A container with helium is sealed by a movable heavy piston. Initially, the pist

Question

A container with helium is sealed by a movable heavy
piston. Initially, the piston is in equilibrium.
The piston is then slowly lifted a distance L and held until
the thermal equilibrium with the surroundings is reestablished.
After that, the container is thermally insulated
from the surroundings and piston is released. Eventually,
it comes to rest. What is the final position of the piston
with respect to its initial position? Neglect the outside air
pressure on the piston and the heat capacities of the piston
and the container.

I think it should begin with a energy balance negleting KE, U, and Q but where from there im not sure.

Explanation / Answer

Since the text specifies that first the piston moves and "the thermal equilibrium with surroundings is reestablished" means that the first transformation is AT THE SAME TEMPERATURE. (The exterior is an infinite energy reservoir, which means it keeps the same temperature).

P0*V0 =P1*V1

If L0 is the initial position we have
P0*L0*S =P1*(L0+L)*S
(L0+L)/L0 =P0/P1

or equivalent
(P1/P0) =L0/(L0+L)

Now the countainer is insulated. It means the second transformation is adiabatic. Since at the end all the weight of the piston will again push on the gas (as initially) the final pressure will be the same as the initial pressure P0. Final state is denoted by the index 2 below.

P1*V1^gamma = P2*V2^gamma
P1*V1^(3/5) = P0*V2^(3/5)
gas is helium (ideal monoatomic) it means gamma = 5/3

P1*L^(5/3) =P0*L2^(5/3)
L2 =L*(P1/P0)^(3/5)

L2 = L* [L0/(L0+L)]^3/5

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