Vapour Pressure as a Function of Temperature of Ether

Vapour Pressure as a Function of Temperature of Ether Xin Wang Title To measure the enthalpy of vaporization and the boiling point of diethyl ether by cooling the ether down and continuously recording a series of different temperature readings and their corresponding vapour pressure values. Abstract Vapour pressure p of ether under a series of different temperatures T were measured and three repeats were done. The three graphs of ln(p) against 1/T were plotted, all of them showed a nice linear relationship between ln(p) and 1/T . Then the gradient which was equal to à ¯Ã‚ Ã‚ ²Hvap/R was used, and hence the average value of à ¯Ã‚ Ã‚ ²Hvap of three repeats was decided as the final result . The boiling point was calculated by using à ¯Ã‚ Ã‚ ²Hvap and the final value used of constant C was also the mean of three repeats’ constants obtained. Error analysis and improvements both focused on the leaking of apparatus and the validities of measurements obtained. Introduction Thermodynamics is a very important part of chemistry and chemical studies. It involves many studies about energy such as the transfer of energy and the conservation of energy. For this experiment which is to study the vapor pressure as a function of temperature, the thermodynamic part which should be focused on is that a system would tends to make its Gibbs Free Energy which is the energy free to do work minimum at a constant temperature and pressure. In this experiment, a single and pure compound is used and the study is about the equilibrium between its liquid phase and its gaseous phase at a constant pressure and temperature. If the pure compound liquid is placed inside a closed container, under different temperature and different pressure, there will always be a equilibrium between the liquid phase and gaseous phase. This thermodynamic equilibrium will be at different position due to the temperature and pressure Therefore, the region of the temperature and pressure where the compound changes from liquid to gases or reversibly has been defined as phase boundaries, which can be found in a graph of pressure against temperature. To be more specific, the temperature is the temperature of the system, but the pressure mentioned is the vapour pressure. The vapour pressure is defined as the pressure of the vapour which is in equilibrium with its liquid in the closed container, and it does vary with the temperature. According to the second law of thermodynamics which indicates â€Å"At equilibrium, the chemical potential of a substance is the same throughout a sample, regardless of how many phases are present.† [1] , a equation for the two chemical potentials of two phases can be obtained, which shows the chemical potential of the liquid is equal to the chemical potential of the gases in equilibrium in the closed container. To make this simpler, the chemical potential â€Å" µÃ¢â‚¬  which is the molar Gibbs free energy [1] is introduced. Therefore, at a constant pressure and temperature,  µ is equal to the first derivative of G with respect to n, where G is the Gibbs free energy and n is the number of moles. As phase boundary is represented by the line which separates the liquid phase region and gaseous phase region in a graph of vapour pressure against temperature, therefore the gradient of the phase boundary which is written as dp/dT can be considered. In the closed container, the overall chemical potential of the system is always constant by considering conservation of energy. Therefore, the change in  µ of liquid phase should always be equal to the change in  µ of gaseous phase when temperature and pressure change, which can be expressed as d µ(liquid) = d µ(gases)[1]. The fundamental equation â€Å"dU = TdS PdV† [2] can be used to derive another equation of d µ. Because of H = U + pV [3], therefore dH = dU + d(pV) = dU + pdV + Vdp is derived. As known, G = H -TS and