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anonymous
 4 years ago
Thermodynamics
anonymous
 4 years ago
Thermodynamics

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm on it. Give me a few minutes!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If I am to approach this problem in the scope of your class, I need to know if you are familiar with (1) enthalpy of formation, (2) Gibb's Free Energy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you solve this with gibbs free energy? I would like to see how it's done if it's not to much trouble.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sure. I'll assume you are familiar with the derivation and implications of the Gibbs function. We know that Gibbs free energy represent a maximization of the work, since all systems tend to minimize free energy. This being said, let's assume the air bag, after combustion, returns to ambient conditions. Therefore, we are removing all the heat and work from the system and using these in the most useful way, i.e. entropy generated will be zero. However, we still need to consider the entropy at the ambient conditions, since entropy is typically defined from a reference point of 0 K. We know that the maximum work is equal to \[W = G_R^o  G_P^o\] From a handle table found here:courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf We can see that \[g_f^o = 93.8 \left [ {\rm kJ \over kmol} \right]{\rm ~ for ~NaN_3}\]\[g_f^o = 0 \left [ {\rm kJ \over kmol} \right]{\rm ~ for ~Na~ and~ N_2}\] From dimensional analysis, we can observe that, \[26.2 [{\rm g]~ NaN_3} = 0.4 [{\rm kmol]~ NaN_3}\]Therefore, the maximum work is \[W = 0.4*93.8\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This way seems like it could be much faster than figuring out the enthalpy of formation and then determining the solution from there. It would be great to read up on but the web is being censored :( great answer though.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using Gibbs is super easy if we are considering a particular combustion case where a few assumptions are made: 1) Products are returned to ambient conditions. 2) We want reversible work. 3) We have adiabatic combustion. (All the heat is available to do work, which is part of assumption 2) Enthalpy of formation is great for combustion cases where a perfect work isn't being considered. In either case, I find enthalpy of formation to be more useful. Especially in combination with exergy for combustion analysis.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0easmore, I don't think that is the process I'm supposed to follow.
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