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Topic: thermodynamics & ideal gas  (Read 7437 times)

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Offline Donaldson Tan

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thermodynamics & ideal gas
« on: January 01, 2005, 09:35:16 PM »
I am not sure how to solve this:

a stoichiometric mixture of methane and air explodes in a closed module. the specific internal energy liberated by the explosion is 2730kJ/kg. The heat capacity at constant volume of all species before and after the explosion is 1.08 kJ/kg.K and is independent of the temperature. you may assume the mixture behaves ideally.

a. estimate the rise in temperature
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given it's a close system, du = q + w
given it's a closed module, i assume it's a constant volume reactor, hence w = 0
du = q => q = -2730kJ/kg
given q = mc(dT), q/m = c(dT)
2730 = (1.08)dT
dT = 2730/1.08 = 2527.778K
temperature rise = dT =  2530K

b. given initial condition is 1bar, 288K, find the immediate pressure  after the reaction
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T1 = 288K, P1 = 1bar
T2 = 288 + dT = 2815.78 = 2820K, P2 = ???

PV = nRT = mRT/M where m: mass of gas and M: molar mass of gas
given m remains constant throughout the process, but not M. In another words, I can't use P1/T1 = P2/T2 to solve.

Please advise..
« Last Edit: January 02, 2005, 01:48:07 AM by geodome »
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Offline Donaldson Tan

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Re:thermodynamics
« Reply #1 on: January 01, 2005, 09:52:26 PM »
I had given additional thoughts regarding the problem.

given the pre-reaction mixture is in stoichiometric ratio,
the reaction: CH4 + 2O2 => CO2+ 2H2O and
molar composition of air is 21% O2 and 79% N2,
then the pre-reaction mixture would contain 10% CH4, 19% O2, 71% N2.

post-reaction mixture contains zero methane and O2.
taking a basis of 100kmol for the pre-reaction mixture, the composition of the post-reaction mixture would be
nitrogen = 71kmol
carbon dioxide = 20kmol
water = 40kmol
total number of moles = 131kmol
percentage composition of post-reaction mixture:
54% N2, 15% carbon dioxide, 31% water
« Last Edit: January 01, 2005, 10:09:38 PM by geodome »
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Offline Donaldson Tan

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Re:thermodynamics
« Reply #2 on: January 01, 2005, 10:02:10 PM »
for (b),
assume constant volume and ideal gas behavior,

PV = nRT = mRT/M
V = mRT/MP
mRT1/M1P1 = mRT2/M2P2
T1/M1P1 = T2/M2P2

given P1,T1,T2 are all known, and M1, M2 can be derived since the molar composition of the mixture before and after the explosion is known, the unknown pressure P2 can be work out

Please advice..

« Last Edit: January 01, 2005, 10:14:28 PM by geodome »
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Offline Mitch

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Re:thermodynamics
« Reply #3 on: January 01, 2005, 10:45:40 PM »
I really don't know, but I don't think this is a plug and chug problem. It feels like integration will be involved for part b.
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Re:thermodynamics
« Reply #4 on: January 01, 2005, 11:10:25 PM »
Your calculations for pre and post reaction kmoles are incorrect.
Actually, the kmoles are the same pre and post, because the stoichiometry of combustion is 3 moles of gas produce 3 moles of new gas, the rest like nitrogen remaining constant.
Therefore, it would seem that P1/T1 = P2/T2  would work at first glance.

Offline Donaldson Tan

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Re:thermodynamics
« Reply #5 on: January 01, 2005, 11:24:52 PM »
OK. I made a mistake in my percentage composition.

pre-reaction mixture:
10% CH4, 20% O2, 70% N2
molar mass = (.1X16) + (.2X32) + (.7X28) = 27.6kg/kmol

post-reaction mixture (assume all reactants use up):
10% CO2, 20% H2O, 70% N2
molar mass = (.1X44) + (.2X18) + (.7X28) = 27.6kg/kmol

since M1 = M2, therefore P1/T1 = P2/T2. Is this correct?

btw is it always true that the molar mass of a mixture before and after a chemical reaction, provided the reactants are in stoichiometric ratio?
« Last Edit: January 01, 2005, 11:27:55 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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Re:thermodynamics
« Reply #6 on: January 01, 2005, 11:42:00 PM »
"M1 = M2, therefore P1/T1 = P2/T2. Is this correct?"
I think so..

2)I think not.
   The number of moles pre and post can be different in some reactions. Since the total mass pre and post has to be the same, then the weighted average "molar mass"  must be different.

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