Chapter 6, Problem 6.119QE

Interpretation Introduction

Interpretation:

The pressure change in a tire inflated to $32\text{ psi}$ at $90°\text{F}$ when the temperature declines to $\text{32 }°\text{F}$ has to be calculated.

Concept Introduction:

For two gases that differs in the pressure, volume and temperature the combined gas law expression is given as follows:

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

Here,

$P_1$ denotes the pressure of gas 1.

$V_1$ denotes volume of gas 1.

$T_1$ denotes temperature of gas 1.

$P_2$ denotes the pressure of gas 2.

$V_2$ denotes volume of gas 2.

$T_2$ denotes temperature of gas 2.

Pounds per square inch or $\text{psi}$ is the pressure relative to the pressure in vacuum, hence any pressure measurement done at sea level in air must have $\text{15 psi}$ added to it.

Answer to Problem 6.119QE

The pressure change in a tire inflated to $32\text{ psi}$ at $90°\text{F}$ when the temperature declines to $\text{32 }°\text{F}$ is $5\mathrm{psi}$.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(°\text{C}\right)=\frac{5}{9}\left[T\left(°\text{F}\right)-32\right]$        (1)

Here,

$T\left(°\text{F}\right)$ denotes the temperature in Fahrenheit.

$T\left(°\text{C}\right)$ denotes the temperature in Celsius.

Substitute $90°\text{F}$ for $T\left(°\text{F}\right)$ in equation (1).

  $\begin{array}{c}T\left(°\text{C}\right)=\frac{5}{9}\left[90°\text{F}-32\right]\\ =32.22°\text{C}\end{array}$

Substitute $\text{32 }°\text{F}$ for $T\left(°\text{F}\right)$ in equation (1).

  $\begin{array}{c}T\left(°\text{C}\right)=\frac{5}{9}\left[\text{32 }°\text{F}-32\right]\\ =0°\text{C}\end{array}$

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (2)

Here,

$T\left(\text{K}\right)$ denotes the temperature in kelvins.

$T\left(°\text{C}\right)$ denotes the temperature in Celsius.

Substitute $32.22°\text{C}$ for $T\left(°\text{C}\right)$ in equation (2).

  $\begin{array}{c}T\left(\text{K}\right)=32.22°\text{C}+273\text{ K}\\ =305.22\text{ K}\end{array}$

Substitute $\text{0 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (2).

  $\begin{array}{c}T\left(\text{K}\right)=0°\text{C}+273\text{ K}\\ =273\text{ K}\end{array}$

The conversion factor to convert pounds per square inch to $\text{atm}$ is as follows:

  $1\text{ atm}=\text{15 psi}$

Hence $32\text{ psi}$ is the pressure inflated and any pressure measurement done at sea level in air must have $\text{15 psi}$ added to it therefore net pressure in the tire initially is as follows:

  $\begin{array}{c}{P}_{\text{intial}}=32\text{ psi}+\text{15 psi}\\ =47\text{ psi}\end{array}$

If the volume of inflation is same then combined gas law can also be written as follows:

  $\frac{P_{\text{intial}}}{T_{\mathrm{i}\text{ntial}}}=\frac{P_{\text{final}}}{T_{\text{final}}}$        (3)

Here,

$P_{\text{intial}}$ denotes the initial pressure.

$P_{\text{final}}$ denotes the final pressure.

$T_{\mathrm{i}\text{ntial}}$ denotes the initial temperature.

$T_{\text{final}}$ denotes the final temperature.

Substitute $47\text{ psi}$ for $P_{\text{intial}}$, $305.22\text{ K}$ for $T_{\mathrm{i}\text{ntial}}$ and $273\text{ K}$ for $T_{\text{final}}$ in equation (3).

  $\frac{47\text{ psi}}{305.22\text{ K}}=\frac{P_{\text{final}}}{273\text{ K}}$

Rearrange the expression to calculate the value of $P_{\text{final}}$.

  $\begin{array}{c}{P}_{\text{final}}=\left(273\text{ K}\right)\left(\frac{47\text{ psi}}{305.22\text{ K}}\right)\\ =42.0385\text{ psi}\end{array}$

The expression to calculate the change in pressure is given as follows:

  $\Delta P=P_{\text{intial}}-P_{\text{final}}$        (4)

Substitute $47\text{ psi}$ for $P_{\text{intial}}$ and $42.0385\text{ psi}$ for $P_{\text{final}}$ in equation (4).

  $\begin{array}{c}\Delta P=P_{\text{intial}}-P_{\text{final}}\\ =47\text{ psi}-42.0385\text{ psi}\\ =4.9615\\ \approx 5\mathrm{psi}\end{array}$