A rigid insulated closed tank consists of two compartments of equal volume separated by a valve. Initially one contains air at 145 psia and temperature of 130 F, while second is completely empty(vaccum). the valve is opened and remains so until both are in pressure and temperature equilibrium. determine final pressure and temperature of air assuming each compartment is 20ft^3.

I know I will be using dU=dQ-dW , I think one of the terms if 0 but don't know what else.

Both dQ and dW are zero. Therefore dU = 0 That means the temperature does not change. Pressure decreaseds by a factor of two because of the doubled volume of the gas.

Enthalpy also doesn't change because PV = constant along with U.

I am pretty sure the temperature will decrease, there should be a heat transfer. Anyone verify this?

Don't be too sure of yourself.

You are wrong. Gas expanding into a vacuum enclosure does not drop in temperature once the fluid motion stops and equilibrium is established. The walls are not pushed out. No work is done and no heat is transfered in. U does not change. T is proportional to U, and independent of pressure. So T does not change, either.

Your teacher should confirm this.

Thank you, seems to make sense, sorry was sure I heard differently in class.

It is true that a gas that is expanding cools, as long as it is doing work on some gas into which it is expanding. But when expanding into a sealed evacuated enclosure, it has no gas other than itself to do work against. Eventually the expansion motion has to stop and the directed kinetic energy of the expanding (temporarily cooled) gas becomes random again, and this reheats it to the orginal temperature.

To solve this problem, you can consider the following steps:

Step 1: Determine the initial state of the air in the first compartment.
Given:
Pressure in the first compartment (P1) = 145 psia
Temperature in the first compartment (T1) = 130 °F

Step 2: Determine the initial state of the second compartment.
Given:
The second compartment is completely empty (vacuum), so there is no air present.

Step 3: Calculate the final state of the air after equilibrium.
Since the valve is opened and remains so until pressure and temperature equilibrium is reached, we can assume that the air from the first compartment will expand into the second compartment until equilibrium is achieved.

Note: Since both compartments have equal volumes (20 ft^3), we can assume that the final pressure and temperature will be the same in both compartments.

Step 4: Apply the first law of thermodynamics (dU = dQ - dW).
We can use the first law of thermodynamics to determine the change in internal energy (dU) of the air in the first compartment.

dU = dQ - dW

Here, dU represents the change in internal energy, dQ represents the heat transfer, and dW represents the work done.

Since the process is adiabatic (no heat transfer), dQ is zero.

Therefore, the equation becomes:

dU = -dW

Step 5: Determine the work done (dW).
The work done during the expansion process can be calculated using the formula:

dW = PdV

Here, P represents the pressure, and dV represents the change in volume.

Since the volume change is from 20 ft^3 to 0 ft^3 (as the air expands completely into the second compartment), the change in volume (dV) is -20 ft^3.

Step 6: Calculate the final pressure and temperature of the air.
Using the ideal gas law equation:

P1V1/T1 = P2V2/T2

Here, P1 and T1 are the initial pressure and temperature in the first compartment, P2 and T2 are the final pressure and temperature in both compartments, and V1 and V2 are the initial and final volumes.

Since V2 = 0 ft^3, we can substitute V1 = 20 ft^3 and V2 = 0 ft^3 into the equation:

P1V1/T1 = P2V2/T2

P1V1/T1 = P2(0 ft^3)/T2

P1V1/T1 = 0

Therefore, the final pressure of the air in both compartments is 0 psia.

Since the final pressure is 0 psia, we can't determine the final temperature because the equation (P1V1/T1 = P2V2/T2) is undefined.

Hence, we can conclude that the final pressure of the air is 0 psia, and the final temperature cannot be determined.