Understanding of Molar Mass of Air Research Paper (Coursework Sample)

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Instructor
Date
The molar mass of air
Introduction
The 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 pressure
V= volume of the gas
n= the number of gas moles
R=the ‘’ constant of the gas’’ which is equal to 0.082057L atm /K mol
T= gas temperature measured in Kelvin
The equation above gives us a very pertinent and general description 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.
Results
The molecular weight of the air data sheet
3.44g1.78g3.45g Trial1 trial2 trial 3 average Part 1: M air filled balloon=
0.6781.671.72g
M He filled balloon=
1.72g1.380.691.74
Ϫm=
1.00gatm1.00gatm1.00gatm Part two
Pressure p=
1.58L1.605L1.605LVolume v=
24OC24OC23.90C
Temperature T=
297.05oK 297.05o K 297.15oK
Calculation of the moles n of the gas in the balloons applying the values for the P.V and T above for each of the trials
Trial one
n= PVRT
1.00g×1.67L0.082atm×297.5°K
n= 0.0683
Trial two
n=PVRT 1.00gatm×1.605L0.082atm×297.15°K
n=0.665
Trial three
n=PVRT
1.00gatm×1.585L0.082×297.15°K
n=0.0656
Average
Trial two+ trial two+ trial three 3
0.0683+0.0665+0.06563
Answer = 0.668
The 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=nftlHe
M air = ∆m+nHe MHen air
1.383+(0.0668×)0.0668= 24.268g/mol
Below is the comparison of the molecular weight of gas to the theoretical weight of the dry air as given by the TA
acceptable value-experimental value acceptable value×100%=error
28.96-24.26828.96×100=16.20%
Discussion
To find the molecular weigh...