Activity Set 5 1. Consider Charles’ data and see what absolute zero temperature you get: See Excel Sheet for graph of Volume vs. Temperature. The graphs give that the temperature and volume are directly proportional with R squared values of 1 and 0.991. Since we know the structure of a molecule, we know that there are electron repulsions when two molecules approach each other (as well as nuclear repulsions should the nuclei approach). Even if there were not, a decrease in pressure would not physically compress two molecules that were touching as a lack of temperature is a lack of energy not a force. This sets a physical limit on the decrease in volume. Since there is obviously a physical limit to the decrease in volume with temperature decrease, and there is a linear relationship between volume and temperature, there must exist such a temperature that no molecule in the space contains any thermal energy whatsoever and does not translate, rotate, or vibrate and occupies only the space necessary for the electronic orbitals themselves (absolute zero). This is the value on the x‐axis when graph is extrapolated to the volume that would only account for the volume of the molecules themselves (at ‐273.15 C) Calculation: To calculate absolute zero from this graph, first define what the minimum possible volume is. The hydrogen atom has a diameter of about 1 angstrom. Treating the molecules like double boxes of 1 angstrom length as an approximation gives: 2 angstroms3 *(6.022*1023 molecules/mol )(0.0447 mols) = 5.38*1022 angstroms cubed=5.384*10‐14 cm3≈ 0 cm3 So we can solve essentially for zero volume. From the linear line fit we have: Y=0.0037x + 0.991 Where y is the volume and x is the temperature in degrees Celsius. Setting the volume equal to zero yields: 0=0.0037x+0.991 X=‐0.991/0.0037 X= ‐267.84 Degrees Celsius 2. Can you imagine any reason why Avogadro was right? Why is V proportional to n and not dependent on density of the gas or the mass of the molecules in question? This problem can be conceptualized by understanding that pressure is a measure of the intensities and the frequencies of collisions against the sides of a container. Take the volume to be a measure of the average path length a molecule would need to travel in order to collide with the side of the container. Any given volume has the same average path length regardless of the shape of the container. The temperature is a measure of the average thermal energy of the molecules which determines the intensities with which they hit. Taking the pressure and the temperature to be the same, the intensities with which the molecules are hitting remains constant (temperature constant) and the frequencies with which they hit must also remain constant (pressure constant), then the addition of any new molecules would increase the total number of collisions by an amount proportional to the amount of molecules added. In order to counteract the new total number of collisions by the addition of new molecules, so that the average intensities and frequencies of collisions remain constant (pressure and temperature still remain constant) then the only option available would be to increase the path length by the same proportion as the addition of the new molecules. This proportional increase in path length directly translates into a proportional increase in volume. The mass of the molecules added would not matter, as the average thermal energy (temperature) must remain the same and the frequencies of the collisions must remain the same (pressure). A larger molecule added would have the same energy, so it would hit less often, but with more force equating to the same pressure. It would not depend upon the density because given the mathematical equation: V Mw*n/V
V2 Mw*n And because as previously stated, since average thermal energy is constant as well as the sum of the intensities and frequencies of collisions, volume is not dependent upon the mass. For the problem with the two chambers of a cylinder separated by a disk, if the disk were to be removed and the temperature and the pressure were to remain constant, the piston would move inward to restore the previous available volume removing the disk created slightly more total available volume . This occurred because no parameter of the ideal gas law was truly altered. The new environment has twice as much volume with twice as many molecules/atoms. This still results in the same pressure and temperature. If reality were such that the pressure was dependent upon the density of the gas, then the two containers couldn’t possibly exist at similar volumes, pressures, and temperatures before removing the disk, because their densities are not the same, and mixing the two would cause some sort of change in pressure/and or temperature/ and or volume since these two containers couldn’t possible both exist with the same pressure, volume, and temperature before the mixing. 