Let's break this down step by step: --- ## **Step 1: Find the mass of air needed to raise the pressure from 0 psig to 35 psig** ### **Given:** - Initial pressure, \( P_1 = 0 \) psig (gauge) = 14.7 psia (absolute) - Final pressure, \( P_2 = 35 \) psig = \( 35 + 14.7 = 49.7 \) psia - Volume, \( V = 3 \) ft³ - Temperature, \( T = 60^\circ F = 520 \) °R (use \( T(°R) = T(°F) + 459.67 \)) - Assume air behaves as an ideal gas: - \( R = 53.35 \) ft·lbf/(lbm·°R) (for air) --- ### **Step 1a: Calculate the mass of air at initial and final states using the ideal gas law** \[ PV = mRT \Rightarrow m = \frac{PV}{RT} \] #### **Initial state (after leak):** \[ P_1 = 14.7 \text{ psia} \] \[ T_1 = 60^\circ F = 60 + 459.67 = 519.67 \approx 520^\circ R \] \[ m_1 = \frac{14.7 \times 3}{53.35 \times 520} \] Let's calculate: \[ 53.35 \times 520 = 27,742 \] \[ 14.7 \times 3 = 44.1 \] \[ m_1 = \frac{44.1}{27,742} = 0.00159 \text{ lbm} \] --- #### **Final state (after filling to rated pressure):** \[ P_2 = 49.7 \text{ psia} \] \[ m_2 = \frac{49.7 \times 3}{53.35 \times 520} \] \[ 49.7 \times 3 = 149.1 \] \[ m_2 = \frac{149.1}{27,742} = 0.00537 \text{ lbm} \] --- #### **Air mass to add:** \[ \Delta m = m_2 - m_1 = 0.00537 - 0.00159 = 0.00378 \text{ lbm} \] --- ### **Boxed Final Answer (Part 1):** \[ \boxed{m_{-} = 0.0038 \text{ lbm}} \] (Rounded to 2 significant digits) --- ## **Step 2: Find the new pressure after temperature increases and volume expands** ### **Given:** - Initial state: \( P_2 = 49.7 \) psia, \( V_2 = 3 \) ft³, \( T_2 = 520^\circ R \) - Final state: \( T_3 = 110^\circ F = 110 + 459.67 = 569.67 \approx 570^\circ R \), \( V_3 = 3 \times 1.05 = 3.15 \) ft³ - Mass of air remains constant (\( m_2 \)), since the leak is plugged. --- ### **Use the ideal gas law ratio:** \[ \frac{P_2 V_2}{T_2} = \frac{P_3 V_3}{T_3} \] \[ P_3 = P_2 \frac{V_2}{V_3} \frac{T_3}{T_2} \] \[ P_3 = 49.7 \times \frac{3}{3.15} \times \frac{570}{520} \] Calculate step by step: - \( \frac{3}{3.15} = 0.9524 \) - \( \frac{570}{520} = 1.096 \) - \( 49.7 \times 0.9524 \times 1.096 \approx 49.7 \times 1.044 \approx 51.88 \) psia --- ### **Convert back to psig:** \[ P_{3,\text{psig}} = P_3 - 14.7 = 51.88 - 14.7 = 37.18 \text{ psig} \] --- ### **Boxed Final Answer (Part 2):** \[ \boxed{P = 37 \text{ psig}} \] --- ## **Summary:** - **Mass of air to add:** \(\boxed{0.0038~\text{lbm}}\) - **Final pressure after expansion and heating:** \(\boxed{37~\text{psig}}\) Let me know if you have any questions or need further clarifications!

