Find the acceptable concentration of CO

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The acceptable hourly average concentration of carbon monoxide (CO) is 35 mg/m^3, and the task is to convert this to ppm at -30 °C and 0.92 atm. The discussion emphasizes the importance of understanding the relationship between temperature, pressure, and concentration when performing the conversion. Participants note that the original question lacks clarity regarding whether the concentration is expressed in volume/volume or weight/volume, complicating the calculation. It is suggested that the concentration should be treated as 35 mg/m^3 under standard conditions, and adjustments for the new temperature and pressure should follow. The conversation highlights the need for precise definitions and context in chemistry problems to ensure accurate calculations.
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Homework Statement


Based on the “National Ambient Air Quality Objectives”, the acceptable hourly average concentration of carbon monoxide (CO) is 35 mg/m^3.

Find the acceptable concentration of CO in ppm if the temperature is -30 °C and pressure is 0.92 atm.

Express the concentration as a percent by volume.

Homework Equations


PV = nRT (maybe)

The Attempt at a Solution


Hello, everyone!

Could someone please help me understand how to do the problem above?

So far, I get the following.:
101.325 kPa * V = (35E-3 g / 28.01 g) mol * 8.31 kPa L / (mol K) * (-30 + 273.15) K

V = 0.0249180252056537913 L

Am I on the right track? If so, what do I do next? If not, then what must I do?

Any input would be greatly appreciated!
 
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Write the ppm definition.
 
Right equation, otherwise not really. We often get questions supposedly of chemistry here which are really problems of arithmetic that students learned years before starting chemistry.

You are given a concentration in mg/m-3 at one T and P. You want to know in the first place the mg /m-3 at another T, P. That is per unit volume - the volume in question, that is the m3 doesn't change. From your equation in 2 what remains the same between the two situations is P/R, can you see? That is the main bit of non-arithmetical knowledge. The other is knowing what is really meant by T in this context. :oldwink:

Work out the new mg/m-3 then what that is in ppm.
 
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epenguin said:
You are given a concentration in mg/m-3 at one T and P.

Not exactly, original T and P are not given.

epenguin said:
Work out the new mg/m-3 then what that is in ppm.

As original statement doesn't contain any additional information, I am ready to assume it is to be understood as "35 mg/m3 at any conditions".

It doesn't say anything about whether is it ppm v/v or w/w, which makes it ambiguous and impossible to answer.
 
The way I read the question is 35 mg / m^3 at standard conditions is the legislative rule, and that what is sought is the adaption of the legislative rule to the new conditions ( -30 °C and 0.92 atm pressure).

Edit:
As for the ppm formula that you're asking for, according to this link ( https://sciencing.com/calculate-ppm-5194302.html ), it's weight / volume * 10^6.
 
s3a said:
The way I read the question is 35 mg / m^3 at standard conditions is the legislative rule, and that what is sought is the adaption of the legislative rule to the new conditions ( -30 °C and 0.92 atm pressure).

I am not convinced about the STP part, but you can definitely try this way. Note, that 35 mg of CO at STP means some easy to calculate volume in a well defined volume of air (1 m3). If you change T, P you change it for both gases. Will the ratio of volumes CO/air change? Will the molar ratio of CO/air change? Will the mass ratio of CO/air change? Does it make any sense to make these calculations?

As for the ppm formula that you're asking for, according to this link (...), it's weight / volume * 10^6.

Weight over volume - what units for weight, what units for volume? As long as they are not well defined there is no way to have an unambiguous result.

Compare http://www.chembuddy.com/?left=concentration&right=ppm-ppb-ppt

I still think assuming it is 35 mg per cubic meter of air regardless of T, P is the only way to add sense to the question.
 
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