Chapter 19, Problem 76CP

(a)

To determine

The distance, the piston will move down when dog steps onto it.

Answer to Problem 76CP

The distance, the piston will move down when dog steps onto it is $\underset{\_}{2.38\text{mm}}$.

Explanation of Solution

Write the equation for the distance that will move down.

  $\left(\Delta h\right)=h_i-h^{\prime }$                                                                                                 (I)

Here, $\left(\Delta h\right)$ is the difference in height, $h_i$ is the equilibrium height and $h^{\prime }$ is the height when the dog on the piston.

Write the expression of ideal gas law,

  $PV=P_0{V}_0$                                                                                                   (II)

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

Write the expression for the final pressure,

  $P=P_0+\frac{m_{p}g}{A}$                                                                                           (III)

Here, $m_p$ is the mass of the piston, $g$ is the gravitational acceleration and $A$ is the area.

Write the expression for the initial volume.

  $V_0=A{h}_0$                                                                                                  (IV)

Here, $h_0$ is the height without piston and dog.

Write the expression for the final volume.

  $V=A{h}_i$                                                                                                 (V)

Write the expression for the equilibrium height by using (II), (III), (IV) and (V).

  $h_i=\frac{h_0}{1+m_{p}g/P_{0}A}$                                                                                 (VI)

Write the expression for the height with the dog mass on the piston by using (II), (III), (IV) and (V).

  $h^{\prime }=\frac{h_0}{1+\left(m_p+M\right)g/P_{0}A}$                                                                                (VII)

Here, $M$ is the mass of dog.

Write the expression for the cross-sectional area.

  $A=\pi r^2$                                                                                                         (VIII)

Here, $r$ is the radius.

Rewrite the expression for the distance that will move down from equation (I) by using (VI) (VII) and (VIII).

  $\left(\Delta h\right)=\left(\frac{h_0}{1+m_{p}g/P_0\pi r^2}\right)-\left(\frac{h_0}{1+\left(m_p+M\right)g/P_0\pi r^2}\right)$                                      (IX)

Conclusion:

Substitute $50.0\text{cm}$ for $h_0$, $20.0\text{kg}$ for $m_p$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $1\text{atm}$ for $P_0$, $40.0\text{cm}$ for $r$, $25.0\text{kg}$ for $M$ in Equation (IX) to find $\left(\Delta h\right)$.

  $\begin{array}{c}\left(\Delta h\right)=\left(\frac{\left\{\left(50.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}}{1+\left(20.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)/\left[\left(1\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right]\pi {\left[\left(40.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\right)-\\ \left(\frac{\left\{\left(50.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}}{1+\left(20.0\text{kg}+25.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)/\left[\left(1\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right]\pi {\left[\left(40.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\right)\\ =\left(2.38\times {10}^{-3}\text{m}\right)\left(\frac{1\times {10}^3\text{mm}}{1\text{m}}\right)\\ =2.38\text{mm}\end{array}$

Thus, the distance, the piston will move down when dog steps onto it is $\underset{\_}{2.38\text{mm}}$.

(b)

To determine

The required temperature to raise the piston.

Answer to Problem 76CP

The required temperature to raise the piston is $\underset{\_}{21.4\text{°C}}$.

Explanation of Solution

Write the equation from the ideal gas law.

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

Here, $V^{\prime }$ is the final volume and $T^{\prime }$ is the temperature.

Rewrite the equation for the final time from equation (X) by using (IV), (V), (VI), (VII) and (VIII).

  $T=T_i\left[\frac{1+\left(m_p+M\right)g/P_0\pi r^2}{1+m_{p}g/P_0\pi r^2}\right]$                          (XI)

Conclusion:

Substitute, $20.0\text{°C}$ for $T_i$, $20.0\text{kg}$ for $m_p$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $1\text{atm}$ for $P_0$, $40.0\text{cm}$ for $r$, $25.0\text{kg}$ for $M$ in Equation (XI) to find $T$.

  $\begin{array}{c}T=\left(20.0\text{°C}+273\right)\\ \left[\frac{1+\left(20.0\text{kg}+25.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)/\left[\left(1\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right]\pi {\left[\left(40.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}{1+\left(20.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)/\left[\left(1\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\right]\pi {\left[\left(40.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\right]\\ =\left(294.4-273\right)°\text{C}\\ =21.4\text{°C}\end{array}$

Thus, the required temperature to raise the piston is $\underset{\_}{21.4\text{°C}}$.