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.
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.
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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.
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.
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