Chapter 16, Problem 73P

(a)

To determine

The amount of height will rise the piston when the temperature is raised to $250°\text{C}$.

Answer to Problem 73P

The amount of height will rise the piston when the temperature is raised to $250°\text{C}$ is $\underset{\_}{0.169\text{m}}$.

Explanation of Solution

Consider the Figure shown below.

Write the ideal gas equation for the initial and final state of the cylinder when the amount of gas remains constant.

    $\frac{P_0{V}_0}{T_0}=\frac{P^{\prime }{V}^{\prime }}{T^{\prime }}$        (I)

Here, $P_0$ is the initial pressure, $V_0$ is the initial volume, $T_0$ is the initial temperature, $P^{\prime }$ is the final pressure, $V^{\prime }$ is the final volume, and $T^{\prime }$ is the final temperature.

When the piston compresses the spring by a height $h$, then the spring force is increases by $F=kh$. So the external pressure on the piston is $\frac{kh}{A}$. That is the final pressure is $P^{\prime }=P_0+\frac{kh}{A}$. And the volume will increases by $Ah$ when the piston rises. That is the final volume is $V^{\prime }=V_0+Ah$.

Use $V_0+Ah$ for $V^{\prime }$, and $\frac{kh}{A}$ for $P^{\prime }$ in the equation (I), and solve.

    $\begin{array}{l}\frac{P_0{V}_0}{T_0}=\frac{\left(P_0+\frac{kh}{A}\right)\left(V_0+Ah\right)}{T^{\prime }}\\ \left(P_0+\frac{kh}{A}\right)\left(V_0+Ah\right)=P_0{V}_0\left(\frac{T^{\prime }}{T_0}\right)\end{array}$        (II)

Conclusion:

Substitute, $1.013\times {10}^5\text{N}/{\text{m}}^2$ for $P_0$, $2.00\times {10}^5\text{N}/{\text{m}}^{3}h$ for $\frac{kh}{A}$, $5.00\times {10}^{-3}{\text{m}}^3$ for $V_0$, $\left(0.010{\text{m}}^2\right)h$ for $Ah$, $523\text{K}$ for $T^{\prime }$, and $293\text{K}$ for $T_0$ in the equation (II), and solve for $h$.

    $\begin{array}{l}\left(1.013\times {10}^5\text{N}/{\text{m}}^2+2.00\times {10}^5\text{N}/{\text{m}}^{3}h\right)\left(5.00\times {10}^{-3}{\text{m}}^3+0.010{\text{m}}^{2}h\right)=\\ \left(1.013\times {10}^5\text{N}/{\text{m}}^2\right)\left(5.00\times {10}^{-3}{\text{m}}^3\right)\left(\frac{523\text{K}}{298\text{K}}\right)\end{array}$

Solve the above equation for positive root only.

    $\begin{array}{c}2000h^2+2013h-397=0\\ h=\frac{-2013+2689}{4000}\\ =0.169\text{m}\end{array}$

Therefore, the amount of height will rise the piston when the temperature is raised to $250°\text{C}$ is $\underset{\_}{0.169\text{m}}$.

(b)

To determine

The pressure of the gas at $250°\text{C}$.

Answer to Problem 73P

The pressure of the gas at $250°\text{C}$ is $\underset{\_}{1.35\times {10}^5\text{Pa}}$.

Explanation of Solution

Conclusion:

Substitute, $1.013\times {10}^5\text{N}/{\text{m}}^2$ for $P_0$, $2.00\times {10}^3\text{N}/\text{m}$ for $k$, $0.169\text{m}$ for $h$, and $0.010{\text{m}}^2$ for $A$ in the equation $P^{\prime }=P_0+\frac{kh}{A}$.

    $\begin{array}{c}{P}^{\prime }=1.013\times {10}^5\text{N}/{\text{m}}^2+\frac{\left(2.00\times {10}^3\text{N}/\text{m}\right)\left(0.169\text{m}\right)}{0.010{\text{m}}^2}\\ =1.35\times {10}^5\text{Pa}\end{array}$

Therefore, the pressure of the gas at $250°\text{C}$ is $\underset{\_}{1.35\times {10}^5\text{Pa}}$.