Phase change question

[mleach2]

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Would someone mind taking a look at this question, please? Thank you!

If a 22 g chunk of dry ice (CO2) is placed in an empty 1 liter bottle and tightly corked, what would be the final pressure in the bottle after all the dry ice has evaporated and the temperature has reached 27 degrees C?

The answer is 13.3.

 

[casper22]

PV = nRT

n will be 22/44 (MW of CO2)

Thus P = (22/44*0.0821*300)/1

Are you sure the answer is 13.3... I am getting 12.3...

What I am not doing write here...anybody???

 

[wantVCUdental]

Would someone mind taking a look at this question, please? Thank you!

misread Q...

 

[casper22]

I think u mean V1 = 11.2 (22.4/2)....Correct??

Then the math comes out to be 12.3....

 

[wantVCUdental]

I think u mean V1 = 11.2 (22.4/2)....Correct??

Then the math comes out to be 12.3....

+1, also go the same answer 12.3

 

[MowgliR]

Where is this question from?

 

[MowgliR]

I think u mean V1 = 11.2 (22.4/2)....Correct??

Then the math comes out to be 12.3....

What are your referring to?

 

[mleach2]

Where is this question from?

It's from the Ace The DAT service

 

[mleach2]

I think u mean V1 = 11.2 (22.4/2)....Correct??

Then the math comes out to be 12.3....

That's what I got too, but the answer choice was 13.3. Could be that the answer choice is wrong. I wouldn't advise using the AceTheDAT service. Thanks for your help.

 

[dat2010]

Hi
Just saw your post.
Possible solution.
Since 22g of dry CO2 was given then that is 22/36 = 0.6 moles of gas when completely evaporated.
if 1 mole of gas occupies 22.4 litres therefore 0.6 moles will occupy 13.7 litres and since the volume is sealed into one liter space, (knowing that P/V is inversely proportional) and then also taking the temperature and R into consideration. You will have a long calculation in your hands. But using concept based logics which is true.
Note your dry ice became so at 5atm and would become gas again at about 1 atm thus you should be okay to assume that you new pressure if volume is not restricted is 1atm however......
You will find that since volume is decreased 13.7-fold times (ie expected 13.7 litres compared to a can-sealed/restricted to a 1 litre space ) then the pressure on the container would increase 13.7-fold times hence you have 13.3 as your approximate answer.

 

[DQLEUCD]

Hi
Just saw your post.
Possible solution.
Since 22g of dry CO2 was given then that is 22/36 = 0.6 moles of gas when completely evaporated.
if 1 mole of gas occupies 22.4 litres therefore 0.6 moles will occupy 13.7 litres and since the volume is sealed into one liter space, (knowing that P/V is inversely proportional) and then also taking the temperature and R into consideration. You will have a long calculation in your hands. But using concept based logics which is true.
Note your dry ice became so at 5atm and would become gas again at about 1 atm thus you should be okay to assume that you new pressure if volume is not restricted is 1atm however......
You will find that since volume is decreased 13.7-fold times (ie expected 13.7 litres compared to a can-sealed/restricted to a 1 litre space ) then the pressure on the container would increase 13.7-fold times hence you have 13.3 as your approximate answer.

Your mole caculation is off therefore your answer is off by 1 atm

P1V1/T1n1 = P2V2/T2n2 = the combined gas laws formula where the 1 series is for STP and 2 series is your unknown. Solve for P2 and you get 12.3 atm