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## Homework Statement

The figure at the bottom shows a cylindrical tank of diameter D with a moveable 3.00-kg circular disk sitting on top. The disk seals in the gas inside but is able to move without friction. The gas inside is at temperature T. The height of the disk is initially at h = 4.00 m above the bottom of the tank. When a lead brick of mass 9.00 kg is gently placed on the disk, it moves downward, compressing the gas and decreasing the volume. Assuming the temperature of the gas is held constant, at what distance from the bottom of the will the disk come to rest?

mass_{disk}=3.00 kg

mass_{Pb}=9.00 kg

m_{T}=12.00kg

h_i=4.00m

Isothermal \therefore T_i=T_f

V_i > V_f ; P_i<P_f

## Homework Equations

\P = \rho*\g*\h

P_1*V_1*T_2=P_2*V_2*T_1

W=nRT*ln(\frac{V_f}{V_i})=m*g*d

## The Attempt at a Solution

\[/B]delta P = -\rho*g*\int_{y}^{4} = -0.5*\rho*g*(y^2-16)

F_{disk+Pb}=(m_{d}+m{Pb})/g=127.53N

P_1*V_1=P_2*V_2

P_1*A_1*H_1=P_2*A_2*H_2 (A_1=A_2)

H_2=4*\frac{P_1}{P_2}

I'm not really sure where to proceed here since the initial pressure nor the final pressure nor any of the volumes are given so I'm generally extremely confused on how to proceed to find the final distance the piston would rest at. I know since that by Boyles law as Volume decreases, Pressure increases at constant Temperature. I'm just not sure how we factor any of that in without being given volume or area.