2 cylinders, pushing gas from one to another

  • #1
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Homework Statement



I managed to do part (i) and first part to (ii):

Homework Equations



U = Q + W




The Attempt at a Solution




for the first part:
T/Tf = 1/(2 - y)

for (ii):
W = yVP

Given: (3/2)R = Q/(nΔT),

Q = (3/2)nR(ΔT)

Therefore U = Q + W

U = (3/2)nR(ΔT) + yVP

But then,

isn't U already equal to (3/2)nR(ΔT) since its a monoatomic gas?

Then that gives y = 0, so i'm utterly confused.


Here's what the answer writes:

W = yVP

H = (3/2)nR(Tf - T) (does H here mean U?)

Does this imply that heat transferred Q, equals to zero? so W = U?
But there's no indication that this is an adiabatic process? How can we assume that?
 

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Answers and Replies

  • #2
I like Serena
Homework Helper
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Hi unscientific! :smile:

I haven't worked through your entire problem yet, but I think I can answer your questions.

Q=(3/2)nR(ΔT) is not true in general (for ideal monatomic gasses).
What it true, is U=(3/2)nRT or ΔU=(3/2)nR(ΔT) for ideal monatomic gasses.

H would be "enthalpy" defined as H=U+PV.
But that formula would not be right. In this case it should be: H = (5/2)nR(Tf - T).
So perhaps they did intend U, or else you made a typo.

And no, it is not an adiabatic process. The problem specifies that it is an isobaric process.

For reference, here's a table with formulas:
http://en.wikipedia.org/wiki/Table_...ions#Equation_Table_for_a_monatomic_Ideal_Gas
Look at the column for isobaric processes.
 
  • #3
1,728
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Hi unscientific! :smile:

I haven't worked through your entire problem yet, but I think I can answer your questions.

Q=(3/2)nR(ΔT) is not true in general (for ideal monatomic gasses).
What it true, is U=(3/2)nRT or ΔU=(3/2)nR(ΔT) for ideal monatomic gasses.

H would be "enthalpy" defined as H=U+PV.
But that formula would not be right. In this case it should be: H = (5/2)nR(Tf - T).
So perhaps they did intend U, or else you made a typo.

And no, it is not an adiabatic process. The problem specifies that it is an isobaric process.

For reference, here's a table with formulas:
http://en.wikipedia.org/wiki/Table_...ions#Equation_Table_for_a_monatomic_Ideal_Gas
Look at the column for isobaric processes.

I understand how ΔU=(3/2)nR(ΔT), but i simply don't understand how they simply equate
U = W, in the process ignoring Q, as U = Q + W...
 
  • #4
I like Serena
Homework Helper
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I understand how ΔU=(3/2)nR(ΔT), but i simply don't understand how they simply equate
U = W, in the process ignoring Q, as U = Q + W...
Where do they do this?

You're right, you can't simply equate U=W.
It is only true in an adiabatic process.
 
  • #5
ehild
Homework Helper
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It is not a quasi-equilibrium process as the gas in both containers are not in equilibrium during the motion of the piston: there is a pressure difference between the containers and a gas flow. The process is not isobaric as the pressure of the whole gas is not defined.

The gas was in equilibrium before opening the valve and reached equilibrium at the end when the temperature and the pressure became equal in both containers.

In the initial state, PV=nRT, in the final one, P(2-y)V=nRTf.

I think it was meant that the walls were adiabatic, not allowing heat transfer. The final temperature would be the same as the initial one otherwise.

So the change of internal energy is equal to the external work done on the whole system, PVy.

The change of internal energy is 3/2 nR(Tf-T).

So you have three equations:

PV=nRT
PV(2-y)=nRTf
3/2nR(Tf-T)= PVy
.

From these, you can find the numerical value of both y and Tf/T.

ehild
 

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