[Lachln]

4.0 grams dry ice (carbon dioxide) was dropped into a flash cotnaining air at 100kPa, and a stopper was inserted. When all the dry ice had sublimed and the temperature had been brought back to its initial value of 15 degrees Celsius, what would be the total pressure in the flash?

100 kPa= 0.986923267 atm
4.0 grams CO2 = 0.0901 moles CO2

I guess you need to use the gas laws in some way.

I'm just stuck trying to answer the question without the volume of the flask.

Please help?

[Borek]

I don't see how to solve the problem with the information given.

[Lachln]

I don't see how to solve the problem with the information given.

It's driving me crazy.

Boyle's law states that the volume is inversely proportional to the pressure.
So P=1/V
0.986923267=1/V
V=1.01325L

Is that right? Or do I have to consider the value of K?


Thank-you for replying! It's disheartening when no one replies.

[Borek]

So P=1/V
0.986923267=1/V
V=1.01325L

Is that right? Or do I have to consider the value of K?

No idea what you mean by K, if 1 (as in P=K/V) then yes - you don't know the value of K, so you can't calculate the original volume. There is not enough data. IMHO original question can't be solved.

Thank-you for replying! It's disheartening when no one replies.

Bear in mind it is middle of the night is US and early morning in Europe, where most of our members reside. Add to that we visit the forum in our free time. Getting an answer in 4 hours is not a bad result

[billnotgatez]

If we rewrote the question
-----
4.0 grams dry ice (carbon dioxide) was dropped into a flask containing Nitrogen at 100kPa, and a stopper was inserted. When all the dry ice had sublimed and the temperature had been brought back to its initial value of 15 degrees Celsius, what would be the total pressure in the flask?
-----
Could we solve the problem?

Note:
I changed the word     air     to     Nitrogen.
As an aside
I also changed the typos but that appears not to be important to the questions results.
But, when you have typos it makes us wonder if you presented the question without missing something.

[Borek]

No, you have not changed anything. Identity of the gas doesn't matter.

[billnotgatez]

I see what I missed now
You know the original pressure of the flask before Carbon dioxide
You know the original temperature of the flask before Carbon dioxide
But, you do not know the weight or moles of air ( or my substitute of Nitrogen) before Carbon dioxide.
And, you do not know the volume of the flask.

My guess is that @Lachln missed putting the flask size when they retyped the question.


Edit: I fixed typo of the word pressure being typed twice
I meant temperature in the following sentence.

[Lachln]

My guess is that @Lachln missed putting the flask size when they retyped the question.

Nope - that's the exact question (except for the typos). I've been puzzled about it all day

I also changed the typos but that appears not to be important to the questions results.
But, when you have typos it makes us wonder if you presented the question without missing something.

I'm a high school student and, like I said before, I couldn't get this question off my mind. I sneakily posted this topic during my accounting lesson so it was a bit of a rushed effort.

Chemistry before accounting!

Thanks for your responses! This is the second time I've needed to ask a question and you've been consistently helpful. Expect a few more questions (hopefully not too many) next year as I enter into year 12 (my final year)  .

[billnotgatez]

@Lachln
Did your question sheet have any other question that had flask size that may have been assumed in this question?

As an aside
Do you think you could have solved the problem given a flask size like 100 ml or 1 L?

[billnotgatez]

@Lachln
Were you given the answer to your original question?

[Lachln]

@Lachln
Did your question sheet have any other question that had flask size that may have been assumed in this question?

As an aside
Do you think you could have solved the problem given a flask size like 100 ml or 1 L?

Well, I do have the answers. The answers say the total pressure is 209kPa. This means 109kPa must be exerted by the CO₂.

Using PV=nRT
(109/101.325)*V=(4/44.01)×0.08206×(15+273.15)
V=1.9977L

Basically two litres if you account for the fact that 0.08206 is only approx.

[Borek]

Well, I do have the answers. The answers say the total pressure is 209kPa. This means 109kPa must be exerted by the CO₂.

Using PV=nRT
(109/101.325)*V=(4/44.01)×0.08206×(15+273.15)
V=1.9977L

Basically two litres if you account for the fact that 0.08206 is only approx.

So this is an answer based on the assumption it is all in a 2 L flask. Make the flask 4 L and the answer will be around 150 kPa, make the flask 1 L and the answer will be around 300 kPa - neither of these is "correct" in terms of data given in the original question.

In other words: whoever asked the question failed to give all the necessary information, making the problem impossible to solve. Playing with numbers in hope it will change something is just a a waste of time.

[Borek]

I have split the thread, please don't add new problems to old threads, but start new threads.

http://www.chemicalforums.com/index.php?topic=77495.0