- #1
ForcesofChaos
- 4
- 0
(I Solved It Thanks)
The problem:
a cylinder contains 3g of liquid octane and an initial o2 pressure of 1atm. the cylinder is fitted with a piston and initially occupies a volume of 10L at 298 K. The octane is completely consumed by the reaction. Assume The piston is massless and frictionless and that the cylinder neither absorbs nor releases heat. Calculate the final volume occupied by the cylinder.
now what i have so far
C8H18 + 25O2 ----> 8 CO2 + 9H2O
this is the reaction after converting 3 g of octane to moles new reaction
.0263 <octane> + .409 O2 (.329 mols react to consume all octane) --> .2104 C02 + .2367 H2O <theres .0799 leftover mols of O2 from the reaction .409 - .329>
ok i assume that pressure is constant. because after the reaction the piston is going to move to equalize the pressure.
so i need to find the final temperture or maybe there's a way to find volume without doing temperture. i calculated enerergy released which is delta H = -5470kj if this helps any.
if you have any idea how to finish this problem it would be helpful you don't have to solve it i guess i just need to know what to use to figure it out I've played with q = mc delta T <q = delta H @ constant P> but i dnt knw if it will work and I am not sure exactly how to use it. I've worked a long time on this and i have had no luck
anything will help maybe. Thanks
i turn it in tomorrow so if i don't get any ideas by then he will explain it for us but i won't get the credit oh well its extra credit.
Thanks again.
The problem:
a cylinder contains 3g of liquid octane and an initial o2 pressure of 1atm. the cylinder is fitted with a piston and initially occupies a volume of 10L at 298 K. The octane is completely consumed by the reaction. Assume The piston is massless and frictionless and that the cylinder neither absorbs nor releases heat. Calculate the final volume occupied by the cylinder.
now what i have so far
C8H18 + 25O2 ----> 8 CO2 + 9H2O
this is the reaction after converting 3 g of octane to moles new reaction
.0263 <octane> + .409 O2 (.329 mols react to consume all octane) --> .2104 C02 + .2367 H2O <theres .0799 leftover mols of O2 from the reaction .409 - .329>
ok i assume that pressure is constant. because after the reaction the piston is going to move to equalize the pressure.
so i need to find the final temperture or maybe there's a way to find volume without doing temperture. i calculated enerergy released which is delta H = -5470kj if this helps any.
if you have any idea how to finish this problem it would be helpful you don't have to solve it i guess i just need to know what to use to figure it out I've played with q = mc delta T <q = delta H @ constant P> but i dnt knw if it will work and I am not sure exactly how to use it. I've worked a long time on this and i have had no luck
anything will help maybe. Thanks
i turn it in tomorrow so if i don't get any ideas by then he will explain it for us but i won't get the credit oh well its extra credit.
Thanks again.
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