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1 Some intercontinental jet airliners use kerosene as fuel - CIE - A-Level Chemistry - Question 1 - 2011 - Paper 1
Question 1
1 Some intercontinental jet airliners use kerosene as fuel. The formula of kerosene may be taken as C₁₄H₃₀. (a) To which homologous series of compounds does kerosen... show full transcript
Worked Solution & Example Answer:1 Some intercontinental jet airliners use kerosene as fuel - CIE - A-Level Chemistry - Question 1 - 2011 - Paper 1
Step 1
Step 2
Step 3
Calculate the mass, in tonnes, of C₁₄H₃₀ burnt on a flight from Beijing to Paris.
Answer
To calculate the mass of C₁₄H₃₀ burnt:
egin{align*}
ext{Total distance} &= 8195 ext{ km} \
ext{Kerosene burnt per km} &= 10.8 ext{ kg} \
ext{Total kerosene burnt} &= 8195 imes 10.8 ext{ kg} \
&= 88.506 ext{ kg} \
&= rac{88.506}{1000} ext{ tonnes} \
&= 0.0885 ext{ tonnes} \
ext{(rounded to one decimal place: 88.5)}
ext{Therefore, approximately } 88.5 ext{ kg is burned, or } 0.0885 ext{ tonnes.}
ext{[1 tonne = 1000 kg]}
ext{So the final answer is} ext{88.5 kg}.
\end{align*}
Step 4
Use your equation in (b) to calculate the mass, in tonnes, of CO₂ produced during this flight.
Answer
To find the mass of CO₂ produced, we first calculate based on the kerosene burnt:
egin{align*} ext{Molar mass of C}{14} ext{H}{30} &= (14 imes 12 + 30 imes 1) = 198 ext{ g/mol} \ ext{From (b), the reaction produces:} \ 2 ext{ C}{14} ext{H}{30} ightarrow 28 ext{ CO}_2 \ ext{Thus, } rac{88.5 g}{198 g/mol} imes rac{28}{2} = 275.3 g \ &= rac{275.3 g}{1000} ext{ tonnes} \ &= 0.2753 ext{ tonnes of CO}_2 ext{ produced during the flight.} ext{(rounded to one decimal place: 275.3)} \end{align*}
Step 5
Use the general gas equation PV = nRT to calculate the amount, in moles, of air in the tyre at sea level.
Answer
Using the equation:
,
we can identify:
- P = 6 × 10⁵ Pa
- V = 710 ext{ cm}³ = 710 imes 10^{-6} ext{ m}³
- R = 8.31 ext{ J/(mol K)}
- T = 20 + 273 = 293 ext{ K}
Now substituting in the values:
n = rac{PV}{RT} = rac{(6 imes 10^5 ext{ Pa})(710 imes 10^{-6} ext{ m}^3)}{(8.31 ext{ J/(mol K)})(293 ext{ K})}
Calculating gives:
ightarrow 0.175 ext{ moles}$$Step 6
Calculate the pressure inside the tyre at a height of 10000m.
Answer
At 10,000 m, we can use the equation:
P = rac{nRT}{V} where:
- n = 0.175 ext{ moles} (from part (d))
- R = 8.31 ext{ J/(mol K)}
- T = 5 + 273 = 278 ext{ K}
- V = 710 imes 10^{-6} ext{ m}³
Substituting in the values: P = rac{(0.175)(8.31)(278)}{710 imes 10^{-6}} = 56940.165 ext{ Pa} This rounds to approximately 56940 Pa.