Jump to content
Sign in to follow this  
Catwalker

Chemistry II- This week: Gas Volume Problem

Recommended Posts

First, 1.35 mol of CO is not .04g; you already know that 1 mol = 28.01g.

 

Second, this is a problem where DIMENSIONAL ANALYSIS IS YOUR FRIEND YES IT IS DO IT PUT UNITS IN YOUR EQUATION.

 

Whoops. 37.9g of CO seems a little high to me.

 

I guess I don't know what you mean by dimensional analysis. If I did, I'd probably do it. Explain?

 

Also, I thought about dividing the number of molecules I have by 1 million and then dividing that by 800 to find the number of molecules of CO. Divide that by Avogadro's number to find the moles and then grams turn out to be 4.74 x 10-8 which seems right to me.

Share this post


Link to post
Share on other sites

The basis of dimensional analysis is that the units (dimensions) on both sides of an equation must be an identical.

 

This prevents us saying things like "6kg = 4 m/s"

 

It's saved my arse many times in physics tests.

 

By way of illustrative example:

 

You cant remember the equation for velocity.

 

Is it velocity = distance/time or is it velocity = time/distance.

 

Well, the units of velocity are meters per second (m/s). The units of time are seconds (s) and the units of distance are meters (m).

 

From this we can logically deduce that velocity must be distance/time because the units on both sides of the equation will be the same m/s = m/s.

 

If we tried to say that velocity = time/distance we would have m/s = s/m which is clearly inconsistent.

 

 

This is the basis of dimensional analysis, a tool that can be applied across many disciplines.

 

In your situation, you need to keep track of the units (dimensions) that you are using and make sure that you're being consistent.

Share this post


Link to post
Share on other sites

I understand now.

 

The gas equation uses liters. I had to convert from cubic meters to liters to use it. I used a site on the internet that I've been using for weeks for the conversion and have never had a problem. I'm not sure what the problem is now.

Share this post


Link to post
Share on other sites

This is one of the last questions I'll have to do for this class. It seems like it should be easy, but I'm not sure what route I should take to solve it.

 

A concentration of 8.00 x 102 ppm by volume of CO (28.01 g/mol) is considered lethal to humans. Calculate the minimum mass of CO in grams that would become a lethal concentration in a closed room that is 17.6m long, 8.8m wide and 2.64m high. The temperature and pressure are 20.0 C and 756mm Hg, respectively.

 

The first thing is to figure out the volume of the room: 408.88 m3 or 408880L.

 

Also, we're looking at .0008% of the gas will be CO.

 

I've thought about using PV=nRT to figure out how many moles of gas might be in the room. In this case, (.995 x 408880)/n(.082057 x 293), n = 16921.4 moles of gas in the room. (That seems absurdly high.) Then we use Avogadro's number to figure out how many molecules are in the room: 1.019 x 1028. Then I should just have to find out what .0008% of that is to determine the answer: 8.15204 x 1023 molecules of CO, which works out to 1.35 moles of CO. That is .04833g.

 

That seems like it might be right, half of a half of a gram of CO. Is it?

 

 

I would use the density of air at stp for the problem, because it seems not quite right without it? They don't give you any other density so I'd have to assume to use that? I looked it up and this is the Yahoo answer for the density of air at stp: 1.297kg/m^3

 

 

So this is how I'd do the problem:

 

Volume of the room you got: 408.88 m3

 

Which I'd then use to find the mass of air in the room: (1.297kg/m^3)*(408.88 m3) = X kg air (because I don't have a calculator on me and I'm too lazy to use the computer one)

 

Then I'd use the mass of air I just found to find the mass of CO: (X kg air)(1000 g/1kg) (8.00 x 102CO/1000000air) = Z g CO

 

 

EDIT:

OH WAIT, I missed the beginning of the problem, re-working it now.....can't do it that way (obviously uber easy that way) since they gave you pressure and stuff. Need to wake up before I try to do problems...

Share this post


Link to post
Share on other sites

My computer is having issues on me and then it blinked in the corner that Windows is installing updates, and now its saying there are multiple security problems or something >.< so I'm going to be brief and hope that this post makes it!

 

if you convert your pressure to kPa you can solve directly for the density of air at your temperature and pressure using the method outlined on this page:

 

http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html

 

 

 

or instead:

They also have a molar mass of air, which you could use to get from you moles of air that you calculated to get to mass of air, and then use the final step that I included in the previous post.

