91Ó°ÊÓ

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

$p=\frac{mg}{\mathrm{Ï„}\mathrm{Ï„}{r}^2}$

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 $\mathrm{Ï„}{\mathrm{Ï„}}^2{h}_1$. Therefore,

$p_1=\frac{m_{1}g}{\mathrm{Ï„}\mathrm{Ï„}{r}^2}$

For second case when the mass is 9 kg .

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

$p_2=\frac{m_{2}g}{\mathrm{Ï„}\mathrm{Ï„}{r}^2}$

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

$$\begin{gathered} p_1{V}_1=p_2{V}_2 \\ \frac{m_{1}g}{{\mathrm{Ï€°ù}}^2}×{\mathrm{Ï€°ù}}^2{\mathrm{h}}_1=\frac{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{g}}{{\mathrm{Ï€°ù}}^2}×{\mathrm{Ï€°ù}}^2{\mathrm{h}}_2 \\ {\mathrm{m}}_1{\mathrm{h}}_1=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{h}}_2 \\ {\mathrm{h}}_2=\frac{{\mathrm{m}}_1{\mathrm{h}}_1}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)} \end{gathered}$$

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

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

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