If all the available chemical energy in a full tank of gasoline(assume pure isoo

Question

If all the available chemical energy in a full tank of gasoline(assume pure isooctane) in a Mini Cooper is somehow converted tothe potential energy of the vehicle in the earth's gravitationalfield, how high above the earth will it go? Assume that thegasoline is converted quantitively to carbon dioxide and watervapor. You will have to look up various data to solve thisproblem.

Mini Cooper Weight = 2546lbs = 1154.85 Kg
Mini Cooper Fuel Tank Capacity: 13.2 gallons = 49.97 L
Earth's Gravity = 9.8 m/s2

So, C8H18 + O2 -> CO2 + H2O

Enthalpy of Formations
C8H18: -250kJ/mol
O2: 0 kJ/mol
CO2: -393.5 kJ/mol
H2O: -241.82 kJ/mol

Please help! Thanks!

Explanation / Answer

Isooctane — Density:0.688 g/ml, Liquid molar mass 114.22 49.97 L *(0.688 g /mL)*(1000 mL/1L)*(1 mol/114.22) = 300.992471moles isooctane C8H18 + (25/2) O2 -> 8CO2 + 9H2O delH = 9*-241.82 + 8*-393.5 - (25/2)*0 -1*-259          NoteHformation_isooctane = -259 kJ/mol delH = -5065.38 kJ (per mol C8H18 combusted) total energy available 300.992471 moles isooctane*(5065.38 kJ/mol isooctane) = 1524641.24kJ Work against gravitational field = mass*g*height = availableenergy height = (available energy / [mass*g]) height = [1524641.24 kJ *(1000 J/1kJ) ]/ [1154.85 kg *9.8m/s2] height = 1.34 x 103 meters

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