 

 

I don't think you can do mol CO/ 1000000 mol air like you said you were going to do, because I thought ppm was only good for mass/mass which isn't the same as mole/mole ? I'm not entirely for sure on this, but I'm fairly certain you should always convert to masses then use ppm to convert to another substance in your concentration. Which is why you need to solve for either the density of air, or use that molar mass of air average value to convert from moles to mass before using ppm.

 

 

I hope that helps and doesn't make life more confusing for you!

 

I need to work on my computer and figure out what I've done to it now... *sigh*

Share this post


Link to post
Share on other sites

This isn't as hard as people are making it. The key is that your lethal concentration is ppm by volume.

 

This is a very important distinction. It means that 8e2/1e6 gives you the VOLUME of the total CO divided by the total room volume. This means that you have 327.104L (I'm assuming that your room volume is correct) of CO. Now use the ideal gas law to figure out the moles.

 

pV=nRT

 

(756 mmHg)(327.104 L)=n(62.363 L*mmHg/mol/K)(293 K)

 

n=(756 mmHg)(327.104 L)*(1/62.363 K*mol/L/mmHg)/(293 K)

 

n=13.53 mol*(28.01 g/mol)

 

n=379.08g CO

 

EDIT - Cat, after looking over your solution, I'd note that it also works. However, you made one minor error - 800 ppm is 0.08% of total volume, not 0.0008%. It looks like your decimal point wasn't moved after you converted from a decimal to a percentage.

Share this post


Link to post
Share on other sites

or you could do it Rollin's easier way and get the right answer.

 

 

(which makes sense why you using moles of CO/1000000 moles of air WILL work even though its not mass like I thought it had to be---because moles of gas are directly proportional to volume of gas)

Share this post


Link to post
Share on other sites

This is the exact answer I posted to this problem. The professor gave me a 100%. Tell me if this is right. It's pretty much what I put as my solution earlier, but this just doesn't feel right, especially since my answer is different than RD's.

 

A concentration of 8.00 x 102 ppm by volume of CO (28.01 g/mol) is considered lethal to humans. Calculate the minimum mass of CO in grams that would become a lethal concentration in a closed room that is 17.6m long, 8.8m wide and 2.64m high. The temperature and pressure are 20.0 C and 756mm Hg, respectively.

 

Volume of the room: 408.88 m3 or 408880L.

 

.08% of the gas will be CO.

 

Use PV=nRT to determine the moles of gas in the room. (.995 x 408880)/n(.082057 x 293), n = 16921.4 moles of gas in the room.

 

Then we use Avogadro's number to figure out how many molecules are in the room: 1.019 x 1028

 

Multiply that by .08% to find the number of CO molecules = 6.52 x 1020

 

Use Avogadro's number to determine how many moles: .001 moles

 

This works out to 3.9 x 10-6 grams of CO.

Share this post


Link to post
Share on other sites

This is the exact answer I posted to this problem. The professor gave me a 100%. Tell me if this is right. It's pretty much what I put as my solution earlier, but this just doesn't feel right, especially since my answer is different than RD's.

 

A concentration of 8.00 x 102 ppm by volume of CO (28.01 g/mol) is considered lethal to humans. Calculate the minimum mass of CO in grams that would become a lethal concentration in a closed room that is 17.6m long, 8.8m wide and 2.64m high. The temperature and pressure are 20.0 C and 756mm Hg, respectively.

 

Volume of the room: 408.88 m3 or 408880L.

 

.08% of the gas will be CO.

 

Use PV=nRT to determine the moles of gas in the room. (.995 x 408880)/n(.082057 x 293), n = 16921.4 moles of gas in the room.

 

Then we use Avogadro's number to figure out how many molecules are in the room: 1.019 x 1028

 

Multiply that by .08% to find the number of CO molecules = 6.52 x 1020

 

Use Avogadro's number to determine how many moles: .001 moles

 

This works out to 3.9 x 10-6 grams of CO.

He gave you credit because you only made one error and that was a multiplication mistake. You understood the concept.

 

1.019e28 * 0.0008 = 8.13e24, not 6.52e20.

 

When working with exponents, it's worth sanity checking the result. Something that's 10-4 times 1028 should be on the scale of 1024, and if it isn't then it's time to do some checking. I'd skip that entire step though, you're putting avagadro's number in and then taking it right back out.

 

16921.4 mol * (6.22e23 g/mol) * 0.0008 / (6.22e23 g/mol) = 0.0008 * 16921.4 mol = 13.53 mol

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×