Chapter 16, Problem 74P

A cylinder that has a 40.0-cm radius and is 50.0 cm deep is filled with air at 20.0°C and 1.00 atm (Fig. P10.74a). A 20.0-kg piston is now lowered into the cylinder, compressing the air trapped inside as it takes equilibrium height h_i (Fig. P16.74b). Finally, a 25.0-kg dog stands on the piston, further compressing the air, which remains at 20°C (Fig. P16.74c). (a) How far down (Δh) does the piston move when the dog steps onto it? (b) To what temperature should the gas be warmed to raise the piston and dog back to h_i?

To determine

The height which the piston moves down when the dog steps onto the piston.

Answer to Problem 74P

The height at which the piston move down when the dog steps onto the piston is $\underset{\_}{2.38\text{mm}}$.

Explanation of Solution

Write the expression for Boyle’s law, for the initial and final state since the temperature is constant.

    $PV=P_0{V}_0$        (I)

Here, $P$ is the pressure after the compression, $P_0$ is the initial pressure, $V$ is the volume after the compression, and $V_0$ is the initial volume.

Use $A{h}_i$ for $V$, and $A{h}_0$ for $V_0$ in the equation (I), and solve for $P$.

    $\begin{array}{c}PA{h}_i=P_{0}A{h}_0\\ P=P_0\left(\frac{h_0}{h_i}\right)\end{array}$        (II)

Here, $h_i$ is the height after the first compression, and $h_0$ is the initial height.

When the piston moves down, then the external pressure on the piston is $\frac{m_{\text{p}}g}{A}$. That is the final pressure is $P=P_0+\frac{m_{\text{p}}g}{A}$.

Equate the equation (II) with the equation $P=P_0+\frac{m_{\text{p}}g}{A}$.

    $\begin{array}{c}{P}_0+\frac{m_{\text{p}}g}{A}=P_0\left(\frac{h_0}{h_i}\right)\\ P_0\left(1+\frac{m_{\text{p}}g}{P_{0}A}\right)=P_0\left(\frac{h_0}{h_i}\right)\\ h_i=\frac{h_0}{1+\frac{m_{\text{p}}g}{P_{0}A}}\end{array}$        (III)

Replace the mass $m_{\text{p}}$ with total mass $m_{\text{p}}+M$ in the equation (III), to find final height of the piston.

    $h^{\prime }=\frac{h_0}{1+\frac{\left(m_{\text{p}}+M\right)g}{P_{0}A}}$        (IV)

Here, $M$ is the mass of the dog.

The piston moves downward when the dog steps on the piston. It can be expressed as,

    $\Delta h=h_i-h^{\prime }$        (V)

Use equation (III) and (IV) in (V).

    $\Delta h=\frac{h_0}{1+\frac{m_{\text{p}}g}{P_{0}A}}-\frac{h_0}{1+\frac{\left(m_{\text{p}}+M\right)g}{P_{0}A}}$        (VI)

Conclusion:

Substitute, $50.0\text{m}$ for $h_0$, $20.0\text{kg}$ for $m_{\text{p}}$, $9.80\text{m}/{\text{s}}^2$ for $g$, $1.013\times {10}^5\text{Pa}$ for $P_0$, $\pi {\left(0.400\text{m}\right)}^2$ for $A$, and $25.0\text{kg}$ for $M$ in the equation (VI), to find $\Delta h$.

    $\begin{array}{c}\Delta h=\frac{50.0\text{m}}{1+\frac{\left(20.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)}{\left(1.013\times {10}^5\text{Pa}\right)\pi {\left(0.400\text{m}\right)}^2}}-\frac{50.0\text{m}}{1+\frac{\left(20.0\text{kg}+25.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)}{\left(1.013\times {10}^5\text{Pa}\right)\pi {\left(0.400\text{m}\right)}^2}}\\ =2.38\text{mm}\end{array}$

Therefore, the height at which the piston move down when the dog steps onto the piston is $\underset{\_}{2.38\text{mm}}$.

To determine

The temperature at which the gas should be warmed to raise the piston and dog back to $h_i$.

Answer to Problem 74P

The temperature at which the gas should be warmed to raise the piston and dog back to $h_i$ is $\underset{\_}{21.4°\text{C}}$.

Explanation of Solution

Write the expression for Charles’s law, for the initial and final state since the pressure is constant.

    $\frac{V}{T}=\frac{V^{\prime }}{T_i}$        (VII)

Here, $T$ is the temperature at which the gas should be warmed to raise the piston and dog back to $h_i$, $T_i$ is the temperature after the first compression.

Use $A{h}_i$ for $V$, and $A{h}^{\prime }$ for $V^{\prime }$ in the equation (VII), and solve for $T$.

    $\begin{array}{c}\frac{A{h}_i}{T}=\frac{A{h}^{\prime }}{T_i}\\ T=T_i\left(\frac{h_i}{h^{\prime }}\right)\end{array}$        (VIII)

Use equation (III) and (IV) in the equation (VIII).

    $\begin{array}{c}T=T_i\left(\frac{\frac{h_0}{1+\frac{m_{\text{p}}g}{P_{0}A}}}{\frac{h_0}{1+\frac{\left(m_{\text{p}}+M\right)g}{P_{0}A}}}\right)\\ =T_i\left(\frac{1+\frac{\left(m_{\text{p}}+M\right)g}{P_{0}A}}{1+\frac{m_{\text{p}}g}{P_{0}A}}\right)\end{array}$        (IX)

Conclusion:

Substitute, $293\text{K}$ for $T_i$, $50.0\text{m}$ for $h_0$, $20.0\text{kg}$ for $m_{\text{p}}$, $9.80\text{m}/{\text{s}}^2$ for $g$, $1.013\times {10}^5\text{Pa}$ for $P_0$, $\pi {\left(0.400\text{m}\right)}^2$ for $A$, and $25.0\text{kg}$ for $M$ in the equation (IX), to find $T$.

    $\begin{array}{c}T=293\text{K}\left[\frac{1+\frac{\left(20.0\text{kg}+25.0\text{kg}\right)9.80\text{m}/{\text{s}}^2}{\left(1.013\times {10}^5\text{Pa}\right)\pi {\left(0.400\text{m}\right)}^2}}{1+\frac{20.0\text{kg}\times 9.80\text{m}/{\text{s}}^2}{\left(1.013\times {10}^5\text{Pa}\right)\pi {\left(0.400\text{m}\right)}^2}}\right]\\ =294.4\text{K}\\ \text{=}294.4\text{K}-273\\ =21.4°\text{C}\end{array}$

Therefore, The temperature at which the gas should be warmed to raise the piston and dog back to $h_i$ is $\underset{\_}{21.4°\text{C}}$.