Problem 27 - College Physics

Question

An automobile tire is filled to an absolute pressure of 3.0 atm at a temperature of \(30^{\circ} \mathrm{C}\). Later it is driven to a place where the temperature is only \(-20^{\circ} \mathrm{C} .\) What is the absolute pressure of the tire at the cold place? (Assume that the air in the tire behaves as an ideal gas and the volume is constant.)

Step-by-Step Solution

Answer

The absolute pressure of the tire at the cold place is approximately 2.51 atm.

Step 1: Understand the Ideal Gas Law

The ideal gas law is given by the equation \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. Here, the volume \( V \) remains constant, and the number of moles \( n \) does not change.

Step 2: Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula \( T(K) = T(°C) + 273.15 \). Initially, the temperature is \(30^{\circ} \text{C}\), which is \(30 + 273.15 = 303.15 \text{ K}\). The final temperature is \(-20^{\circ} \text{C}\), which is \(-20 + 273.15 = 253.15 \text{ K}\).

Step 3: Apply the Pressure-Temperature Relationship

According to the ideal gas law, when volume and moles are constant, pressure is directly proportional to temperature: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Here, \( P_1 = 3.0 \text{ atm} \) and \( T_1 = 303.15 \text{ K} \). We need to find \( P_2 \) when \( T_2 = 253.15 \text{ K} \).

Step 4: Solve for the Final Pressure

Rearrange the relationship \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) to find \( P_2 \): \( P_2 = P_1 \times \frac{T_2}{T_1} \). Substituting the values, we get \( P_2 = 3.0 \times \frac{253.15}{303.15} \approx 2.51 \text{ atm} \).

Step 5: Verify the Calculation

Check the calculations for accuracy by ensuring each arithmetic operation is correct. \( 3.0 \times \frac{253.15}{303.15} \) simplifies to \( 2.51 \text{ atm} \). This confirms the calculations are accurate, and the pressure under the new conditions is determined.

Key Concepts

Pressure-Temperature Relationship

In the world of physics and chemistry, understanding the behavior of gases under varying conditions is crucial. One key principle that describes this behavior is the pressure-temperature relationship, often referred to in the context of the Ideal Gas Law. When we talk about this relationship, we are focusing on how pressure and temperature are interdependent, provided other factors like the volume and amount of gas (moles) remain constant.

According to the Ideal Gas Law, when the volume and moles remain constant, the pressure of a gas is directly proportional to its temperature when measured on an absolute scale like Kelvin. This means that as the temperature of a gas increases, so does its pressure, assuming the volume stays the same. Conversely, if the temperature decreases, the pressure will also decrease.

The mathematical expression for this relationship is captured in the formula:

• \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)

where \( P_1 \) and \( P_2 \) represent the initial and final pressures, respectively, and \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin. By rearranging this formula, you can solve for any unknown in a given problem, such as the final pressure as in our exercise.

Absolute Pressure Calculation

Calculating absolute pressure is a fundamental skill in physics, particularly when dealing with gas laws. Absolute pressure is the actual pressure exerted by a gas, inclusive of the atmospheric pressure. It's an important measure, distinguishing it from gauge pressure, which only measures pressure above the local atmospheric pressure.

In our exercise, the absolute pressure of the tire initially was given as 3.0 atm at a particular temperature. To calculate the absolute pressure at a lower temperature, we need to apply the pressure-temperature relationship from the Ideal Gas Law. Using the equation:

• \( P_2 = P_1 \times \frac{T_2}{T_1} \)

allows us to find the new absolute pressure \( P_2 \) at the reduced temperature. We substitute the known values to determine the pressure difference caused by the temperature change.

Temperature Conversion to Kelvin

Temperature conversion is an essential step in gas law calculations, as the equations require temperatures to be expressed in Kelvin rather than Celsius. The Kelvin scale is based on the absolute zero, the point at which particles theoretically stop moving.

To convert from Celsius to Kelvin, which is often written in the form \( T(K) = T(°C) + 273.15 \), you add 273.15 to the Celsius measurement. This straightforward conversion allows temperatures to be used accurately in formulas related to the Ideal Gas Law.

For instance, in the tire exercise, the temperature was initially provided in Celsius. We converted temperatures from 30°C to 303.15 K and from -20°C to 253.15 K. By working with Kelvin, it becomes much simpler to apply calculations involving ratios, as Kelvin is a direct measure of the kinetic energy within the gas.