Engineeringa Better Air Bag
DataTable 1: Model Air Bag
Activity 
Data and Calculations 
Volume of 6 × 9 inch bag 
1.2 liters 
Room temperature in Kelvin 
The temperature in Celsius was 22°C. In Kelvin, this was (22+273)= 295K 
Room pressure in atm 
The barometric pressure was 30.12 inches. In atmospheres (atm) this was = 1.01 atm 
^{Moles of CO}2 ^{required to inflate }bag at room temperature and pressure 
The Ideal Gas Law is given by PV=nRT where P = pressure (1.01 atm), V = volume (1.2 L), R = Gas constant (0.0821 L atm mol^{1} K^{1}), T = temperature (295K), n = number of moles of CO­_{2}. Rearranging the equation gives n = PV/RT n = 0.05 moles of CO_{2 } 
Balanced equation for the ^{reaction of NaHCO}3 ^{and CH}3^{COOH to produce}^{ }^{CO}2 
The reaction follows the equation: NaHCO_{3 (s) }+ CH_{3}COOH _{(aq) }> CO_{2 (g) }+ H_{2}O _{(l) }+ CH_{3}COONa _{(aq) } 
^{Mass of NaHCO}3 ^{needed for the }reaction (84.0 g/mol) 
The molar ratio in the equation above is 1:1 for each of the components. Therefore there will be the same number of moles for each component. The number of moles of NaHCO_{3 }is thus 0.05 moles. The mass will therefore be: Mass = molar mass x number of moles Mass = (84.0 g/ Mass = 4.2 g 
Volume of vinegar required (0.833 M acetic acid) 
Since the molar ratio is 1:1, the number of moles of acetic acid that will react is 0.05 moles. The volume will be given by: Volume = number of moles / Molarity Volume = 0.05 Volume = 0.06 L 
DataTable 2: Model Air Bag
Trial # 
^{NaHCO}3 (grams) 
Vinegar (mL) 
Observations 
1 
4.2 g 
60 mL 
The bag inflates the slowest. 
2 
4.5 g 
65 mL 
The bag fills up with CO_{2 }at a quicker rate compared to the initial experiment that used 4.2 g of NaHCO_{3 }and 60 mL of vinegar. 
3 
4.7 g 
70 mL 
On the three experiments carried out, this bag inflates the quickest. 
DataTable 3: 80L DriverSide AirBag
Activity 
Calculations 
^{Moles of CO}2 ^{required to }inflate 80L driverside air bag at room temperature and pressure 
The Ideal Gas Law is given by PV=nRT where P = pressure (1.01 atm), V = volume (80 L), R = Gas constant (0.0821 L atm mol^{1} K^{1}), T = temperature (295K), n = number of moles of CO­_{2}. Rearranging the equation gives n = PV/RT n = 3.303 moles of CO_{2} 
Balanced equation for ^{the reaction of NaHCO}3 ^{and CH}3^{COOH to}^{ }^{CO}2 
The balanced equation is: NaHCO_{3 (s) }+ CH_{3}COOH _{(aq) }> CO_{2 (g) }+ H_{2}O _{(l) }+ CH_{3}COONa _{(aq) } 
^{Mass of NaHCO}3 ^{needed }for the reaction (84.0 g/mol) 
The molar ratio in the equation above is 1:1 for each of the components. Therefore there will be the same number of moles for each component. The number of moles of NaHCO_{3 }is thus 3.303 moles. The mass will therefore be: Mass = molar mass x number of moles Mass = (84.0 g/ Mass = 277.45 g 
Volume of vinegar required (0.833 M acetic acid) 
Since the molar ratio is 1:1, then the number of moles of acetic acid that will react is 3.303 moles. The volume will be given by: Volume = number of moles / Molarity Volume = 3.303 Volume = 3.965 L 
Activity 
Calculations 
^{Moles of CO}2 ^{required to }inflate 160L front passengerside air bag at room temperature and pressure 
The Ideal Gas Law is given by PV=nRT where P = pressure (1.01 atm), V = volume (160 L), R = Gas constant (0.0821 L atm mol^{1} K^{1}), T = temperature (295K), n = number of moles of CO­_{2}. Rearranging the equation gives n = PV/RT n = 6.672 moles of CO_{2} 
Balanced equation for the ^{reaction of NaHCO}3 ^{and CH}3^{COOH to}^{ }^{CO}2 
The equation is: NaHCO_{3 (s) }+ CH_{3}COOH _{(aq) }> CO_{2 (g) }+ H_{2}O _{(l) }+ CH_{3}COONa _{(aq) } 
^{Mass of NaHCO}3 ^{needed }for the reaction (84.0 g/mol) 
The molar ratio in the equation above is 1:1 for each of the components. Therefore there will be the same number of moles for each component. The number of moles of NaHCO_{3 }is thus 6.672 moles. The mass will therefore be: Mass = molar mass x number of moles Mass = (84.0 g/ Mass = 560.45 g 
Volume of vinegar required (0.833 M acetic acid) 
Because the molar ratio of the reaction is 1:1, the number of moles of acetic acid that will react is 6.672 moles. Therefore, the volume of vinegar will be given by: Volume = number of moles / Molarity Volume = 6.672 Volume = 8.00 L 
Engineeringa Better Air Bag Questions