then dG = dH d(TS) = dH TdS SdT is obtained. By combining the two derived equations and the fundamental equation mentioned above, dG = -SdT + Vdp can then be obtained finally. Due to the definition of  µ, equation d µ = -SdT + Vdp can be derived. If d µ = -SdT + Vdp is substituted into d µ(liquid) = d µ(gases), [V(gases) V(liquid)]dp = [S(gases) S(liquid)]dT can be derived. Therefore, dp/dT = à ¯Ã‚ Ã‚ ²S/à ¯Ã‚ Ã‚ ²T where à ¯Ã‚ Ã‚ ²S is defined as the the entropy of phase transition which can be equal to S(gases) S(liquid) and à ¯Ã‚ Ã‚ ²T is defined as the volumes of phase transition which can be equal to V(gases) V(liquid). Due to the perfect gas law which is pV = nTR, where R is the gas constant which is equal to 8.314 J K-1 mol-1 and T is the temperature in kelvins. For  µ which is molar, therefore n = 1 and then pV = RT, so V = RT/p is obtained for gases. The assumption that à ¯Ã‚ Ã‚ ²V = V(gases) V(liquid) ≈ V(gases) [2] can be made as the volume of liquid in the closed system is much smaller than the volume of gases, and hence à ¯Ã‚ Ã‚ ²V = RT/p can be derived. For à ¯Ã‚ Ã‚ ²S, as à ¯Ã‚ Ã‚ ²Hvap = Tà ¯Ã‚ Ã‚ ²S can be derived due to the equilibrium, so à ¯Ã‚ Ã‚ ²S = à ¯Ã‚ Ã‚ ²Hvap /T is then derived. By substituting the two new equations for à ¯Ã‚ Ã‚ ²S and à ¯Ã‚ Ã‚ ²T into dp/dT = à ¯Ã‚ Ã‚ ²S/à ¯Ã‚ Ã‚ ²T, dp/dT = pà ¯Ã‚ Ã‚ ²Hvap /RT2 can be obtained. Then do the integration for the equation dp/dT = pà ¯Ã‚ Ã‚ ²Hvap /RT2 , ln(p) = -à ¯Ã‚ Ã‚ ²Hvap/RT + C can be derived, where C is a constant. For the equation ln(p) = -à ¯Ã‚ Ã‚ ²Hvap/RT + C, a graph of ln(p) against 1/T can be plotted to find à ¯Ã‚ Ã‚ ²Hvap which is the enthalpy of vaporization can obtained from the gradient of the graph due to the linear relationship between ln(p) and 1/T. Therefore, à ¯Ã‚ Ã‚ ²Hvap = -gradient Ãâ€" R To use the equation to find the boiling point of the compound used, due to the definition of boiling point, just substitute pressure = 1 atmosphere ≈ 1 bar into the equation to find the corresponding T which is therefore the boiling point. Therefore, ln(1) = 0 = -à ¯Ã‚ Ã‚ ²Hvap/RTboil + C, and hence Tboil = à ¯Ã‚ Ã‚ ²Hvap/RC can be used for calculation. Equipment Methylated spirits, cardice, ether(diethyl ether), tweezers, a large dewar flask, a smaller dewar flask, wasted dewar flasks, clamps, experimental vacuum set-up with air pump connected, stopwatch, electrical mixer, thermometers, funnels, beakers Experiment Turned on the digital vacuum gauge and then took down the bulb A from the apparatus set, some ether was added into the bulb and then the bulb was reconnected to the apparatus. Next, prepared methylated spirits was added to a large dewar, then enough cardice was transferred from the the main container into a small container and added to the dewar with methylated spirits by using a tweezer. Enough cardice was added to make the methylated spirits reach a very low temperature which was about -78à ¢Ã¢â‚¬Å¾Ã†â€™. After the temperature of the methylated spirits was low enough, some of the mixture was poured into a smaller dewar and then the cold trap was submerged by the mixture in the smaller dewar which was clamped and therefore fixed to continuously cold the cold trap. Then the air pump was turned on, the Tap 1 which was vacuum to Ether in the bulb A and the Tap 3 which was vacuum to Atmosphere were both closed while the Tap 2 which was vacuum to Transducer was open. After the cold trap was cold enough, open Tap 1 to boil the ether for seconds to make sure all the air inside those tubes were pumped out, and then closed Tap 1 and measure the rate of leaking by closing Tap 2 as well and then measuring the amount of pressure increased by looking at the digital vacuum gauge in one minute by using a stopwatch. The thermometer and the electrical mixer were well setted up around the bulb A as a thermostat bath, then submerged the bulb A with ether inside into the remaining mixture in the large