Chapter 17, Problem 54P

(a)

To determine

The final gauge pressure of the tire.

Explanation of Solution

Given:

Initial gauge pressure is $200\text{kPa}$ .

Initial temperature of tire is $20°\text{C}$ .

Final temperature of tire is $20°\text{C}$ .

The atmospheric pressure is $101\text{kPa}$ .

Formula used:

Write the expression for ideal-gas law.

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Rearrange the above expression in term of $P_2$ and simplify.

  $P_2=\frac{V_1{T}_2}{V_2{T}_1}{P}_1$ ........ (1)

Here, $P_1$ is initial pressure, $V_1$ is initial volume, $T_1$ is initial temperature, $P_2$ is final pressure, $V_2$ is final volume and $T_2$ is final temperature of gas.

Write the expression for gauge pressure.

  $P_{2,\text{g}}=P_2-P_{\text{atm}}$ ........ (2)

Here, $P_{2,\text{g}}$ is the gauge pressure, $P_{\text{atm}}$ is the atmospheric pressure.

Calculation:

Calculate the initial pressure of gas.

  $\begin{array}{c}{P}_1=200\text{kPa}+101\text{kPa}\\ =301\text{kPa}\end{array}$

Convert the initial temperature from $°\text{C}$ to $\text{K}$ .

  $\begin{array}{c}{T}_1=\left(20+273\right)\text{K}\\ =293\text{K}\end{array}$

Convert the final temperature from $°\text{C}$ to $\text{K}$ .

  $\begin{array}{c}{T}_2=\left(50+273\right)\text{K}\\ =323\text{K}\end{array}$

Substitute $V_1$ for $V_2$

  for $P_1$ , $293\text{K}$ for $T_1$ and $323\text{K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{P}_2=\frac{V_1\left(323\text{K}\right)}{V_1\left(293\text{K}\right)}\left(301\text{kPa}\right)\\ =332\text{kPa}\end{array}$

Substitute $332\text{kPa}$ for $P_2$ and $101\text{kPa}$ for $P_{\text{atm}}$ in equation (1).

  $\begin{array}{c}{P}_{2,\text{g}}=332\text{kPa}-101\text{kPa}\\ =231\text{kPa}\end{array}$

Conclusion:

Thus, the final gauge pressure of tire is $231\text{kPa}$ .

(b)

To determine

The final gauge pressure of the tire.

Explanation of Solution

Given:

Initial gauge pressure is $200\text{kPa}$ .

Initial temperature of tire is $20°\text{C}$ .

Final temperature of tire is $20°\text{C}$ .

The atmospheric pressure is $101\text{kPa}$ .

The increase in volume is $10$ percent.

Formula used:

Write the expression for ideal-gas law.

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

The initial and final volume of gas is equal.

Rearrange the above expression in term of $P_2$ and simplify.

  $P_2=\frac{V_1{T}_2}{V_2{T}_1}{P}_1$ ........ (1)

Here, $P_1$ is initial pressure, $V_1$ is initial volume, $T_1$ is initial temperature, $P_2$ is final pressure, $V_2$ is final volume and $T_2$ is final temperature of gas.

Write the expression for gauge pressure.

  $P_{2,\text{g}}=P_2-P_{\text{atm}}$ ........ (2)

Here, $P_{2,\text{g}}$ is the gauge pressure, $P_{\text{atm}}$ is the atmospheric pressure.

Calculation:

Calculate the initial pressure of gas.

  $\begin{array}{c}{P}_1=200\text{kPa}+101\text{kPa}\\ =301\text{kPa}\end{array}$

Convert the initial temperature from $°\text{C}$ to $\text{K}$ .

  $\begin{array}{c}{T}_1=\left(20+273\right)\text{K}\\ =293\text{K}\end{array}$

Convert the final temperature from $°\text{C}$ to $\text{K}$ .

  $\begin{array}{c}{T}_2=\left(50+273\right)\text{K}\\ =323\text{K}\end{array}$

Calculate the final volume in term of initial volume.

  $\begin{array}{c}{V}_2=V_1+10\%\text{ of}{V}_1\\ =1.1V_1\end{array}$

Substitute $1.1V_1$ for $V_2$

  for $P_1$ , $293\text{K}$ for $T_1$ and $323\text{K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{P}_2=\frac{V_1\left(323\text{K}\right)}{1.1V_1\left(293\text{K}\right)}\left(301\text{kPa}\right)\\ =302\text{kPa}\end{array}$

Substitute $302\text{kPa}$ for $P_2$ and $101\text{kPa}$ for $P_{\text{atm}}$ in equation (1).

  $\begin{array}{c}{P}_{2,\text{g}}=302\text{kPa}-101\text{kPa}\\ =201\text{kPa}\end{array}$

Conclusion:

Thus, the final gauge pressure of tire is $201\text{kPa}$ .