Chapter 13, Problem 60A

(a)

To determine

New pressure in the tire.

Answer to Problem 60A

New pressure of the tire is $3.2\times {10}^{5}Pa$ .

Explanation of Solution

Given:

Gauge pressure $=30.0psi$ , where psi is one pound per square inch $=6.90\times {10}^{3}Pa$

Initial pressure, $P_i=$ $\text{(6}{\text{.90 × 10}}^{\text{3}}\text{Pa / psi) = 3}{\text{.08 × 10}}^{\text{5}}\text{ Pa}$

Initial volume, $V_i=$ $0.55m^3$

Final volume, $V_f=$ $0.58m^3$

Initial temperature, $T_i=$ $280K$

Final temperature, $T_f=$ $310K$

Formula used:

Using Ideal gas law:

  $\frac{P_i{V}_i}{T_f}=\frac{P_f{V}_f}{T_f}$ or $P_f=\frac{P_i{V}_i{T}_f}{T_i{V}_f}$

Where, $P_i=$ Initial pressure, $P_f=$ Final pressure, $V_i=$ Initial volume, $V_f=$ Final volume, $T_i=$ Initial temperature and $T_f=$ final temperature

Calculation:

Apply the formula, $P_f=\frac{P_i{V}_i{T}_f}{T_i{V}_f}$ , and put the given values:

  $\begin{array}{l}{\text{P}}_{\text{2}}\text{ = }\frac{\text{(3}{\text{.08 ×10}}^{\text{5}}\text{Pa)(0}{\text{.55 m}}^{\text{3}}\text{)(310 }\text{K)}}{\text{(280}\text{K)(0}\text{.58}{\text{m}}^{\text{3}}\text{)}}\\ {\text{P}}_{\text{2}}\text{= 3}{\text{.2×10}}^{\text{5}}\text{Pa}\end{array}$

Conclusion: The new pressure of the tire is $3.2\times {10}^{5}Pa$ .

(b)

To determine

The new gauge pressure.

Answer to Problem 60A

The new gauge pressure is $31psi$ .

Explanation of Solution

Given:

Gauge pressure $=30.0psi$ , where psi is pound per square inch $=6.90\times {10}^{3}Pa$

Initial pressure, $P_i=$ $\text{(6}{\text{.90 × 10}}^{\text{3}}\text{Pa / psi) = 3}{\text{.08 × 10}}^{\text{5}}\text{ Pa}$

Initial volume, $V_i=$ $0.55m^3$

Final volume, $V_f=$ $0.58m^3$

Initial temperature, $T_i=$ $280K$

Final temperature $T_f=$ $310K$

Formula used:

From Ideal gas law:

  $\frac{P_i{V}_i}{T_f}=\frac{P_f{V}_f}{T_f}$ or $P_f=\frac{P_i{V}_i{T}_f}{T_i{V}_f}$

Where, $P_i=$ Initial pressure, $P_f=$ Final pressure, $V_i=$ Initial volume, $V_f=$ Final volume, $T_i=$ Initial temperature and $T_f=$ final temperature

Calculation: The new gauge pressure is:

  $P_{gauge}=\frac{(30.0psi)(0.55m^3)(310K)}{(280K)(0.58m^3)}$

  $P_{gauge}=31psi$

Conclusion: Hence, $31psi$ is the new gauge pressure.