dewar and clamped the large dewar to make it fixed. Then continuously add cardice into the large dewar and the smaller dewar to cool down the ether and keep the cold trap still cold. When the temperature of the mixture in the large dewar reached about -78à ¢Ã¢â‚¬Å¾Ã†â€™, the large dewar was removed and Tap 2 was closed with Tap 1 open. The mixture in the large dewar was the pored into a wasted dewar and then the dewar was allowed to warm up. Then the ether in the bulb A was allowe d to warm up to room temperature and cold tap water in a beaker was used to submerge the bulb A to help warm up. After a enough long time was used for warming up, the thermometer was touched to the bulb A to measure the temperature of the ether at room temperature and this temperature reading and the corresponding pressure reading were both recorded as the first pair of data. The empty large dewar had also been warmed up already, fresh methylated spirits was then added to the dewar and the bulb A was submerged into the fresh mixture with the thermostat bath set also submerged in the mixture. The thermostat bath was started by turning on the electrical mixer, at the same time, Tap 3 and Tap 2 were still closed while Tap 1 was open. For obtaining data sets, a small amount of cardice was added each time to make sure the temperature only went down by 2 to 3 degrees or kelvins. The temperature reading and its corresponding pressure reading were both recorded after adding the cardice and waiting for both reading to be steady. The experiment was finished until the temperature had reached about -55 degrees . After finishing the experiment, the electrical mixer was then turned off and the Tap 1 and the Tap 2 were closed but the Tap 3 was then opened. The air pump was switched off after that. Next, both the large dewar and the smaller dewar were removed, and the methylated mixture in both dewars were poured back into the the container of methylated spirits using a funnel. The experiment had been repeated for three times and therefore three data sets were obtained. And all the repeats were done by using the same apparatus under the same condition. Results During the experiments, the raw data recorded for temperature was in degree and pressure was in mbar. See Table 1. For 1/T, Temperature was changed into the unit of kelvin. For ln(p), pressure was changed into the unit of bar. See Appendix A, Appendix B and Appendix C. Run 1 Run 2 Run 3 T /degrees p /mbar T /degrees p /mbar T /degrees p /mbar 15 661 8 511 14 582 9 547 5 448 4 432 7 476 3 393 1 385 4 424 -1 342 -2 316 2 378 -4 296 -6 268 -2 330 -7 253 -10 216 -4 291 -12 205 -14 182 -7 250 -15 174 -18 150 -10 220 -18 150 -21 126 -12 193 -21 132 -25 102 -14 175 -25 105 -27 91 -17 155 -31 80 -31 73 -20 132 -34 66 -35 59 -22 116 -37 53 -38 46 -26 95 -41 43 -42 37 -29 80 -44 38 -45 30 -33 65 -47 32 -48 26 -36 53 -51 26 -50 22 -40 41 -55 23 -52 20 -42 35 -60 18 -57 15 -44 32 -59 13 -48 25 -62 11 -52 20 -54 17 -56 14 -59 12 Table 1. Raw data from the experiments Figure 1. ln(p) against 1/T for Run 1 Figure 2. ln(p) against 1/T fro Run 2 Figure 3. ln(p) against 1/T for Run 3 Rate of leaking of pressure of the apparatus = 2 mbar/min To obtain à ¯Ã‚ Ã‚ ²Hvap, the average value of gradients of three repeats which were obtained from Figure 1, Figure 2 and Figure 3 was used to calculate the final result à ¯Ã‚ Ã‚ ²Hvap. This was the same for the constant C, therefore the mean value of constants C of three repeats was used as the final result of C. The average gradient value = [(-3381.1) + (-3048.6) + (-3287.6)] à · 3 = -3239.1 K à ¯Ã‚ Ã‚ ²Hvap = -gradient Ãâ€" R = -3239.1 Ãâ€" 8.314 = 26929.8774 J mol-1 Constant C = (11.329 + 10.087 + 10.969) à · 3 = 10.795 Boiling point = Tboil = à ¯Ã‚ Ã‚ ²Hvap/RC = 26929.8774/(8.314 Ãâ€" 10.795) = 300.0555813 K ≈ 26.9 à ¢Ã¢â‚¬Å¾Ã†â€™ (to 3 s.f.) Data Analysis The theoretical value of the enthalpy of vaporization of diethyl ether is 27140 