In an evacuated enclosure, a vertical cylindrical tank of diameter D is sealed by a 3.00kg 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. Initially the disk is at rest at a distance 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. 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?

A gas that strictly abides by the ideal-gas law: a gas in which there is typically no molecular attraction.

The term ideal gas is law which is type of equation.

pV = nRT

Here, p is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.

mmThe mass effect on the gas inside the cylinder by a pressure is,

$p=\frac{F}{A}$

Or

$p=\frac{mg}{\tau \tau r^2}$

Here, F is the force, A is the area, m is the mass, g is the acceleration due to gravity, and r is the radius.

For first case when the mass is 3 kg .

Let $m_1$is 3kg and volume $V_1$ is $\tau {\tau }^2{h}_1$. Therefore,

$p_1=\frac{m_{1}g}{\tau \tau r^2}$

For second case when the mass is 9 kg .

Let $m_2$is 9 kg and volume $V_2$ is $\tau \tau r^2{h}_2$.

$p_2=\frac{m_{2}g}{\tau \tau r^2}$

Using ideal gas equation, in equation n , R , T are constant. Therefore,

$$\begin{gathered} p_1{V}_1=p_2{V}_2 \\ \frac{m_{1}g}{{\mathrm{\pi r}}^2}\times {\mathrm{\pi r}}^2{\mathrm{h}}_1=\frac{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{g}}{{\mathrm{\pi r}}^2}\times {\mathrm{\pi r}}^2{\mathrm{h}}_2 \\ {\mathrm{m}}_1{\mathrm{h}}_1=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{h}}_2 \\ {\mathrm{h}}_2=\frac{{\mathrm{m}}_1{\mathrm{h}}_1}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)} \end{gathered}$$

From there put given values and calculate $h_2$ as below.

$$\begin{gathered} h_2=\frac{3kg\times 4m}{\left(3+9\right)kg} \\ =1m \end{gathered}$$

Hence, the distance above the bottom of the tank is 1m .