# Chemistry

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in a cylinder that was 9.0 mm in diameter and 215 mm in length.

I want to know the temperature (in Kelvin) and pressure (in atm) within this tube after combustion assuming that the cork did not “shoot” out of the end, i.e., the pressure was contained within the tube after ignition.

Balanced Reaction: 2 C6H14 + 19 O2  12 CO2 + 14 H2O

We’re going to simplify the problem slightly by assuming that there is no heat/energy loss to the walls of the cylinder (i.e., all the heat produced goes to heating the product gases), the combustion reaction goes to completion, no PV work was performed and the heat capacities (given below) are independent of temperature.

The system before ignition: 1 atm O2 gas and hexane vapor. (O2 is limiting reagent for the combustion reaction which produced H2O(g) and CO2(g) as products.

The heat of combustion is -3891 kJ/mole of hexane
The constant volume heat capacities are…
33.6 J/K-mol for H2O(g)
37.1 J/K-mol for CO2(g)

These are the steps to follow:
(1) compute the cylindrical chamber volume, V
(2) compute the number of moles n in volume initially using n = PV/RT
(3) 2/21 of the total moles initially present are C6H14. Compute that number.
(4) Compute the heat release, assuming the reaction goes to completion. Get that from the number of moiles of hexane and the heat of reaction per mole.
(5) Compute the final temperature such that the heat release equals the heat transferred to the reaction products H2O and CO2.

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