J mol-1 at 25 à ¢Ã¢â‚¬Å¾Ã†â€™ [4] , therefore, percentage of error of à ¯Ã‚ Ã‚ ²Hvap = [(27140 26929.8774) à · 27140] Ãâ€" 100% ≈ 0.774% (to 3 s.f.) The theoretical value of the boiling point of diethyl ether is 34à ¢Ã¢â‚¬Å¾Ã†â€™ [5], therefore, the percentage of error of boiling point = [(34 26.9) à · 34] Ãâ€" 100% ≈ 20.9% (to 3 s.f.) Uncertainty in T = (0.25 à · 1) Ãâ€" 100% = 25% Uncertainty in 1/T = (0.25 à · 12) Ãâ€"100% = 25% Uncertainty in p = (0.5 à · 11) Ãâ€" 100% ≈4.55% (to 3 s.f.) Uncertainty in ln(p) = (0.5 à · 11) Ãâ€" 100% ≈4.55% (to 3 s.f.) Uncertainty in à ¯Ã‚ Ã‚ ²Hvap = uncertainty in gradient = 25% + 4.55% = 29.55% Uncertainty in constant C = 29.55% Uncertainty in boiling point Tboil = 29.55% + 29.55% = 59.1% The values of R2 which are coefficients of determination were all shown to be fairly close to 1, therefore, the linear correlation between ln(p) and T was shown to be very nice. The percentage of error of boiling point is much larger than the percentage of error of à ¯Ã‚ Ã‚ ²Hvap, and the uncertainty of boiling point is even larger due to the uncertainties of 1/T and ln(p). This might come from two main sources of errors. One is the leaking of the experimental apparatus and the other is not high enough validities of temperature and pressure measurements obtained due to other errors like human errors. For the leaking of the apparatus, the rate of leaking was measured as 2 mbar/min. To be more specif, the leaking was caused by the incompletely sealed connections between those glass tubes. The time from starting to take down measurements to finishing the experiment was about half an hour, this has led to a large uncertainty in the pressure measurements. Therefore, the readings of pressure at the lower temperature are much less accurate and smaller than the actual value that they should be, this shifted the gradient to a higher value, but the more serious effect was on the constant C which is the intercept with the axis of ln(p). As pressure was first recorded in mbar, when changing it into the unit of bar, the value was less than 1, therefore ln(p) is less than zero, a decrease in the pressure value can lead to a large change in the value of ln(p). So the constant C is much higher than the actual value. Although the percentage of error of à ¯Ã‚ Ã‚ ²Hvap is small, but this does not mean the experiment was very accurate because its uncertainty is quite large, and this low percentage error might due to the combination of different errors. For not high enough validities of temperature and pressure measurements obtained. The human error of recording temperature has the largest effect, and it has been indicated by the uncertainty of temperature reading which is 25%, this uncertainty is quite large enough to lead to inaccurate temperature readings and therefore the large uncertainties of the enthalpy of vaporization and boiling point. Another error causing low validities is the time used to wait until the temperature and pressure readings to be steady, which can definitely lead to more leaking of pressure. Although the experiment was about 30-minute long, but some readings were recorded before they became steady, and this would make he readings recorded higher than the actual ones. Improvements To reduce the error caused by the leaking of the apparatus, more advanced apparatus much be used. The leaking in these experiments was mostly caused by the incompletely sealed connections between those glass tube. Therefore, a set-up which has all glass tubes well connected without any crevice would be an ideal choice, this kind of set-up should be an entirety and can effectively avoid leaking. If the problem of leaking is solved, the time for waiting until the readings to become steady each time can be as long as possible to make sure the measurements are as accurate as they can. To reduce the human error on taking down the readings of temperature and pressure, computers