Answer:
a. 2562 L
b. 4082 L
c. 413,6 kJ
d. 0,115 kWh
e. 1,3 cents
f. 0,34 %
Explanation:
a. To calculate the volume we should obtain the moles of Octane to know the reactant moles and produced moles. Then, with ideal gas law obtain the change in volume:
3,784 L ≡ 3784 mL octane × (0,703 g / 1 mL) = <em>2660 g Octane</em>
<em> density</em>
2660 g octane × ( 1 mol / 114,23 g octane) = <em>23,29 mol octane</em>
<em>molar mass of C₈H₁₈</em>
With the combustion reaction of octane we can know how many moles are produced from 23,29 mol of octane, thus, in (1) :
2 C₈H₁₈ (l) + 25 O₂(g) ---> 18 H₂O(g) + 16 CO₂ (g) <em>(1)</em>
23,29 mol octane × ( 25 mol O₂ / 2 mol octane) = <em>291,1 mol O₂ -reactant moles-</em>
23,29 mol octane × ( 18 + 16 produced mol / 2 mol octane) = 395,9 moles produced
Ideal gas formula says:
V = nRT/ P
Where:
n = Δmoles number (produced-reactant) → 104,8 moles
R = Ideal gas constant → 0,082 atm·L/mol·K
T = Temperature → 25°C, 298,15 K
P = Pressure → 1 atm
Thus, replacing in the equation:
ΔV = 2562 L
b. To calculate the gas volume we should use the same values of ideal gas formula just changing the temperature value for 475 K -Because the produced moles of gas and presure are the same and R is constant.
Thus, the volume of produced gases is:
ΔV = 4082 L
c. The work, w, is equal to -pressure times Δ Volume:
w = - P×ΔV
The pressure is 1 atm and ΔV in the system is 4082 L
So, w = 4082 atm·L (101,325 J / 1 atm·L) = 413,6 kJ
d. As kJ is equal to kWs, 413,6 kJ ≡ 413,6 kWs × ( 1 hour / 3600 s) =
0,115 kWh
e. In Seattle 1kWh cost 11,35 cents. So, 0,115 kWh cost:
0,115 kWh × (11,35 cents/ 1kWh) = 1,3 cents
f. The energy calculated in part C, <em>413,6 kJ</em> is due to the work done by the system in gas expansion but total of heat produced in (1) are 1,2 ×10⁵ kJ. Thus, the proportion of work in gas expansion in total energy in combustion of octane is:
413,6 kJ / 1,2×10 ⁵ kJ × 100 = 0,34 %
I hope it helps!
