Question

The $50kg$ circular piston shown in FIGURE P18.58 floats on $0.12mol$of compressed air.

a. What is the piston height $h$if the temperature is $30°C$?

b. How far does the piston move if the temperature is increased by $100°C$?

Short answer

a. The height of the piston is $\mathbf{24}\mathbf{}\mathit{c}\mathit{m}$

b. The piston moves$\mathbf{7}\mathbf{}\mathit{c}\mathit{m}$

Step-by-step solution

Given Parameters 

Given information are;

Weight of the cylinder : $50kg$

Circular piston floats on :$0.12mol$of compressed air

Used Theory :

Theideal gas law is given by

$PV=nRT$

Where$R=8.31J/mol.K$ is the universal gas constant.

Piston pressure is given by :

$p=p_{atm}+\frac{mg}{A}$

Finding the piston height 

(a) The gas takes the shape of a cylinder, that is, its area is given by

$$\begin{gathered} A=\pi \left(\frac{d^2}{4}\right) \\ =\pi {\left(\frac{0.10}{2}\right)}^2=7.85x{10}^{-3}m² \end{gathered}$$

Now get the gas pressure as follows:

$$\begin{gathered} p=p_{atm}+\frac{mg}{A} \\ =1.013x{10}^{5}Pa+\frac{(50kg)(9.8m/s²)}{7.85x{10}^{-3}m²} \\ =1.63x{10}^{5}Pa \end{gathered}$$

Now, substitute the values of$n,R,T,\mathrm{and}p$to get :

$$\begin{gathered} V=\frac{nRT}{P} \\ =\frac{(0.12mol)(8.31J/molK)(303K)}{(1.63x{10}^{5}Pa)} \\ =1.85x{10}^{-3}m³ \end{gathered}$$

The volume of gas is $V=Ah$. Where $h$is the height side of the piston. Use the volume value of the cylinder to get the height .

$$\begin{gathered} h=\frac{V}{A} \\ =\frac{1.85x{10}^{-3}{m}^3}{7.85x{10}^{-3}{m}^2} \\ =0.24m \\ =\mathbf{24}\mathbf{}\mathit{c}\mathit{m} \end{gathered}$$

Determining of piston's movement if the temperature is increased

(b) Use equation $V=\frac{nRT}{P}$to get the relationship between altitude and temperature.

$$\begin{gathered} V=\frac{nRT}{P} \\ \Rightarrow hA=\frac{nRT}{P} \\ \Rightarrow h=\frac{nRT}{AP} \end{gathered}$$

The final temperature of Kelvin is $T₂=130°C+273=403K$.

Now, substitute the values of $n,R,T_2,A,\mathrm{and}p$

$$\begin{gathered} h_2=\frac{nR{T}_2}{Ap} \\ =\frac{(0.12mol)(8.31J/molK)(403K)}{(7.85x{10}^{-3}m²)(1.63x{10}^{5}Pa)} \\ =0.31m \\ =31cm \end{gathered}$$

The distance traveled by the piston is the difference between the initial height and the final height.

$$\begin{gathered} d=h₂-h₁ \\ =31cm-24cm \\ =\mathbf{7}\mathit{c}\mathit{m} \end{gathered}$$