Let’s consider an actual automotive airbag:

Using the air pressure you recorded in step 3 on page 8, calculate the moles of nitrogen required to fill a 160 liter passengerside air bag. Show all calculations.
The Ideal Gas Law is given byPV=nRT whereP = pressure (1.01 atm), V = volume (160 L), R = Gas constant (0.0821L atm mol^{1}K^{1}),n = number of moles of N­_{2}.Assuming the nitrogen gas is released at a temperature of 22°C(295K), the Ideal Gas Law can be rewritten as n= PV/RT
n= 6.672 moles of N_{2}

Using the reaction from page 5 of the procedure, calculate the grams of sodium azide required to produce the required amount of nitrogen. Show all calculations.
The reaction in an airbag thatinvolving sodium azide that leads to the releasing of nitrogen gasfollows the equation:
2NaN_{3(s) }>2Na _{(s) }+3N_{2 (g) }
The molar ratio of NaN_{3}to N_{2 }is2:3. Therefore, for every 3 moles of N_{2}there are 2 moles ofNaN_{3}.However, since the reaction led to the production of 6.672 moles ofN_{2},then the number of moles of NaN_{3}will be given by:
Moles of NaN_{3}= (2×6.672)/3
Moles of NaN_{3}= 4.448 moles.
Since NaN_{3}has a molar mass of65 g/mol, then the mass of NaN_{3}required to fill the160 L airbag will be given by:
Mass = Molar mass x Number ofmoles
Mass = (65 g/mol)/ 4.448 mol
Mass = 14.61 g of NaN_{3}

Do you think air bag manufacturers design the bags to inflate to a) slightly under atmospheric pressure, b) approximately equal to atmospheric pressure or c) over atmospheric pressure? Explain the reasoning for your choice. If you use any sources of information to answer this question, cite them using APA format.
Airbag manufacturers designairbags to inflate to a maximum pressure that is under theatmospheric pressure so that it can easily deflate when impacted by aperson inside a car, and by doing so absorb as much force as possiblefrom the impact and reduce chances of serious injury to the person.If it inflated to an atmospheric pressure higher than the atmosphericpressure then it would not deflate after being impacted by a personinside the car, and it would therefore not absorb any impact forcesfrom the person’s body, thereby translating to serious injuriessustained by the person (Barbat, Li, & Prasad, 2013).

Describe at least 2 factors that you think air bag manufacturers take into account, if any, when designing systems for different parts of the world. Explain why these factors might be important. If you use any sources of information to answer this question, cite them using APA format.
Airbag manufacturers consider thespeed with which the airbag will deploy and fill up with N_{2}gas. When a caraccident occurs, a driver can hit the steering wheel in around 23milliseconds (Deng, et al., 2013). Therefore, airbag manufacturerschoose those chemical compounds that when set off by the car’scrash sensor can combust or react to rapidly generate gases withwhich to fill the airbag in a time that does not exceed thisduration. In the case of a person sitting on the passenger side ofthe car the duration before he hits the dashboard is longer since thedistance to the dashboard is higher than that to the steering wheel.In such a case airbag designers often place bigger airbags on thepassenger side than on the driver’s side to achieve the same result(Deng, et al., 2013).
Another key aspect considered byairbag manufacturers in their design is the manner with which theenergy absorbed from a person’s impact with an airbag can bedissipated. To quickly absorb and dissipate as much energy aspossible following an impact, airbag manufacturers design theirairbags to have small vents through which the gases can escape afterimpact. As the gases escape the airbag deflates, thereby creating asoft cushion that slows down the person’s momentum (Deng, et al.,2013).
References
Barbat, S., Li, X., & Prasad,P. (2013). Bumper and Grille Airbags Concept for Enhanced VehicleCompatibility in Side Impact: Phase II. TrafficInjury Prevention,14(1), 3039.
Deng, X., Potula, S., Grewal, H.,Solanki, K., Tschopp, M., & Horstemeyer, M. (2013). Finiteelement analysis of occupant head injuries: Parametric effects of theside curtain airbag deployment interaction with a dummy head in aside impact crash. AccidentAnalysis and Prevention,232241.
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