Chapter 17, Problem 80P

(a)

To determine

To Find: The height of the gas column.

Explanation of Solution

Given:

Number of moles of gas, $n=0.10\text{mol}$

At STP, volume is $V=2.24\text{L}$

Mass of the piston, $m=\text{1}\text{.4 kg}$

Formula Used:

Ideal gas law:

  $PV=nRT$

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

Calculations:

Pressure inside the cylinder:

  $\begin{array}{l}{P}_{in}=P_{atm}+\frac{mg}{A}\\ P_{in}=\frac{nRT}{V}\\ V=hA\\ \frac{nRT}{hA}=P_{atm}+\frac{mg}{A}\\ h=\frac{nRT}{A{P}_{atm}+mg}\\ h=\frac{\left(P_{atm}V\right)}{A{P}_{atm}+mg}\end{array}$

Substitute the values:

  $\begin{array}{l}h=\frac{\left(P_{atm}V\right)}{A{P}_{atm}+mg}\\ V=\left(\text{2}\text{.4}\text{m}\right)\left(A\right)\\ 2.24\times {10}^{-3}{\text{m}}^{\text{3}}=\left(\text{2}\text{.4}\text{m}\right)\left(A\right)\\ A=9.33\times {10}^{-4}{\text{m}}^{\text{2}}\\ h=\frac{\text{2}\text{.4}\text{m}}{\text{1+}\frac{\text{1}\text{.4kg×9}{\text{.81m/s}}^{\text{2}}}{\text{9}{\text{.33×10}}^{\text{-4}}{\text{m}}^{\text{2}}\text{×1}{\text{.01×10}}^{\text{5}}\text{Pa}}}\\ h=2.1\text{m}\end{array}$

Conclusion:

Thus, the height of the gas column is 2.1 m.

(b)

To determine

To Find:The frequency of vibration of piston.

Explanation of Solution

Given:

Number of moles of gas, $n=0.10\text{mol}$

At STP, volume is $V=2.24\text{L}$

Mass of the piston, $m=\text{1}\text{.4 kg}$

Formula Used:

Frequency of vibration of piston can be obtained by:

  $f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Here, m is the mass and kis the stiffness constant.

Calculations

In equilibrium position:

  $P_{in}A-mg-P_{atm}A=0$

Let the displacement from the equilibrium position be x .

  $P_{in}\text{'}A-mg-P_{atm}A=m{a}_x$

  $P_{in}\text{'}A-P_{in}A=m{a}_x$

From ideal gas law:

  $\begin{array}{l}{P}_{in}\text{'}V\text{'}=P_{in}V\\ P_{in}\text{'}\left(V+Ax\right)=P_{in}V\\ P_{in}\text{'}=P_{in}\frac{V}{\left(V+Ax\right)}\\ P_{in}\text{'}=P_{in}\frac{Ah}{\left(Ah+Ax\right)}\\ P_{in}\text{'}=P_{in}A\left(\frac{1}{1+\frac{x}{h}}\right)\\ P_{in}A{\left(1+\frac{x}{h}\right)}^{-1}-P_{in}A=m{a}_x\\ x<<h\\ P_{in}A\left(1-\frac{x}{h}\right)-P_{in}A=m{a}_x\\ -\frac{nRT}{Ah}\left(\frac{Ax}{h}\right)=m{a}_x\\ a_x=-\frac{nRT}{m{h}^2}x\\ \end{array}$

Condition of S.H.M:

  $\begin{array}{l}{a}_x=-\frac{k}{m}x\\ \frac{k}{m}=\frac{nRT}{m{h}^2}\\ f=\frac{1}{2\pi }\sqrt{\frac{nRT}{m{h}^2}}\end{array}$

Substitute the values and solve:

  $\begin{array}{l}f=\frac{1}{2\pi }\sqrt{\frac{0.1\text{mol}\times 8.314\text{J/mol×K}\times 300\text{K}}{1.\text{4}\text{kg}\times {\left(\text{2}\text{.1}\text{m}\right)}^2}}\\ f=1.\text{0}\text{Hz}\end{array}$

Conclusion:

Thus, the frequency of vibration of piston is 1.0 Hz.