**1. Introduction** Understanding the behavior of gases within a tire under different conditions requires knowledge of fundamental thermodynamic principles, particularly those related to the ideal gas law. The ideal gas law relates pressure (P), volume (V), temperature (T), and mass (m) of a gas through the equation \( PV = mRT \), where R is the specific gas constant. In this problem, the initial state of the tire's air is known after a leak reduces the pressure to zero gauge pressure, and the tire volume remains constant. Since the gas is compressed or expanded without heat transfer (assuming ideal conditions), the process involves calculating the change in mass needed to restore the pressure to the rated value, considering the temperature remains constant during re-inflation. Later, the tire experiences temperature and volume changes due to external conditions, affecting the internal pressure. Applying the ideal gas law again allows us to determine the new pressure after heating and expansion. **Explanation:** This introduction emphasizes the significance of the ideal gas law and thermodynamic principles for calculating the quantities of air involved and the subsequent pressure changes in response to temperature and volume variations. These concepts form the foundation for solving the problem systematically. --- **2. Presentation of Relevant Formulas and Representation of Given Data** **Relevant Formulas:** - **Ideal Gas Law:** \[ PV = mRT \] where: - \( P \) = absolute pressure (psia) - \( V \) = volume (ft³) - \( m \) = mass of gas (lbm) - \( R \) = specific gas constant for air (\(\mathrm{ft \cdbf \cdot \mathrm{lbf} / (lbm \cdot °R)} \)) - \( T \) = absolute temperature (°R) - **Mass Calculation:** \[ m = \frac{PV}{RT} \] - **Pressure after temperature and volume change:** \[ P_3 = P_2 \times \frac{V_2}{V_3} \times \frac{T_3}{T_2} \] **Given Data:** | Parameter | Value | Notes | |--------------|---------|----------------------------------------------| | Initial gauge pressure after leak | 0 psig | corresponds to 14.7 psia (absolute) | | Final rated pressure | 35 psig | corresponds to 49.7 psia (absolute) | | Tire volume | 3 ft³ | constant during initial filling | | Temperature after leak | 60°F | convert to °R for calculations | | Temperature after heating | 110°F | convert to °R | | Volume expansion | 5% | new volume \( V_3 = 1.05 \times V_2 \) | **Constants:** - \( R_{air} \approx 53.35 \ \mathrm{ft \cdot lbf / (lbm \cdot °R)} \) - Conversion: \( T(°R) = T(°F) + 459.67 \) **Explanation:** The formulas derive directly from the ideal gas law, which is fundamental in thermodynamics for relating pressure, volume, temperature, and mass of gases. Proper conversion to absolute units is necessary for accuracy, and the data provides all the parameters needed for calculations. --- **3. A Detailed Step-by-Step Solution** **Step 1: Calculate the initial mass of air after leak and before filling** - Convert temperature to absolute units: \[ T_1 = 60^\circ F + 459.67 = 519.67^\circ R \] - Calculate initial mass \( m_1 \): \[ m_1 = \frac{P_1 V}{R T_1} = \frac{14.7 \times 3}{53.35 \times 520} \] \[ = \frac{44.1}{27,742} \approx 0.00159 \text{ lbm} \] **Explanation:** This mass represents the amount of air remaining after the leak, at zero gauge pressure (14.7 psia absolute) and 60°F. --- **Step 2: Calculate the mass of air needed to restore the rated pressure** - Convert rated gauge pressure to absolute: \[ P_2 = 35 + 14.7 = 49.7 \text{ psia} \] - Calculate the mass at rated pressure: \[ m_2 = \frac{49.7 \times 3}{53.35 \times 520} = \frac{149.1}{27,742} \approx 0.00537 \text{ lbm} \] - Find the additional mass required: \[ \Delta m = m_2 - m_1 = 0.00537 - 0.00159 = 0.00378 \text{ lbm} \] **Explanation:** This is the amount of air that must be supplied by the compressor to bring the tire from zero gauge pressure to its rated pressure, assuming temperature and volume are unchanged. --- **Step 3: Determine the pressure after heating and volume expansion** - Convert the new temperature: \[ T_3 = 110^\circ F + 459.67 = 569.67^\circ R \] - Calculate new volume: \[ V_3 = 1.05 \times V_2 = 1.05 \times 3 = 3.15 \text{ ft}^3 \] - Use the ratio form of the ideal gas law to find the new pressure: \[ P_3 = P_2 \times \frac{V_2}{V_3} \times \frac{T_3}{T_2} \] \[ = 49.7 \times \frac{3}{3.15} \times \frac{570}{520} \] Calculate each factor: \[ \frac{3}{3.15} \approx 0.9524 \] \[ \frac{570}{520} \approx 1.096 \] Putting it all together: \[ P_3 = 49.7 \times 0.9524 \times 1.096 \approx 49.7 \times 1.044 \approx 51.88 \text{ psia} \] - Convert back to gauge pressure: \[ P_{gauge} = 51.88 - 14.7 = 37.18 \text{ psig} \] **Explanation:** The increase in temperature and volume expansion causes the internal pressure to rise above the original rated value. This calculation quantifies the new pressure considering the thermodynamic effects. --- ### **Summary:** - The **mass of air** that needs to be supplied by the compressor to restore the tire to its rated pressure is approximately **0.0038 lbm**. - After heating the tire to 110°F and allowing the volume to expand by 5%, the **pressure inside the tire** increases to approximately **37 psig**. **This comprehensive approach ensures a clear understanding of the thermodynamic principles involved and their application to practical problems related to tire pressures and gas behavior.**