Problem 61 - University Physics with Modern Physics

Question

An automobile tire has a volume of 0.0150 \(\mathrm{m}^{3}\) on a cold day when the temperature of the air in the tire is \(5.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.02 atm. Under these conditions the gauge pressure is measured to be 1.70 atm (about 25 \(\mathrm{lb} / \mathrm{in.} .\) ). After the car is driven on the highway for 30 min, the temperature of the air in the tires has risen to \(45.0^{\circ} \mathrm{C}\) and the volume has risen to 0.0159 \(\mathrm{m}^{3}\) . What then is the gauge pressure?

Step-by-Step Solution

Answer

The final gauge pressure is approximately 2.175 atm.

Step 1: Convert Temperatures to Kelvin

Convert the initial temperature from Celsius to Kelvin: \( T_1 = 5.0^{\circ} C + 273.15 = 278.15 \ K \). Convert the final temperature from Celsius to Kelvin: \( T_2 = 45.0^{\circ} C + 273.15 = 318.15 \ K \).

Step 2: Calculate Initial Absolute Pressure

The initial gauge pressure is given as 1.70 atm, and the atmospheric pressure is 1.02 atm. Therefore, the initial absolute pressure is \( P_1 = 1.70 + 1.02 = 2.72 \ \text{atm} \).

Step 3: Use Ideal Gas Law (Initial State)

Using the ideal gas law \( PV = nRT \), where \( n \) and \( R \) are constants, express the initial state as \( P_1 V_1 = nRT_1 \), where \( V_1 = 0.0150 \ m^3 \).

Step 4: Use Ideal Gas Law (Final State)

Similarly, express the final state as \( P_2 V_2 = nRT_2 \), where \( V_2 = 0.0159 \ m^3 \) and the temperature is \( T_2 = 318.15 \ K \).

Step 5: Set Up Equation for Final Pressure

Since \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) by the ideal gas law ratio, substitute the known values: \[ P_2 = \frac{P_1 V_1 T_2}{V_2 T_1} \].

Step 6: Substitute and Solve for \( P_2 \)

Substitute \( P_1 = 2.72 \ \text{atm}, \ V_1 = 0.0150 \ m^3, \ V_2 = 0.0159 \ m^3, \ T_1 = 278.15 \ K, \) and \( T_2 = 318.15 \ K \) into the equation: \[ P_2 = \frac{2.72 \times 0.0150 \times 318.15}{0.0159 \times 278.15} \]. This results in \( P_2 \approx 3.195 \ \text{atm} \).

Step 7: Find Final Gauge Pressure

The final absolute pressure \( P_2 \) is 3.195 atm. The gauge pressure is the absolute pressure minus the atmospheric pressure: \( 3.195 - 1.02 = 2.175 \ \text{atm} \).

Key Concepts

Gauge Pressure

When talking about pressures in an automobile tire, it’s crucial to understand the concept of gauge pressure. Gauge pressure measures how much pressure is exerted by the tire over and above the surrounding atmospheric pressure. This is why gauge pressure can sometimes appear lower than absolute pressure.

For example, if the atmospheric pressure is 1.02 atm and the gauge pressure is 1.70 atm, it means that the actual pressure inside the tire is higher by adding both values. So, the tire's absolute pressure is 2.72 atm. Always remember:

• Gauge pressure = absolute pressure - atmospheric pressure.
• It does not consider any external pressure exerted by the atmosphere.

This is why a flat tire, at an atmospheric condition, often reads zero gauge pressure even though there is atmospheric pressure acting on it.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the air above a particular point, typically at sea level. It’s the pressure exerted by the atmosphere and is a crucial parameter in many physics calculations, including those involving the ideal gas law.

In the context of the automobile tire problem, the atmospheric pressure is given as 1.02 atm. This means:

• It’s the baseline or "zero" point for gauge pressure readings.
• All gauge pressure measurements are calculated over this basic atmospheric level.

Atmospheric pressure can vary based on altitude and weather conditions, but for most practical problems, the local value is used as a constant to simplify calculations.

Absolute Pressure

Absolute pressure includes both the gauge pressure and the atmospheric pressure. It reflects the total pressure acting within a closed system, like an automobile tire. This configuration is fundamental to solve problems using the ideal gas law.

In our exercise, the tire has an initial absolute pressure calculated from both the gauge pressure and the atmospheric pressure as 2.72 atm. Remember:

• Absolute pressure = gauge pressure + atmospheric pressure.
• It is the actual pressure experience, not just the pressure above atmospheric levels.

This distinction is crucial, especially when performing calculations that need the total amount of pressure exerted on gases, as described in the ideal gas law.

Temperature Conversion

Temperature conversion is a key step in dealing with the ideal gas law, particularly when measuring temperature changes in gases. The Celsius temperature scale must often be converted into Kelvin because the Kelvin scale provides an absolute measure of temperature necessary for gas law calculations.

To convert Celsius into Kelvin, one needs to add 273.15 to the degree Celsius value. For instance:

• The initial temperature of 5.0 °C converts to 278.15 K.
• The final temperature of 45.0 °C converts to 318.15 K.

Kelvin should always be used in thermodynamic equations as it does not allow for negative values, eliminating problems of undefined operations in gas laws.

Volume Change

Understanding volume change is essential when solving problems using the ideal gas law. Volume change in gases occurs due to temperature and pressure differences; it plays a crucial role in accurately calculating final pressures.

In the exercise, the tire's initial volume is 0.0150 m³, and it increases to 0.0159 m³ when the car is driven. This change influences the behavior of gases, as described by the ideal gas law. Remember:

• Increases in temperature often cause increases in volume if pressure remains constant.
• Changes in volume, pressure, and temperature are interdependent.

Always ensure to use consistent units and accurately measure all changes in physical conditions to apply formulas like the ideal gas law accurately.