Chapter 16, Problem 25P

(a)

To determine

The tire pressure.

Answer to Problem 25P

The tire pressure is $\underset{\_}{4.00\times {10}^5\text{Pa}}$.

Explanation of Solution

Write the expression for ideal gas equation for the initial condition.

    $P_i{V}_i=nR{T}_i$        (I)

Here, $P_i$ is the initial pressure, $V_i$ is the initial volume, $T_i$ is the initial temperature, $n$ is the number of moles, and $R$ is the universal gas constant.

Write the expression for ideal gas equation for the initial condition.

    $P_f{V}_f=nR{T}_f$        (II)

Here, $P_f$ is the final pressure, $V_f$ is the final volume, and $T_f$ is the final temperature.

Divide equation (II) by (I), and solve for $P_f$.

    $\begin{array}{c}\frac{P_f{V}_f}{P_i{V}_i}=\frac{nR{T}_f}{nR{T}_i}\\ P_f=P_i\left(\frac{V_i}{V_f}\right)\left(\frac{T_f}{T_i}\right)\end{array}$        (III)

Conclusion:

Substitute, $0.280V_i$ for $V_f$, $1\text{atm}$ for $P_i$, $\left(10°\text{C+273}\text{.15}\right)\text{K}$ for $T_i$, and $\left(40°\text{C+273}\text{.15}\right)\text{K}$ for $T_f$ in the equation (III), to find $P_f$.

    $\begin{array}{c}{P}_f=1\text{atm}\left(\frac{V_i}{0.280V_i}\right)\frac{\left(40°\text{C+273}\text{.15}\right)\text{K}}{\left(10°\text{C+273}\text{.15}\right)\text{K}}\\ =\left(\frac{1\text{atm}}{0.280}\right)\frac{313.15\text{K}}{283.15\text{K}}\\ =3.95\text{atm}\\ \text{=}3.95\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\\ =4.00\times {10}^5\text{Pa}\end{array}$

Therefore, the tire pressure is $\underset{\_}{4.00\times {10}^5\text{Pa}}$.

(b)

To determine

New tire pressure of the car.

Answer to Problem 25P

New tire pressure of the car is $\underset{\_}{4.49\times {10}^5\text{Pa}}$.

Explanation of Solution

Write the expression for ideal gas equation for the car after being driven.

    $P_d{V}_d=nR{T}_d$        (IV)

Here, $P_d$ is the pressure of the tire after the car being driven, $V_d$ is the volume of the tire after the car being driven, and $T_d$ is the temperature of the tire after the car being driven.

Divide equation (IV) by (I), and solve for $P_d$.

    $\begin{array}{c}\frac{P_d{V}_d}{P_i{V}_i}=\frac{nR{T}_d}{=nR{T}_i}\\ P_d=P_i\left(\frac{V_i}{V_d}\right)\left(\frac{T_d}{T_i}\right)\end{array}$        (V)

Conclusion:

Substitute, $\left(1.02\right)\left(0.280V_i\right)$ for $V_d$, $1\text{atm}$ for $P_i$, $\left(85.0°\text{C+273}\text{.15}\right)\text{K}$ for $T_d$, and $\left(10°\text{C+273}\text{.15}\right)\text{K}$ for $T_i$ in the equation (V), to find $P_d$.

    $\begin{array}{c}{P}_d=1\text{atm}\left(\frac{V_i}{\left(1.02\right)\left(0.280V_i\right)}\right)\frac{\left(85.0°\text{C+273}\text{.15}\right)\text{K}}{\left(10°\text{C+273}\text{.15}\right)\text{K}}\\ =\frac{1\text{atm}}{\left(1.02\right)\left(0.280\right)}\frac{358.15\text{K}}{283.15\text{K}}\\ =4.43\text{atm}\\ \text{=4}\text{.43}\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\\ =4.49\times {10}^5\text{Pa}\end{array}$

Therefore, new tire pressure of the car is $\underset{\_}{4.49\times {10}^5\text{Pa}}$.