# 2 cylinders, pushing gas from one to another

• unscientific
In summary, Homework Equations state that in an isobaric process, the change in internal energy is equal to the external work done on the whole system. The problem specifies that it is an isobaric process, and the gas in both containers are not in equilibrium during the motion of the piston.
unscientific

## Homework Statement

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

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?

#### Attachments

• cylinder piston.jpg
48 KB · Views: 453
• piston2.jpg
24 KB · Views: 376
Hi unscientific!

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 like Serena said:
Hi unscientific!

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...

unscientific said:
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.

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

## 1. How does gas move from one cylinder to another?

Gas moves from one cylinder to another through the process of diffusion. When the gas molecules in one cylinder have a higher concentration than in the other cylinder, they will naturally flow from the area of higher concentration to the area of lower concentration.

## 2. What causes the gas to move?

The gas moves due to the differences in gas pressure between the two cylinders. Gas molecules will always move from a higher pressure area to a lower pressure area until the pressure is equalized.

## 3. Can any type of gas be used in this process?

Yes, any type of gas can be used as long as it is in a gaseous state and there is a difference in pressure between the two cylinders.

## 4. Is this process efficient for transferring gas?

The efficiency of this process depends on various factors such as the pressure difference between the cylinders, the size of the cylinders, and the type of gas being transferred. In general, it is a relatively efficient method for transferring gas.

## 5. Are there any risks associated with this process?

There are potential risks associated with handling and transferring gas, such as the risk of gas leaks or explosions. It is important to follow proper safety protocols and use appropriate equipment when conducting this process.

• Introductory Physics Homework Help
Replies
4
Views
814
• Introductory Physics Homework Help
Replies
1
Views
630
• Introductory Physics Homework Help
Replies
1
Views
791
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
22
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
923
• Introductory Physics Homework Help
Replies
1
Views
945
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
1K