Ethanol (ch3ch2oh) has been suggested as an alternative fuel source. ethanol’s enthalpy of combustion is hcomb = –1368 kj/mol. what volume of ethanol (d = 0.789 g/ml) is required to produce the same amount of energy as 1 gallon of gasoline, which can be modeled as isooctane (c8h18) with an energy density of 32.9 kj/ml? (1 gal = 3.785 l; d = 0.703 g/ml)

a. 1.41 gal
b. 0.85 gal
c. 1.00 gal
d. 1.15 gal
d. 0.90 ga

a) V = 1.41 gal

Explanation:

Ethanol:

∴ Hcomb = - 1368 KJ/mol( - ) exothermic reaction

∴ δ = 0.789 g/mL

∴ Mw Ethanol = 46.07 g/mol

⇒ Eδ = (1368 KJ/mol)(mol/46.07 g)(0.789 g/mL) = 23.4285 KJ/mL

isooctane:

∴ Eδ = 32.9 KJ/mL

∴ δ = 0.703 g/mL

⇒ E = (32.9 KJ/mL)(3785 mL) = 124526.5 KJ

Volume of Ethanol:

⇒ (Eδ)(V) = 124526.5 KJ

⇒ V = (124526.5 KJ)/(23.4285 KJ/mL) = 5315.167 mL

⇒ V = (5315.167 mL)( gal/3785 mL) = 1.404 gal ≅ 1.41 gal

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml