Understanding of Molar Mass of Air Research Paper - 825 Words

Understanding of Molar Mass of Air Research Paper (Coursework Sample) Content: NameInstructorDateThe molar mass of airIntroductionThe air that keeps us alive, the gas that is used to preserve soft drinks; balloons we use on fancy occasions are explicit examples of the gases. Gases have their specific physical behaviors which are well described by the scientific laws. The laws have been in experimentation for many years. On the experiment this week would be carrying out an elaborate experiment on the quantitative trait of the gasses. The experiment we have carried out helps us have a good understanding of the physical characteristics of the gases. The experiment is an inquiry of one of the comprehensive gas laws. The gas law could be stated as an equation: PV=Nrt ( Zumdahl and Zumadahl, 2009) . The symbols could be defined as:P= gas pressureV= volume of the gasn= the number of gas molesR=the constant of the gas which is equal to 0.082057L atm /K molT= gas temperature measured in KelvinThe equation above gives us a very pertinent and general des cription of the ideal gas. However the ideal gas may not have absolute physical traits. The equation therefore provides a basis for the experimentation that we performed. When the 4 of the 5 variables for the equations are known the equation may acquire a different arrangement which may make it easy to solve for the unknown value. When describing a gas based on the physical characteristics, it is prudent to take note that gases appear in fluid form. Conventionally, people think that only liquids are fluids, but it should be noted that even the gases are fluids too. For a substance to be referred to as fluid, it must be able to flow and change shape with a lot of ease to fill the container.ResultsThe molecular weight of the air data sheet3.44g1.78g3.45g Trial1 trial2 trial 3 average Part 1: M air filled balloon=0.6781.671.72gM He filled balloon=1.72g1.380.69 1.74m=1.00gatm1.00gatm1.00gatm Part twoPressure p=1.58L1.605L1.605LVolume v=24OC24OC23.90CTemperature T=297.05oK 297.05o K 297.15oKCalculation of the moles n of the gas in the balloons applying the values for the P.V and T above for each of the trialsTrial onen= PVRT1.00g1.67L0.082atm297.5Kn= 0.0683Trial twon=PVRT 1.00gatm1.605L0.082atm297.15Kn=0.665Trial threen=PVRT1.00gatm1.585L0.082297.15Kn=0.0656AverageTrial two+ trial two+ trial three 30.0683+0.0665+0.06563Answer = 0.668The equation m=M air filled balloons=M air filled balloons with m= nftl (here the nftl indicates the molecular weight)The equation can be used find the molecular weight of the gas,nftl air=nftlHeM air = m+nHe MHen air1.383+(0.0668)0.0668= 24.268g/molBelow is the comparison of the molecular weight of gas to the theoretical weight of the dry air as given by the TAacceptable value-experimental value acceptable value100%=error28.96-24.26828.96100=16.20%DiscussionTo find the molecular weigh...