A well-insulated rigid container with a total volume of 8-ft3 is divided into tw
ID: 1717892 • Letter: A
Question
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
Work is done aganist any resisting force
Since it is free to expand, then there is no work done
Also it is given well insulated therefore Q = 0
From first law
dU = 0
U2 = U1
Therefore internal energy remains constant
a) The diff is 0
b)
since it is ideal gas, internal energy depends on only temprature
U2 = U1 implies
T2 = T1
Therefore
P1*V1 = P2*V2
100*4 = P2*8
P2 = 50 psia
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