# Find entropy change when two tanks equalize

• MattHorbacz
In summary, the two tanks have the same pressure and temperature, but the air in tank 1 has more air than the air in tank 2. So the final temperature is higher in tank 1 and the change in entropy is greater.
MattHorbacz

## Homework Statement

Two rigid, insulated tanks are connected with a pipe and valve. One tank has 0.5 kg air at 200 kPa, 300 K and the other has 0.75 kg air at 100 kPa, 400 K. The valve is opened, and the air comes to a single uniform state without any heat transfer. Find the final temperature and the change in entropy of the air.

## Homework Equations

ds=Cv*ln(T2/T1)+R*ln(v2/v1)
PV=nRT
m1cvΔT1=-m2cvΔT2
?

## The Attempt at a Solution

I am pretty lost. I solved for the total volume, mass, and specific volume, but have no idea how to solve for temperature. And I thought that there was no change in entropy without heat transfer. Even showing me what equations to use would be extremely helpful

Edit: just remembered that q_tank1=-q_tank2: m1c_vΔT1=-m2c_vΔT2. Solved for final temp to be 360 K

Last edited:
##\Delta S= Q/T## only holds for reversible processes. Your process is irreversible, so there may be a change in entropy although there is no heat exchange. Namely in your problem, the process is irreversible as the two containers have different temperature. But you may calculate the change of entropy of each of the containers as taken on their own they have a uniform temperature.

I'm not sure if this was correct, but I calculated two ΔS, tank1 initial vs total and also tank2 initial vs total. I used the equation I listed above, then apparently you multiply each value for entropy by the mass then add the values...I think what I did was the right method, but I most definitely messed up somewhere. Is this sort of the right method?

Sounds good to me.

MattHorbacz
The way to approach this problem is to recognize that, since the total volume of the two tanks does not change and no heat enters or leaves the combined system, there is no work done on the combined system, and there is no heat transfer to or from the system. This means that the overall change in internal energy is zero. Also, the final pressures in the two tanks must match. This is enough information to determine the final temperature and pressure. Once you know these, determining the change in entropy is straightforward.

## 1. What is entropy?

Entropy is a thermodynamic property that measures the level of disorder or randomness in a system. It is often referred to as the measure of the system's energy that is unavailable for work.

## 2. How is entropy change calculated?

The change in entropy can be calculated using the equation ΔS = Qrev/T, where ΔS is the change in entropy, Qrev is the reversible heat transfer, and T is the temperature in Kelvin. In the case of equalizing tanks, Qrev would be the heat transfer from one tank to the other.

## 3. What is the significance of equalizing tanks in terms of entropy?

Equalizing tanks can help to maintain a constant level of entropy in a closed system. As the tanks equalize, the heat transfer between them results in a decrease in the overall entropy of the system, helping to maintain a state of equilibrium.

## 4. What factors affect the entropy change when two tanks equalize?

The main factors that affect the entropy change in this scenario are the initial temperature difference between the tanks, the heat transfer rate, and the size of the tanks. A larger initial temperature difference or a faster heat transfer rate would result in a larger change in entropy.

## 5. How does the second law of thermodynamics apply to the equalization of two tanks?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. In the case of equalizing tanks, the transfer of heat from one tank to the other results in a decrease in the overall entropy of the system, but the total entropy of the tanks and their surroundings will still increase due to the irreversible nature of the process.

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