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Thermodynamics, argh this problem is killing me ;(

  • Thread starter ovoleg
  • Start date
  • #1
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Can anyone help with this problem? I have tried it a dozen times with every single approach that I can think of except I am not getting even close to the answer...

A flask with a volume of 1.50 L, provided with a stopcock, contains ethane gas (C2H6) at 300 K and atmospheric pressure (1.013×10^5 Pa). The molar mass of ethane is 30.1 g/mol. The system is warmed to a temperature of 380 K, with the stopcock open to the atmosphere. The stopcock is then closed, and the flask cooled to its original temperature. a) What is the final pressure of the ethane in the flask? b)How many grams of ethane remain in the flask?
------------------
V=.0015m^3
T1=300k
T2=380k
V1=1.013x10^5 PA

So, using PV=nRT I solve for n getting .0609 mol, this leads me to believe that the volume did not change much(right?)

So
V1P1/T1=V2P2/T2
Assuming V1=V2, P1/T1=P2/T2
Solving for P2 i get 128313 Pa or 1.28x10^5 Pa,

I am lost

Can anyone please help?

I will be in your debt forever :)
 

Answers and Replies

  • #2
2,985
15
Follow the logic of the problem. What happens in step 1?

The system is warmed to a temperature of 380 K, with the stopcock open to the atmosphere.
During this step, the pressure is constant and equal to atmospheric.

Step 2:

. The stopcock is then closed, and the flask cooled to its original temperature.
What happens now? It is CLOSED, which means the pressure is NOT constant, but the VOLUME is constant.

Work from there.

(I am taking thermo right now as well. The CRITICAL thing about thermo is to recognize what the problem is telling you. Once you get out the important information, the rest is trivial.)
 
Last edited:
  • #3
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cyrusabdollahi said:
Follow the logic of the problem. What happens in step 1?



During this step, the pressure is constant and equal to atmospheric.

Step 2:



What happens no? It is CLOSED, which means the pressure is NOT constant, but the VOLUME is constant.

Work from there.

(I am taking thermo right now as well. The CRITICAL thing about thermo is to recognize what the problem is telling you. Once you get out the important information, the rest is trivial.)
I'm still lost :(

I assumed basically what it was telling me

What did I miss in my calculations?
 
  • #4
93
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anyone please :)
 
  • #5
1,119
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The most important thing you did wrong is combining the two steps into one!

0.0609 mol in 1.5L at 300K looks correct to me... but then step one is the heating to 380K with an open stopcock. Try answering part b) first by finding how many moles would you have in your 1.5L container at this temperature (and at atmospheric pressure). Then when you close the stopcock in step two, *this* is the number of moles in the container.
 
  • #6
BobG
Science Advisor
Homework Helper
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cyrus has the best explanation.

In step 1, volume and pressure stayed constant (1 atm and 1.5L) - the temperature and the number of moles changed. For the given final temp (380K) how many moles were left in the container.

In step 2, volume (1.5L) and the number of moles stayed constant (with the number of moles equalling whatever was left at the end of step 1). For the given final temp, what is the pressure.

Edit: In other words, you successfully calculated how many moles you started with, but you didn't calculate how many moles were left once the temperature was raised to 380K.
 
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