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Topic: Increased pressure resulting from pentane reaction  (Read 2979 times)

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Offline zcami2

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Increased pressure resulting from pentane reaction
« on: March 29, 2012, 04:02:21 AM »
Hi again,
In the reaction C5H12(L) + 8O2(G) → 5CO2(G) + 6H2O(G), it is known that were this reaction to occur within a closed system the pressure within the system would increase (e.g. within an engine) however I do not know why.  So my question is why does this occur?  Some reasons that I think that this occurs is due to a decrease in density of the produced molecules relative to the reacted substances.  Is this correct or is the increased pressure a result of the increased heat within the system?  Or both? Or due to some entirely different factors?
Thanks,
Z.C

Offline Borek

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Re: Increased pressure resulting from pentane reaction
« Reply #1 on: March 29, 2012, 04:26:18 AM »
Compare amounts of gases before and after the reaction. PV=nRT, V=const (closed system).
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Offline AWK

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Re: Increased pressure resulting from pentane reaction
« Reply #2 on: March 29, 2012, 04:31:51 AM »
+ heat
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Offline zcami2

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Re: Increased pressure resulting from pentane reaction
« Reply #3 on: March 29, 2012, 06:07:16 AM »
ok,
using the formula PV=nRT, I know that R and V are constant so the change in pressure within the closed system is influenced only by the change in temperature and the number of moles of gas.  I'm not entirely sure how to factor in the influence of the temperature however I know that this aspect would definitely increase as the reaction is exothermic.  As for the total number of moles within the reaction this obviously doesn't change however the number of moles within the gas phase does...  I'm not sure whether to only factor in the number of  moles in the gas phase or all the moles within the reaction.  However if only taking into account the number of moles within a gas state the number of moles of reactant are 256 while the number of moles of product is 328.  So from this should I say that the pressure is influenced by both the increase in the number of moles now in the gas state and the increase in temperature of the system or only the temperature increase of the system?

Thanks for the reply
Z.C

Offline AWK

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Re: Increased pressure resulting from pentane reaction
« Reply #4 on: March 29, 2012, 07:04:39 AM »
However if only taking into account the number of moles within a gas state the number of moles of reactant are 256 while the number of moles of product is 328

???
I can see 9 and 11
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Offline zcami2

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Re: Increased pressure resulting from pentane reaction
« Reply #5 on: March 29, 2012, 07:42:28 AM »
Woops, Sorry. Had a bit of a malfunction there... (haven't done these types of calculations for a while, even though this one is particularly basic)

So yea including only the molecules in the gas phase the number of moles of reactants is 8 and the number of moles of products are 11.  Judging from what you answered however I'm assuming that I do have to take into consideration the ethanol even though it's a liquid.  So 9 : 11.
« Last Edit: March 29, 2012, 07:59:47 AM by zcami2 »

Offline Olympiad_Tutor

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Re: Increased pressure resulting from pentane reaction
« Reply #6 on: March 29, 2012, 07:56:36 AM »
8:11
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Offline fledarmus

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Re: Increased pressure resulting from pentane reaction
« Reply #7 on: March 29, 2012, 08:00:02 AM »
You want the number of moles of molecules in the gas state - gas occupies far more volume than liquid, so the liquid molecules can be ignored, they contribute very little to the pressure.

At the pressures involved in a car engine, the ideal gas law isn't a particularly good approximation, but it will give you a sense of what is happening. Air and atomized gasoline enters the piston at atmospheric pressure, and at bottom dead center, your entire cylinder is full of air  (approximately 80% N2, 20% O2) at one atmosphere and essentially ambient temperature (although the engine gets hot, the air and fuel being drawn in are not in contact with it for very long and transfer of heat to gases is pretty slow - air is a good insulator). Then your piston moves to top dead center and this volume of gas is compressed by whatever your compression ratio is - you can look this up for whatever engine you are interested in. Let's say 10:1 for a reasonable start (and ease of math). That means the final volume of the gas in your cylinder at the top of the stroke is 1/10 of the volume at the bottom of your stroke, and using PV=nRT (n and T are constants at this point), you can see that for the volume to decrease to 1/10, the pressure must increase 10-fold - we are now at 10 atmospheres of pressure. (Of course the temperature does increase when you increase pressure this fast, and for diesel fuels, they increase enough that the fuel actually ignites just as the piston is reaching the top. That way you don't need a spark plug in diesel engines. In gasoline engines, if this happens, you get knocking - ignition before the cylinder reaches top dead center tries to force the piston backwards against the cylinder shaft and can damage the engine. Octane number is a measure of how much you can compress the gas before this happens - the higher the compression ratio, you higher the octane number you have to use to keep the engine from knocking).

So now we have our compressed fuel-air mixture at 10 atmospheres and fairly low temperature, and we light it off with a spark plug. Most of the oxygen (which was 20% of the atmosphere in the cylinder) is used up, and carbon dioxide and water is formed. For every 8 moles of oxygen, we get 5 moles of carbon dioxide and 6 moles of gaseous water - our 8 moles of gas just increased to 11 moles of gas. But for every 2 moles of oxygen that went into the cylinder, we also put in 8 moles of nitrogen, so for the 8 moles of oxygen we are burning, there are 32 moles of nitrogen that aren't - we go from 40 moles of gas going in to 43 moles coming out, an increase of just under 10%. If the number of moles increases by 10% while the volume is held constant, then the pressure must also increase by 10% - our volume of gas is now at 11 atmospheres.

The same analysis can be run on the heat - using the heats of formation of your pentane, carbon dioxide, and water (oxygen is at standard state, so its heat of formation is zero), you can calculate how much heat is being reduced when pentane burns, and use that and the heat capacities of the final gas mixture (nitrogen, carbon dioxide, and water) to determine the increase in temperature inside the cylinder. The volume is still being held constant at this point (top dead center), so whatever the increase in temperature (as a percentage, making sure to use degrees Kelvin), that will be an additional percent increase in pressure.

With those calculations, you can tell which is the larger contributor to your pressure increase.

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