and sensors are suggested. For example, thermocouples [6] which are common temperature sensors used in industry, this type of temperature sensors can directly provide the electrical readings of temperature on the screen of the equipment and the size is fairly nice for using in laboratories. Computers can be used to take down the reading on those electrical equipment such as temperature sensors and the digital vacuum gauge by using the softwares or internal programs. More precautions must be payed attention. Therefore, make sure the ether used is pure enough and the experimental apparatus especially the bulb is clean to prevent from impurities, and during the step of boiling the ether by vacuuming the bulb, take a longer time to boil the ether to make sure that all the air in the tubes can be swept out. Conclusion In conclusion, the graphs have proved that the linear relationship between ln(p) and 1/T was reliable. Therefore, the main equation ln(p) = -à ¯Ã‚ Ã‚ ²Hvap/RT + C can be used to find the enthalpy of vaporization and the boiling point of ether and the assumption which has been made about the volumes is reliable in this case. However, the apparatus and some technics much be improved to reduce errors and then obtain accurate results. References [1] 6. Vapour Pressure a a Function of Temperature, 1st Year Physical Chemistry Laboratory 2013 lab script, p. 2 [2] 6. Vapour Pressure a a Function of Temperature, 1st Year Physical Chemistry Laboratory 2013 lab script, p. 3 [3] Enthalpy. Available from: http://en.wikipedia.org/wiki/Enthalpy [Accessed 26th March 2014] [4] Majer, V.; Wagner, Z.; Svoboda, V.; Cadek, V., Journal of Chemical Thermodynamics, 1980 , vol. 12, # 4 p. 387 392 [5] Gomberg, M., Journal of the American Chemical Society, 1923 , vol. 45, p. 398 398 [6] TemperatureSensors. Available from: https://controls.engin.umich.edu/wiki/index.php/TemperatureSensors [Accessed 26th March 2014] Appendix A Run 1 Temperature/degrees Pressure/mbar 1/T (T in kelvin) ln(p) (p in bar) 15 661 0.003470415 -0.414001439 9 547 0.003544214 -0.603306477 7 476 0.003569516 -0.742337425 4 424 0.003608154 -0.858021824 2 378 0.003634381 -0.972861083 -2 330 0.003687996 -1.108662625 -4 291 0.0037154 -1.234432012 -7 250 0.00375728 -1.386294361 -10 220 0.003800114 -1.514127733 -12 193 0.003829217 -1.64506509 -14 175 0.003858769 -1.742969305 -17 155 0.003903963 -1.864330162 -20 132 0.003950227 -2.024953356 -22 116 0.003981684 -2.154165088 -26 95 0.004046126 -2.353878387 -29 80 0.004095843 -2.525728644 -33 65 0.004164064 -2.733368009 -36 53 0.00421674 -2.937463365 -40 41 0.004289084 -3.194183212 -42 35 0.004326195 -3.352407217 -44 32 0.004363954 -3.442019376 -48 25 0.004441483 -3.688879454 -52 20 0.004521818 -3.912023005 -54 17 0.004563085 -4.074541935 -56 14 0.004605112 -4.268697949 -59 12 0.004669624 -4.422848629 Appendix B Run 2 Temperature/degree Pressure/mbar 1/T (T in kelvin) ln(p) (p in bar) 8 511 0.00355682 -0.671385689 5 448 0.003595182 -0.802962047 3 393 0.00362122 -0.933945667 -1 342 0.003674444 -1.072944542 -4 296 0.0037154 -1.217395825 -7 253 0.00375728 -1.37436579 -12 205 0.003829217 -1.5847453 -15 174 0.003873717 -1.74869998 -18 150 0.003919263 -1.897119985 -21 132 0.003965893 -2.024953356 -25 105 0.004029821 -2.253794929 -31 80 0.004129672 -2.525728644 -34 66 0.004181476 -2.718100537 -37 53 0.004234597 -2.937463365 -41 43 0.00430756 -3.146555163 -44 38 0.004363954 -3.270169119 -47 32 0.004421844 -3.442019376 -51 26 0.004501463 -3.649658741 -55 23 0.004584002 -3.772261063 -60 18 0.004691532 -4.017383521 Appendix C Run 3 Temperature/degree Pressure/mbar 1/T (T in kelvin) ln(p) (p in bar) 14 582 0.0034825 -0.541284831 4 432 0.003608154 -0.839329691 1 385 0.003647638 -0.954511945 -2 316 0.003687996 -1.152013065 -6 268 0.003743215 -1.316768298 -10 216 0.003800114 -1.532476871 -14 182 0.003858769 -1.703748592 -18 150 0.003919263 -1.897119985 -21 126 0.003965893 -2.071473372 -25 102 0.

