Consider a rigid tank with a partition such that there are two equal volumes whi

Question

Consider a rigid tank with a partition such that there are two equal volumes which are known. On one side has water at a given temperature and pressure and known mass. The other side is a vacuum. If the partition is removed, and the tank is perfectly insulated, can the final equilibrium state be determined? Hints: How many independent properties are needed to fix a state? What is the first law for the process described above? Remember we are only interested in the initial and final equilibrium states, not the violent activity that takes place when the partition is removed. Explain you ranswer.

Explanation / Answer

Tank 1

P1 = P

T1 = T

m1 = m

Tank 2

P2 = 0

V2 = V1 = V/2

When the partition is removed

Number of moles in the tank remain same

P1*V1/T1 + P2*V2/T2 = P*V/T

But T = T1

P1/2*T1 = P/T

P = P1/2

Two Independent properties are required to define a state

Perfectly insulated gives dQ = 0

No work is done W = 0

Change in internal energy,dU = 0

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