Equalizing pressure between two vessels

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In summary, two insulated air containers with volumes V1 and V2, pressures p1 and p2, and a shared temperature T are connected with a hose to equalize pressure but without transferring heat. The final temperatures and pressure can be calculated using the equation pV = mRT, but in this case with no heat transfer, the final pressure cannot be determined without knowing the final temperatures. The book states that the final pressure is the same as in the case with heat transfer, but the reason for this is unclear. The calculations for entropy in the case of heat transfer show a decrease, which is not possible, indicating a mistake in the calculations.
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kandelabr
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Homework Statement


There are two insulated air containers, first with a volume V1 and second with V2. In the first there is a pressure p1 and in the second p2. The temperatures are the same in both, thus T. Then both containers are connected with a hose so that pressure equalizes, but no heat is transferred between them.
What are the final temperatures and final pressure?


2. Homework Equations
...

The Attempt at a Solution


If there was heat transfer between containers, the final temperature would be the same in both and even the same as the temperature T in the beginning and because of that I could calculate the final pressure using pV = mRT (R = gas constant for air), where I'd get mass for one container from specific volumes.

But this case with no heat transfer is a bit confusing. I can't get final pressure with the equation used above because I don't know the final temperatures. The book says the final pressure is the same as above, but I don't know why is it so.
 
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  • #2
My results regarding the case where heat transfer occurs are that entropy actually decreases - using this formula:

s = Cvln(pend/p1) + Cpln(vend/v1) +
+ Cvln(pend/p2) + Cpln(vend/v2),

where v1 and v2 are specific volumes before mixing and vend is the final specific volume (the same for pressures).

The actual numbers are:
T = 293 K
p1 = 1 bar
p2 = 20 bar
V1 = 5m3
V2 = 2m3

From that I got
v1 = 0.8403 m3/kg)
v2 = 0.0420 m3/kg)
pend = 6.43 bar
vend = 0.1308 m3/kg

s = -206 J/kgK

I know entropy cannot be decreased with mixing, so what's wrong here?
 
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  • #3


I can provide a response to this content by explaining the principles of pressure equalization and how it relates to the scenario described.

When two containers with different pressures are connected by a hose, the air molecules in each container will start to move towards the other container. This is because air molecules naturally move from areas of high pressure to areas of low pressure, in an attempt to reach equilibrium.

In the scenario described, the two containers are insulated and have the same temperature. This means that there is no heat transfer between them, and the temperature remains constant throughout the process. Therefore, the only factor affecting the movement of air molecules is the pressure difference between the two containers.

As the air molecules move from the higher pressure container to the lower pressure container, the pressure in the first container will decrease while the pressure in the second container will increase. This process will continue until the pressure in both containers is equalized.

At this point, the final pressure in both containers will be the same, and it will be equal to the average of the initial pressures (p1 and p2). This is because the total number of air molecules in both containers remains the same, and the final pressure is a result of the redistribution of these molecules.

The final temperatures in both containers will also be the same as the initial temperature T, as there was no heat transfer between the containers. This can be explained by the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transferred from one form to another.

In conclusion, the final pressure and temperature in both containers will be equalized and will be the average of the initial pressures and temperatures, respectively. This can be calculated using the ideal gas law, pV=nRT, where n is the number of moles of air and R is the gas constant for air.
 

Related to Equalizing pressure between two vessels

What is equalizing pressure between two vessels?

Equalizing pressure between two vessels refers to the process of equalizing the air pressure inside two separate containers or vessels. This is typically done to avoid any imbalances in pressure that can cause damage or instability.

Why is it important to equalize pressure between two vessels?

Equalizing pressure between two vessels is important because it helps to prevent any sudden changes in pressure that can cause damage to the vessels or the contents inside. It also ensures that the vessels are stable and can function properly.

What are the methods for equalizing pressure between two vessels?

There are several methods for equalizing pressure between two vessels, including using a vent valve, connecting the vessels with a pipe, and using a pressure regulator. The method chosen will depend on the specific situation and the vessels involved.

What factors can affect the equalization of pressure between two vessels?

The equalization of pressure between two vessels can be affected by factors such as the size and shape of the vessels, the amount of pressure inside each vessel, and any obstacles that may be present between the two vessels. Temperature and altitude can also play a role in the equalization process.

What precautions should be taken when equalizing pressure between two vessels?

When equalizing pressure between two vessels, it is important to ensure that the vessels are compatible and can handle the equalization process without damage. It is also important to follow proper safety protocols and to monitor the equalization process closely to prevent any potential hazards or accidents.

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