Pressure, Volume, Temperature, etc.

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Homework Help Overview

The problem involves two compartments containing ideal gases at different pressures and temperatures. The gases are allowed to mix after a partition is removed, and the task is to determine the original volumes of the compartments based on given conditions, including molar fractions and total volume.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use the molar fraction and the ideal gas law to find the total pressure and the number of moles, but expresses confusion about the results. Some participants question the assumptions made regarding the total number of moles and the implications of the pressure values. Others suggest revisiting the definitions and relationships between pressure, volume, and temperature.

Discussion Status

Participants are actively engaging with the problem, with some providing alternative approaches and clarifications on the definitions involved. There is a mix of interpretations regarding the calculations and assumptions, particularly about the total number of moles and the behavior of the gases after mixing.

Contextual Notes

There is uncertainty regarding the identities of the gases and whether the system exchanges heat with the environment. The original poster notes that the problem does not specify certain conditions, leading to confusion about the expected outcomes.

Rachael_Victoria
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Ok so I am doing this problem and it is making me insane. The problem states that you have a container, the container has a partition and in compartment A you have an ideal gas at 5 atm and 400K. In compartment B you have another ideal gas at 8 atm and 400K. The partition is removed and the gasses are allowed to mix together. The molar fraction for gas A is found to be 25/43. The total volume of both of the compartments is 29 liters. Find the original volumes of compartments A and B.

So here is how I approached it.

the molar fraction is [Nk/Ntotal] = [Pk/Ptotal].

Using this formula the total number of moles is 43 and using PV=nRT you get the total pressure equal to 48.669 atm. Plugging this number back into the molar fraction formula gives you a partial pressure for gas A of 28.30 atm.

Now it may just be me but from the get go this does not make a lick of sense as you originally had pressure for gas A of 5 atm and pressure for gas B of 8 atm. If the container is not flexible and the total volume of the container is 29 liters I personally would not expect the combined pressure of the two gasses to be more than 3.5 times greater than the sum of their individual pressures. I have gone on to try and figure out what has gone wrong several different ways but am at a total loss. The only thing I can think of is that the container is flexible, but we have not started studying that and it does not mention it anywhere in the text. I am sure it is something glaringly obvious and if anyone would like to point it out to me I would be eternally grateful. My homework is not due until next friday but this is making me insane.
Thanks
Rachael
 
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First of all, you made a mistake finding the number 48.669 using PV=nRT.You don't have n. 25/43 is a ratio ONLY and tells you nothing about how many moles of gas particles in the box.
Hope this answer you question
 
Further Questions Pressure, Temp Volume etc

Thanks for you quick response,
Ok so I looked back in my text and it specifically defines Dalton's Law as Ptotal=[ntotalRT/V] where ntotal is the total number of moles of gas present. the molar ratio is defined as Xk=nk/ntotal=Pk/Ptotal. I would really love it if the total number of moles present was not 43 but nowhere in the question does it give the identity of the gasses. Can you nudge me onto the next step or how you would figure out the total pressure?
Thanks
Rachael
 
Though i haven't slept over the last 17hrs,this problems is far too simple for me to let the fatigue alter my judgements.

Initial state for compartment A:
[tex]p_{1}V_{1}=\nu_{1}RT[/tex](1)
Initial state for compartment B:
[tex]p_{2}V_{2}=\nu_{2}RT[/tex](2)

Divide (1) through (2) and u'll get
[tex]\frac{p_{1}V_{1}}{p_{2}V_{2}}=\frac{\nu_{1}}{\nu_{2}}[/tex] (3)
Plug in the numbers and find
[tex]\frac{5}{8}\frac{V_{1}}{V_{2}}=\frac{25}{18}[/tex](4)
,from where
[tex]V_{1}=\frac{20}{9} V_{2}[/tex](5)

But we know that
[tex]V_{1}+V_{2}=29l[/tex] (6)
,therefore
[tex]V_{1}=20l;V_{2}=9l[/tex] (7)

Voilà.

Daniel.

PS.[tex]\frac{\nu_{1}+\nu_{2}}{\nu_{1}}=\frac{43}{25}\Rightarrow \frac{\nu_{1}}{\nu_{2}}=\frac{25}{18}[/tex] =>RHS from (4).
 
Rachael_Victoria said:
the molar fraction is [Nk/Ntotal] = [Pk/Ptotal].
Correct. (BTW, it is easier to write with subscripts: Nk/Ntotal = Pk/Ptotal)

Using this formula the total number of moles is 43
It just gives the mole fraction as 25/43. It doesn't say the total no. of moles is 43.

I assume the system does not exchange heat with the environment. So the process is adiabatic ([itex]\Delta Q = 0[/itex]). Since there is no work done by the gas either ([itex]\Delta W = 0[/itex]), this process results in no change to the internal energy of the system (So: [itex]\Delta U = 0[/itex])

This means:
[tex]P_AV_A + P_BV_B = P_fV_f[/tex]

and

[tex]n_ART_i + n_BRT_i = (n_A + n_B)RT_f[/tex]

which means that

[tex]T_i = T_f[/tex] so let's just call it T.

[tex]n_A = P_AV_A/RT = (25/18)n_B[/tex] and [tex]n_B = P_BV_B/RT[/tex]

So:

[tex]n_A = (25/18)P_BV_B/RT = P_AV_A/RT[/tex]

[tex]25P_BV_B = 18P_AV_A[/tex]

[tex]25*8*V_B = 18*5*V_A[/tex] where [itex]V_A=29-V_B[/itex]


AM
 

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