In an evacuated enclosure, a vertical cylindrical tank of diameter D is sealed by a 3.00-kg circular disk that can move up and down without friction. Beneath the disk is a quantity of ideal gas at temperature T in the cylinder (Figure 1) . Initially the disk is at rest at a distance of 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, the disk moves downward.
Disk Ideal gas
Q: If the temperature of the gas is kept constant and no gas escapes from the tank, what distance above the bottom of the tank is the disk when it again comes to rest?

Question

19-12-2022
Physics

Answered

In an evacuated enclosure, a vertical cylindrical tank of diameter D is sealed by a 3.00-kg circular disk that can move up and down without friction. Beneath the disk is a quantity of ideal gas at temperature T in the cylinder (Figure 1) . Initially the disk is at rest at a distance of 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, the disk moves downward.
Disk Ideal gas
Q: If the temperature of the gas is kept constant and no gas escapes from the tank, what distance above the bottom of the tank is the disk when it again comes to rest?

Answer :

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Qwseed Qwseed · Answer 1 · 2022-12-24T01:33:43-05:00

To solve this problem, we can apply the principle of conservation of energy to the system consisting of the lead brick, the disk, and the gas.

The initial potential energy of the lead brick is given by its mass multiplied by the acceleration due to gravity multiplied by its initial height above the bottom of the tank: PE_i = mgh_i = (9.00 kg)(9.81 m/s^2)(4.00 m) = 356.74 J.

The final potential energy of the lead brick is given by its mass multiplied by the acceleration due to gravity multiplied by its final height above the bottom of the tank: PE_f = mgh_f.

The initial kinetic energy of the disk is zero, since it is initially at rest. The final kinetic energy of the disk is given by its mass multiplied by the square of its final velocity: KE_f = (1/2)mv^2_f.

The change in the internal energy of the gas is zero, since the temperature of the gas is kept constant and no gas escapes from the tank.

Therefore, the change in potential energy of the lead brick is equal to the change in kinetic energy of the disk:

PE_i - PE_f = KE_f

mgh_i - mgh_f = (1/2)mv^2_f

Solving for the final height of the disk, h_f, we find:

h_f = h_i - (2/m)KE_f

= 4.00 m - (2/9.00 kg)(1/2)(3.00 kg)v^2_f

Since the lead brick is gently placed on the disk, the final velocity of the disk, v_f, will be small. Therefore, we can neglect the kinetic energy of the disk compared to the potential energy of the lead brick, and approximate h_f as:

h_f = h_i - (2/m)KE_f

≈ h_i

= 4.00 m

Therefore, the distance above the bottom of the tank that the disk is when it again comes to rest is approximately 4.00 m.

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