Mary Breckinridge

Jennifer St. Pierre 7/10/2012 Mary Breckenridge NU 120 Michelle R. Edwards MSN, RN Breckenridge School of Nursing Mary Breckenridge was born in 1881 in Kentucky. She was born into an influential family, and for that she enjoyed a privileged childhood as well as getting an education in the U. S and Europe. Mary Breckenridge’s father was the U. S ambassador to Czar Nicholas II of Russia. By the time Mary Breckenridge was 26 years old she had become widowed, as well as losing both of her children at an early age. At this time Mary Breckenridge has decided to dedicate her life in improving the health of women and children. Gina Castlenovo, November 2003. ) Mary Breckinridge became a registered nurse in 1910 and worked at St. Luke’s Hospital in New York. During this time she was as well working in France during World War I, this is where Mary Breckinridge became exposed to new healthcare ideas. Mary Breckenridge stated â€Å"After I had met British nurse-midwives, first in France and then on my visits to London, it grew upon me that nurse-midwifery was the logical response to the needs of the young child in rural America†¦ My work would be for them†. Gina Castlenovo, November 2003. ) Proceeding after World War I Mary Breckenridge went to Columbia University and studied public health. She wanted to conquer the health issues in eastern Kentucky; this area had few roads and absolutely no physicians. Her theory was if she could be successful in such a run down, poor area she could be successful anywhere. Mary Breckenridge got around by traveling horseback and teaching families about their health as well as local lay midwives about birth practices.By doing this she had learned that women lacked prenatal care and gave birth to an average of nine children, this was done by mostly self taught midwives, and farmer’s wives. They relied on traditional beliefs and invasive procedures. (Gina Castlenovo, November 2003. ) Mary Breckinridge believed children’s healthcare should start in the prenatal period (birth-child’s first years) due to a high maternal mortality. When returning to London she became a certified nurse-midwife. She then went to Scotland to observe the work of a community midwifery system.This system served poor and rural areas. The structure was decentralized and was used as a model for the Frontier Nursing Services. Once arriving back in Kentucky in 1925, Mary Breckinridge began the work that would introduce a new type of rural health care in the United States. (Gina Castlenovo, November 2003. ) In 1925 The Frontier Nursing Service (FNS) was established, this was a private charitable organization. The entire serving area was about seven hundred square miles in southeastern Kentucky. Mary Breckinridge raised over six million dollars to support this organization.Many people believed this was due to her influential connection and speaking engagements. The staff was made up of nurse-midwives that w ere trained in England. The staff traveled by horseback and or by foot so that they could provide quality prenatal and childbirth care in the clients’ own home, they functioned as both midwives and family nurses. Clients were able to pay low fees in money and or goods. No one was turned away, and in doing so both maternal and infant mortality rates decreased significantly. (Gina Castlenovo, November 2003. )The Frontier Nursing Service (FNS) registered over sixty four thousand patients since 1925. The FNS as well delivered over seventeen thousand babies with only eleven deaths. One of the nurse-midwife began the first American school of midwifery in New York in 1932. The FNS founded its own school in Hyden Kentucky in 1939. Mary Breckinridge ran the Frontier Nursing Service until she passed away in 1965. (Gina Castlenovo, November 2003. ) The FNS still serves southeastern Kentucky, now with a hospital in Hyden, four rural health clinics, a home health agency, and the FNS Schoo l of Midwifery and family Nursing.Many people from all around the world come to study this particular model of rural and social service delivery. (Gina Castlenovo, November 2003. ) The American College of Nurse Midwives recognizes Breckinridge as â€Å"the first to bring nurse-midwifery to the Untied States† and the Frontier School of Nursing as â€Å"a leader in nurse-midwifery in the Untied States. In 1982 Mary Breckinridge was inducted into the American Nurses Association’s Hall as a tribute for her contributions to the nursing profession in women’s health, community and family nursing, as well as the rural health care delivery. Gina Castlenovo, November 2003. ) Mary Breckinridge had the right Idea about wanting to help the less fortune. In today’s society we have Medicaid. The Medicaid program was designed to provide health coverage for lower-income people, families, pregnant women, children, elderly and people with disabilities. (Medicaid) Reference s Gina Castlenovo, M. M. (November 2003. ). Mary Breckinridge http://www. truthaboutnursing. org/press/pioneers/breckinridge. html. Medicaid. (n. d. ). Medicaidhttp://www. healthcare. gov/using-insurance/low-cost-care/medicaid/. Washington, D. C. 20201.

