Question 3: A well-insulated rigid container with a total volume of 8-ft3 is div

Question

Question 3: A well-insulated rigid container with a total volume of 8-ft3 is divided into two equal volumes by a thin membrane. Initially, one of these chambers is filled with air at 100 psia (lbf/in2) and 80oF while the other is evacuated. The membrane is now ruptured. Assuming air behaves as an ideal gas.

a) Calculate the internal energy change of the air.

b) What is the final pressure in the container?

Hint: As the half of the tank is evacuated, think if there is any work done by the gas during this process.

Explanation / Answer

a) We know that the boundary work and the heat transfer must both be zero because it is a wellinsulated, rigid container. Thus we can write the energy balance equation:

U = Q W

= 0 0

= 0

b) Because we know that internal energy of an ideal gas is related only to temperature, this implies that the temperature must not change after expansion, since the internal energy does not change. This may be counter-intuitive, since it expands, but because it is expanding against a vacuum there is no pressure on it as it expands, and thus it does no work, so does not cool as a result.

With T1=T2 we can write

P2 = P1 X V1 / V2

= 100 X 1/2

= 50 Psia (lbf/in2) ----- ANS

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