3. Show how to rewrite the ideal gas equation to be dimensionless in terms of pressure and concentration, to show that pressure is always proportional to concentration. This can be done by first dividing the ideal gas law by volume on both sides to give: P CRT, where C is concentration in mol/L Molarity. This would change the ideal gas constant dimension to M‐1 atm K‐1 mol‐1. P CT 0.08205746 14 M‐1 atm K‐1 Then we can divide by a reference pressure 1 atm to obtain: Pdimensionless CT 0.08205746 14 M‐1 K‐1 Where the pressure and the constant term lost the atmosphere dimension. The next step is to move the constant to the other side to give: 12.18658240 14 M‐1 K‐1 Pdimensionless CT Then divide by a reference concentration on both sides 1 M to give: 12.18658240 14 K Pdimensionless CdimensionlessT Move the constant back to the other side to give: Pdimensionless CdimensionlessT 0.08205746 14 K‐1 Now every unit in the ideal gas law besides the temperature measured in Kelvin is dimensionless. The constant and the temperature could be written together as a dimensionless temperature unit such as τ. 4. Explain Gay Lussac’s observation and predict the volume of the reaction involved monoatomic atoms. Given the ideal gas equation PV nRT, and taking P,V, and T to be constant throughout the reaction, the n number of mols of reactants was 3 and the n of the products was 2. From the explanation in problem 2, we know the same proportion must hold for the change in volume. The volume must decrease by a factor of 1/3 since n decreased by a factor of 1/3. If both the reactants were taken as monoatomic, the equation would be: 2H O H2O Since the relationship of n is now a decrease from three to one, the volume would also have to decrease by a factor of 2/3. 5. How much mass can a balloon of differing elements carry? The buoyant force of the balloon is equal to the weight of the air that it displaces minus the weight of the balloon itself. Solving for n H2 or He for 1L, 1Atm, and 273K gives: N
1Atm 1L / 0.08205746 14 L atm K
1 mol 1 N 0.0446 mol Mass H2
2*0.0446g Mass H2 0.0892g Mass He
4*0.0446g Mass He 0.179g Mass air .80 28 0.0446
Mass air 1.29g .20 32 0.0446 g 273K Buoyant Force of H2
.00129kg‐0.0000892kg 9.8m/s2
0.0117N buoyant force Buoyant Force of He
0.00129kg‐0.000179kg 9.8m/s2
0.0108N buoyant force Hydrogen and helium have fairly similar buoyant forces for a 1L balloon at 1 atm and 273K. The air is about 80% N2 which has a molecular weight of about 28 the other 20% is O 2 with a molecular weight of about 32. Since argon has an atomic weight of about 40, it would not have a net buoyant force and an argon balloon would not float in air. This is different from a hot air balloon in that the volume will be smaller for a given pressure which the atmosphere will impose since the temperature is less. Otherwise, the principle of buoyancy is the same, hot air balloons just increase the temperature which increases the volume which increases the buoyant force. 6. How are concentration and density related? Concentration is a measure of the number of units be they molecules or atoms or whatever per unit volume. Density is a measure of the amount of mass per unit volume. They are related in that the units in question have an assignable mass per unit such as molecular weight and can be interconverted. In order to solve the problem for number six plug in all values except n use 1 L for volume into the ideal gas law to solve for the number of such molecules in one liter: 256 torr*1 L n 62.36367 11 L Torr K
N
256 torr*1 L / 62.36367 11 L Torr K
N 0.0103 mol 1 mol 1
400K 1 mol 1
400K Now solve for molecular weight using 4.06 g since C 4.06g/L and we solved N for V 1 L. Mw 4.06g/0.0103mol Mw 396g/mol 7. Ideal Gas Law Problems V nRT/P V
1.5mol 0.08205746 14 L atm K
V 33.60L CH2 1mol/33.60L 0.30 M CHe .5mol/33.60L 0.15 M PH2 1Atm* 1/1.5
0.67Atm PHe 1Atm* .5/1.5
0.33Atm 0.67Atm 0.33Atm 1.00Atm 1 mol 1
273K / 1Atm