At&T Wireless: Text Messaging Essay

1. Describe the cost behavior in the wireless industry. What are the implications of this cost behavior for cost-volume-profit (CVP) relationships? The term cost behavior is used to describe whether a cost changes as output changes. In this case the costs are tightly shielded. In order to describe the cost behavior of the industry, we have to study the process that results in cost incurrence. Based on the information in the AT&T case, the industry features a high proportion of fixed costs in relation to acquiring spectrum and building a network. Variable costs are relatively low and, in the case of text messages, are very low. The cost structure in the wireless industry is dominated by fixed costs, so the contribution margin ratio is high. The high fixed costs and large contribution margin ratio result in a relatively high percentage increase in profit. The greater the proportion of fixed costs in a firm’s cost structure, the greater the impact on profit will be from a given percentage change in sales. The wireless industry has high operating leverage, because of high fixed costs and low variable costs. Therefore, the industry has a high ability to generate an increase in net income when sales revenue increases. â€Å"Text messages did not use any extra spectrum – once the carrier had paid the cost of the underlying infrastructure and storage equipment. Any revenue received by the provider on incremental text message usage is almost pure profit†. So, we can assume that, the cost does not change as the output changes. As far as I understand, the costs are incurred when the message is actually sent, and all these cost are very low. 2. What are the key cost drivers? Can a cost driver be used to continually raise prices? Cost drivers 1. Number of text messages per minute 2. Number of cell towers per area covered 3. Number of databases needed for a certain volume of messages 4. Number of customers 5. Number of cell phone plans 6. Number of terabytes 7. Number of devices The choice of the cost driver in this industry is not obvious, and the cost behavior pattern can depend on the cost driver selected. In this case the key cost driver is the number of text messages sent/received per minute. Yes and No. It depends on the cost behavior of the business. If the business has high variable costs and low fixed costs then the organization is more effected by a change in the volume. If the business, like in AT&T’s case, has a high rate of fixed costs with minimal variable costs (we assume that they are at the low end of the range of volume per fixed costs because their profit margin is high) then a change in volume will have little to no effect on the actual costs incurred. 3. What does it cost AT&T to send a text message?[Consider costs of the channel, billing cost, storage cost] Based on this cost, what is AT&T’s profit margin as a percentage of its text messaging business? [Consider per-use pricing and package pricing]How strong a relationship should exist? In Plan Channel cost$.0008641 (.07/81) = using voice messages as a reference which cost $.07 cents a minute. There are 81 text messages/minute transmitted per channel. Billing cost .0017283 (.0008641 x 2) = assume billing costs are twice as much as wireless costs Data base cost.0002857 (10M/35B) = AT&T would carry approximately 1% of worldwide text traffic of 3.5 trillion (35B) in their database which costs ~ 10 million dollars Storage cost .0000003837 (13,430/35B) = Worldwide text storage is 1,343 terabytes. AT&T would carry 1% of this storage =  13.43 terabytes. Cost of storage is $1,000 x 13.43 = $13,430. To get the per text cost divide this by 35B. Total cost per text$.002878 Gross Profit Margin = Gross Profit divided by total revenue $.20 per messageplan of $5 for 200 messagesplan of $15 for 1500 messages (.20 – .00278)/.20(.025-.002878)/.025 (.01 – .002878)/.01 =99%= 88%= 71% 4. How strong a relationship should exist between the prices charged to a customer for a good or service and the cost of providing that good or service? We think that companies should calculate their break-even point for their goods and services and to charge the prices according to that figure. This way, they can make sure that their price covers their expenses. In case of AT&T’s text messaging they are charging much more than the cost. But because there are only four national carriers in United States and they control 90 percent of the market, and text messaging had become widely popular, they can afford high prices. They are considering the demand of the service and pricing the product according to demand and supply. 