Assume the gasoline is burned with 99% efficiency in a car engine, with 1% remaining unburned in the exhaust gases as VOCs (volatile organic compounds). If the engine exhausts 16 kg of gases (MW=30) for each kg of gasoline (MW=100), calculate the fraction of VOCs in the exhaust. Give your answer in parts per million

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cjmejiab cjmejiab · Answer 1 · 2019-11-11T05:31:00+01:00

Answer:

187ppm.

Explanation:

To develop the problem we first need to perform mass conversions to moles. In this way for gases your conversion is given by,

[tex]Mol_{gas} = \frac{m_{gas}}{M_{W}}[/tex]

[tex]Mol_{gas} = \frac{16000g}{30}[/tex]

[tex]Mol_{gas} = 533.3mol[/tex]

For the VOC's same:

[tex]m_{voc} = 0.01*1000=10g[/tex]

then,

[tex]Mol_{voc} = \frac{m_{voc}}{M_{W}}[/tex]

[tex]Mol_{voc} = \frac{10}{100}[/tex]

[tex]Mol_{voc} = 0.1mol[/tex]

We only need to obtain the fraction of Voc's in exhaust. This will be in particles per million, so

[tex]\xi = \fra{Mol_{voc}}{(Mol_{gas}+Mol_{voc})}[/tex]

[tex]\xi = \frac{0.1}{0.1+533.3}*10^6[/tex]

[tex]\xi = 187ppm.[/tex]