theory lecture slide #2). These postulates rationalize the ideal gas law (PV = nrt) at the molecular level. The two postulates that may be the most unrealistic are the first two: 1. Gas particles have no volume, i.e., they are point particles. 2. There are no inter-particle forces - neither attractive nor repulsive. A more realistic model of gas behavior modifies each of the above postulates. The two modified postulates are: 1. Gas particles have volume. If we presume that the particles are hard spheres - like micro billiard balls - they will occupy some non-negligible volume of the container volume. Thus, the empty space, termed, the available volume, is necessarily less than the full volume of the container. So, if V is the volume of the container, the volume available to the gas particles is V minus the unavailable volume due to the spheres. In brief, due to particle size, there is less space available to the gas molecules than the full container volume would imply. 2. There are inter-particle (usually termed intermolecular) forces between the particles. The most important are attractive forces. Thus, gas particles feel each other s presence and pull on each other. We may thus imagine that a molecule in the middle of the sample - if evenly surrounded

5 by its identical neighbors - may not feel any pull at all due to the balancing forces, as you learned in physics. However, a molecule next to the wall only has neighbors beside it and behind. Now the forces will be unbalanced and so there may be some net pull. So, a molecule approaching the wall may hit with less force (why?) and/or be diverted from hitting the wall at all. Let s now consider how the pressure of the gas - as predicted by the ideal gas law - is altered by the two modified postulates above. Of course, pressure is still force/area - this is its definition. We will continue to presume that the microscopic origin of the macroscopic gas pressure is due to the molecules colliding with the container walls. (a) How would the pressure (P) of the gas be altered from its ideal value (P ideal ), due only to modified postulate 1? [The ideal pressure (= nrt/v) is what we would expect if the particles followed the original postulates from kinetic theory.] Please select one of the choices below and then carefully explain your choice. No calculations required or expected. P < P ideal P > P ideal P = P ideal (no change) In an ideal situation, gas particles take up no space, or have no volume. In reality, gas particles do take up some space and have some volume very small volume. Therefore, since they do have volume, there is less available space in a container, which one could conclude that there would be greater pressure in reality than in the ideal situation. So, P is greater than P ideal (b) How would the pressure (P) of the gas be altered from its ideal value (P ideal ), due only to modified postulate 2? [The ideal pressure (= nrt/v) is what we would expect if the particles followed the original postulates from kinetic theory.] Please select one of the choices below and then carefully explain your choice. No calculations required or expected. P < P ideal P > P ideal P = P ideal (no change) In postulate 1, there are no inter-particle forces. Gaas particles are in constant random motion. The elastic collisions of particles with walls is the pressure of the gases. Realistically, there are inter-particle forces which would lower the frequency of particles colliding with walls which would lower the pressure. Therefore, the real pressure is less than the ideal pressure. P is less than P ideal. (3c) Now imagine two real gases X and Y, where each gas is presumed to fulfill both of the modified postulates (molecules have volume and experience intermolecular attractive forces). Gases X and Y are in separate containers of the same volume at the same temperature. Each container contains the same number of moles of gas. A measurement of the pressure of gas X at these conditions is 15.0 atm. A measurement of the pressure of gas Y at these conditions is 11.0 atm. Please answer the following and explain you answers. No calculations required or expected. (i) According to the ideal gas law, P ideal for each gas should be the same. Why?

6 Pressure should be the same, because temperature, moles and volume are all the same. (ii) The calculated ideal pressure is 12.0 atm. Considering each gas separately, which non-ideal factor (molecular volume or attractive forces) is contributing most to the measured pressure? How do you know? If you consider the two gases separate, the pressure of gas X has a higher pressure due to molecular volume change and the pressure due to attractive forces. Since the ideal gas is 12 atm. Gas X has a 3 atm difference while gas y has a 1 atm difference. Therefore, molecular volume contributed most to the changed pressure. 4. A little bit of heat & a honeymoon (a) The specific heat of lead (Pb) is J g -1 C o -1. Calculate this value in cal g -1 C o -1 and cal g -1 K o -1 and cal mol -1 C o -1. (0.128 J/g degrees C )( cal/j) = cal/g degrees C or.031 cal/g degrees C (.031 cal/g degrees C)( 1 degree C/274 degrees K) = 1.13 x 10^-4 cal/g degrees K (.031 cal/g degrees C)( 1 g/6.022 x 10^23/mole) = 5.15 x 10^20 cal/mole degrees C (b) Extra Credit: In honor of Mary-Kate The SI energy unit known as the joule (J) is named after James Prescott Joule. [By the way, according to family records - the name Joule was pronounced jol (long o)!] While Joule was on his honeymoon in Switzerland in 1847, he couldn t forget about his experiments on heat flow and temperature change! He took some precise thermometers along - able to measure temperature differences of a few tenths of a ºC. See what you can find out about Joule s honeymoon experiment and specifically what it had to do with heat flow, temperature change, and waterfalls. What was the purpose of the experiment, i.e., what was Joule trying to prove/show? Put together a few well-chosen sentences - no more than a paragraph. Of course, cite your source(s). While studying a waterfall during his honeymoon, Joule found the temperature of the water at the base of the waterfall higher than the temperature of the water at the top. This proved that the energy of the falling water was being converted into heat. This was a major step in proving the theory of energy conservation, which is known as the First Law of Thermodynamics. I found this source from ulearntoday.com as I stated below. physics_article1.jsp?file=joule