5. Why is the price that AT&T charges to transmit a kilobyte of data via text message so much higher than the price charged to transmit a kilobyte of data via a Smartphone? The fastest growing wireless industry is text messaging. This also reflects the earlier comment that demand for messaging far exceeds supply, therefore driving prices up. Even though it costs less to transmit a text message than data, it is still seen as a very low cost to the average customer. The customer feels like they are getting a good deal because they are using the text messaging more than the data messaging. Part of the wireless industries revenue is from the sale of the devices. Even the cheapest cell phones have the capability to send text messages. Since the company is not making as much of a profit off of the device (cell phone), they are making up the difference in charging more for the text messages. In addition, since voice  messaging is being replaced by data messaging, both text and email, almost all cell phone chains are developing their own version of an IPhone or smart phone. To compete with these other chains, AT&T must charge the lowest price possible for their data to entice customers to buy their products over the competitors. 6. As we move to a service economy, can we expect to have more or fewer businesses with cost behaviors similar to those in the text messaging sector? Explain. I would assume we would see more businesses with cost behaviors similar to the text messaging sector. The reason being is the service industry tends to have more fixed costs that don’t increase linearly with the increase in service. The fixed costs tend to be step-fixed costs, whereas they can maintain services within a certain range up to a point in which they have to increase the fixed costs. The variable costs tend to be minimal since they often don’t have the manufacturing costs of direct materials and direct labor. After a service industry covers the basic operations, less money is needed as sales rise. Once the fixed costs are paid, the expense of processing additional sales is so little that the profits will grow faster than the revenues. The precedence has already been set in multiple service organizations, particularly internet companies. They have shown that once you achieve the hurdle of covering your basic fixed costs, the increased volume of service is very profitable. 7. What should management of wireless firms seek to do now? AT&T should invest in improving the company‘s wireless broadband coverage and its performance. They can improve network coverage by adding cell towers, laying faster fiber-optic cabling, adding capacity to cell sites and upgrading current cell sites to improve internet speeds. By enhancing the network, they can carry a larger volume of data traffic. This will allow them to accommodate more customers and therefore increase their profits.

History of US Immigrants - 1546 Words

Introduction Immigration has always considered as contentious in the United States. More than two hundred years ago Benjamin Franklin concerned that German settler would overwhelm many predominantly British culture of the United States. In mid-nineteenth century an Irish immigrants were scorned as lazy. In the early twentieth century believed that a gesture of new immigrants-Poles, Italians, Russian Jews were too different to ever be assimilated into American life. Today, the fears are used against immigrants from Latin America, but critics are wrong, just as were their counterparts in earlier times. In this report we need to study the relation of diverse people contribution in United States and the countries culture. (McLaughlin, 2006)†¦show more content†¦(Hagan, 1999) The factional fighting that erupted in Mexico after the 1910 revolution and the sheaves of bandit’s proliferation made the camp was a site involved in economic insecurity, political and social. At that time the U.S. industry and the field needed to supply its workers who had gone to the First World War: in this way, Mexican immigrants settled the problem of occupation and safety and earned U.S. capital workforce. The United States government in 1917 legalized the flow setting up special programs to support temporary Mexican labor, program ended in 1921. Cuban Americans Cubans who are migrated to the United States are history backed from the time when the nineteenth Century the Cuban manipulates in the history of the U.S. is often overlooked. The United States official has a record of intervention in affairs of Cuba and in the Spanish-American War. The Cuban population in the U.S., although small compared with that of other Hispanic immigrants, has a high level of education, higher incomes and a high rate of homeownership, announced today the Pew Hispanic Center. The report, 16 pages, based on 2004 data from the Census Bureau and highlights the major features of the Cubans in the United States, both those born in this country and those who emigrated from the island. In total, there are about 1.5 million Cubans in this country, or 4 percent of the Hispanic population, of which 912,686 were born abroad and 636,998 in the UnitedShow MoreRelatedThe Evolution And Relevance Of Immigration1510 Words   |  7 Pagesexception of one group, was an immigrant himself or a descendent of immigrants,†(Kennedy, 2). The United States is a nation of immigrants. There are, however, many Americans who are not accepting of the fact that immigrants were the reason why this country is able to have a population, economy and democracy. To ignore immigrants, is to ignore our history and how America is the way it is now. 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