A vertical cylindrical tank contains 1.80 mol of an ideal gas under a pressure of 0.300 atm at $\mathbf{20}\mathbf{°}\mathit{C}$. The round part of the tank has a radius of 10 cm , and the gas is supporting a piston that can move up and down in the cylinder without friction. There is a vacuum above the piston. (a) What is the mass of this piston? (b) How tall is the column of gas that is supporting the piston?

The term friction may be defined as the force that resists the sliding of one body over another.

Consider the given data as below.

The volume, $V=0.144m^3$

The pressure, $p=0.300atm=0.3039\times {10}^{5}Pa$

Number of moles, $n=1.80mol$

Ideal gas constant, $R=8.314J/mol.K$

Temperature, $T=293K$

The radius, $r=10cm=0.1m$

Using the volume relation of height

$$\begin{gathered} V={\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{h}=\frac{\mathrm{V}}{{\mathrm{\pi r}}^2} \end{gathered}$$

Here, V is the volume, r is the radius, and h is the height.

And the ideal gas 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.

Rearrange the above equation for volume and substitute known values.

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}} \\ =\frac{1.80\times 8.314\times 293}{0.3039\times {10}^5} \\ =0.1442{\mathrm{m}}^3 \end{gathered}$$

Now, the pressure is define by the following formula.

$$\begin{gathered} p=\rho gh \\ =\frac{mg}{V}\times \frac{V}{{\mathrm{\pi r}}^2} \\ =\frac{mg}{{\mathrm{\pi r}}^2} \\ \mathrm{m}=\frac{{\mathrm{p\pi r}}^2}{\mathrm{g}} \end{gathered}$$

Substitute known values in the above formula.

$$\begin{gathered} m=\frac{0.3039\times {10}^5\times 3.14\times {\left(0.1\right)}^2}{9.8} \\ =97.37kg \end{gathered}$$

Hence, the mass of this piston is 97.37 kg

Now the height calculated as

$$\begin{gathered} h=\frac{V}{{\mathrm{\pi r}}^2} \\ =\frac{0.1442}{3.14\times {\left(0.1\right)}^2} \\ =4.59m \end{gathered}$$

Hence, 4.59 m tall is the column of gas that is supporting the piston.