[G O D I V A]

Here is the question:

Every morning I drive my son to school in the car. Each one-way trip takes about 25 minutes. My car’s fuel efficiency rating, according to the manufacturer, is about 30 miles per US gallon of gasoline. Estimate the amount (in kg) of carbon dioxide emission for each one-way trip.

It says that there is no model answer, but that a rough estimate in the same order of magnitude should be able to be found. You will have to find some information on your own, and make some assumptions.

Problem is I don't even know when to begin.  I did some calculations and came to the conclusion that 1 gallon of gasoline roughly emits 9.13 kg of carbon dioxide.  So that's basically 9.13kg of CO2 emitted per 30 miles, but I dont know how far the one way trip is or even if what i did was what is supposed to be done?

[billnotgatez]

and make some assumption

Can you assume they drove 35 mph for the whole trip to school and back?

I found several places using Google that talked about how MPG is determined.

[G O D I V A]

I have no idea what to do next.  Where did you find 35 mph?

[Borek]

Basically - he picked it from the nose

You are asked to make any reasonable assumptions you want. This is one of them. You are suggested time and mpg, obviously you need a link between time and distance, that's where the speed comes to play. If you live in a large city you probably drive much slower during peak hours, but it is up to you to guesstimate the average speed.

[G O D I V A]

I think I have the answer, but I'm not sure.

Here is the question:

Every morning I drive my son to school in the car. Each one-way trip takes about 25 minutes. My car’s fuel efficiency rating, according to the manufacturer, is about 30 miles per US gallon of gasoline. Estimate the amount (in kg) of carbon dioxide emission for each one-way trip.

It says that there is no model answer, but that a rough estimate in the same order of magnitude should be able to be found. You will have to find some information on your own, and make some assumptions.

What I did is:

Assumed 1 US gallon of gasoline weighs 6.4 pounds.  Molar mass of CO₂ is 44, so weight of carbon (M_r 12) would be 44/12 or roughly 3.7.  Assuming the gasoline is 100% octane  then M_r of octane is 114 with carbon make up at 96 so 96/114 is 84%.  Assuming complete combustion; not CO or NO_x then it would be 6.4 * 0.84 * 3.7 = 19.89 pounds.  Divide 2.2 is 9.04 kg so 9.04 kg of CO₂ emitted per US gallon.  Assuming the car goes 50 mph then for 25 minutes thats 20.8 miles so 20.8/30 = 0.693*9.04 = 6.3kg of CO₂ emitted in the 25 minute one way journey. 

However, my professor was talking about how the units were wrong.  miles should be into kilometers, which I dont understand why you would do that.  And that the density of gasoline (because you need the mass) is will be required to find an answer.  However I dont understand how you'd find the volume or even why the density would be needed.  Is what I did the right thing?

[Borek]

Your answer is almost perfecly OK with me, with one minor problem:

so weight of carbon (M_r 12) would be 44/12 or roughly 3.7.

Calculations are OK, but your explanation seems to be misleading. 3.7 is a coefficient to calculate mass of carbon dioxide from mass of burnt carbon.

Other than that IMHO your approach is perfectly correct. Probably assuming that driving your kid to school you drowe 50 mph for 25 minutes is a little bit stretching the reality, but who knows

However, my professor was talking about how the units were wrong.

You have not listed units in many places, so they have to be guessed - that's wrong. But it doesn't mean units as such are wrong. They are all the time OK.

miles should be into kilometers, which I dont understand why you would do that.

And you are right not understanding - there is no need for that. That's just thinking inside of the box. You know the distance in miles, you know mpg (thats in miles) converting to kilometers is additional steps that will just make calcualtion longer, completely not necessary. And in the final result miles cancel out.

And that the density of gasoline (because you need the mass) is will be required to find an answer.

You have used the density - even if not directly - after all 6.4 pound per gallon IS a density unit. Not metric, but perfectly fit for the data you use, as you are given efficiency in mpg, so using gallon as a volume unit is OK (just like using miles was OK), in the end it cancels out.

Mass of the gasoline can be calculated other way around - from the volume and density, but if you know mass of the gallon from the reliable source, looking for density just to make your calculations longer is absurd.

Is what I did the right thing?

Abso-bloody-lutely 

[billnotgatez]

Some people are purest (fanatics) that everything in a problem should be converted to metric first. Some of those people are educators and grade you accordingly. You are at their mercy.

As a side note ( not applicable to your problem) - I on the other had got problems with psi in ideal gas problems and would use 14.6 to convert to apsi plus an R that had psi in it.

Assuming 50 or 35 mph is OK as far as I can tell. Maybe 60 would have been OK and then you would know that speed would get you there at 1 mile per minute. Sometimes I wish I could do 100 when late for class.

After all is said and done, I hope you had fun doing that thought problem.