What Is the Final Pressure in Both Containers After Equalization?

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Homework Help Overview

The problem involves two containers holding an ideal gas at different pressures and temperatures, with the goal of determining the final pressure after the gases equalize. Container A has a pressure of 5x10^5 Pa and a temperature of 300 K, while Container B has a pressure of 10^5 Pa and a temperature of 400 K. The containers are connected, and the temperatures remain constant during the process.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the ideal gas law (PV = nRT) to relate the pressures and volumes of the gases in both containers. There are questions about the relationship between the number of moles in each container and how to express them in terms of known quantities. Some participants explore how to set up equations based on the given pressures, volumes, and temperatures.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the relationships between the variables involved. Some have proposed equations to express the number of moles and how they relate to the pressures, while others express confusion about isolating variables and the implications of the given information.

Contextual Notes

There is a lack of explicit information regarding the volumes of the containers beyond the relationship that Container B's volume is four times that of Container A. Participants are also navigating the constraints of the problem, including the constant temperatures during the equalization process.

vortex193
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Homework Statement


Container A holds an ideal gas at a pressure of 5x10^5 Pa and a temperature of 300 K. It is connected by a thin tube (and closed valve) to container B, with four times the volume of A. Container B holds the same ideal gas at a pressure of 10^5 Pa and a temperature of 400 K. The valve is opened to allow the pressures to equalize. What is the pressure in the two containers? Note: The temperatures in the containers do not go to an equilibrium and both temperatures stay at their original temperatures.

Homework Equations


PV = nRT (R = 8.31 J/mol*K)

The Attempt at a Solution

 
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I hesitated to post because I'm not familiar with this problem.
You might try using your PV = nRT formula. The pressures will equalize so you'll have
P1 = P2
where each P = nRT/V.
You know the R,T and V for each. How will n1 and n2 be related?
 
No relation was given for n1 and n2.
 
Is n the number of moles or molecules? If so, then you know n1 + n2 equals the n1 + n2 from the original configuration, which you can figure out and get an actual number for.
 
n = number of moles. So the final pressure would equal.

P = (n1 +n2)RT/V

but what would T and V be? remember that both containers begin and end with the same temperature.
 
I think you misunderstood me. I said
P1 = P2
n1*R*T1/V1 = n2*R*T2/V2 and n1+n2 = n (known number).
You know R,T1,V1,T2,V2 so your have unknowns n1 and n2 and two equations relating them. Thus you can find n1 and n2. Then you can go back and use P = n1*R*T1/V1 to find P.
 
Actually V1 and V2 is not given. You only know that V2 is 4 times V1.
 
That will work - put in "v1" and "4V1" and cancel the V1's.
 
I don't understand how you can find n1 and n2 if you do not have n.
 
  • #10
n=PV/(RT) for the original container A + PV/(RT) for the original container B
Oh, yes you'll have V1's in there - but surely they cancel out when you put the n1 and n2 into n1*R*T1/V1 = n2*R*T2/V2
 
  • #11
Wow. This question really does not make sense to me. I don't even know how to isolate for either n1 or n2.
 
  • #12
n=PV/(RT) for the original container A + PV/(RT)
=5*V/(R*300) + 1*4V/(R*400)
= .0267*V/R

n = n1 + n2 so n2 = .0267*V/R - n1
sub into n1*R*T1/V1 = n2*R*T2/V2 to get
n1*R*300/V = (.0267*V/R - n1)*R*400/(4V)
Take all the n1 terms to the left:
n1*R*300/V + n1*R*400/(4V) = .0267*V/R*R*400/(4V)
n1*300 + n1*400/(4) = .0267*V/R*400/(4) divide by R, mult by V
n1*(300 + 100) = 2.67*V/R
n1 = .006675*V/R

P1 = n1*R*T1/V1
= .006675*V/R*R*300/V
= 2
I put the pressure in units of 100 000 Pa so this is really 200 000 Pa.
 
  • #13
Wow. Thank you. I really appreciate it. Where is the thanks button on this forum so I can thank you? I'm in Grade 12 and I am sure this is not a high school level question. Thank you once again.
 

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