Question

Air is pumped into an automobile tube upto a pressure of \(200 \mathrm{kPa}\) in the morning when the air temperature is \(22^{\circ} \mathrm{C}\). During the day, temperature rises to \(42^{\circ} \mathrm{C}\) and the tube expands by \(2 \%\) The pressure of the air in the tube at this temperature will be approximately. (A) \(209 \mathrm{kPa}\) (B) \(206 \mathrm{kPa}\) (C) \(200 \mathrm{kPa}\) (D) \(212 \mathrm{kPa}\)

Short answer

The pressure of the air in the tube at the temperature of \(42^{\circ}C\) will be approximately \(209\,\text{kPa}\).

Step-by-step solution

Convert temperatures to Kelvin

To convert Celsius to Kelvin, we add 273.15. Therefore, we have: Initial temperature: \(T_1 = 22^{\circ}C + 273.15 = 295.15 K\) Final temperature: \(T_2 = 42^{\circ}C + 273.15 = 315.15 K\)

Determine initial and final volumes

Let the initial volume be \(V_1\). The volume increases by \(2\%\), so the final volume, \(V_2\), is calculated as follows: \(V_2 = V_1 + 0.02V_1 = 1.02V_1\)

Use the ideal gas law equation

The ideal gas law equation is: \(PV = nRT\) For both initial and final states, the number of moles and the ideal gas constant remain the same. We can write two equations for both states: Initial state: \(P_1V_1 = nRT_1\) Final state: \(P_2V_2 = nRT_2\) Now, we can divide the final state equation by the initial state equation: \(\frac{P_2V_2}{P_1V_1} = \frac{nRT_2}{nRT_1}\) The constants cancel each other out: \(\frac{P_2}{P_1} = \frac{V_2}{V_1} \frac{T_2}{T_1}\) We already know that \(V_2 = 1.02V_1\), so we can substitute that into the equation: \(\frac{P_2}{200\,\text{kPa}} = \frac{1.02V_1}{V_1} \frac{315.15\,K}{295.15\,K}\) Now we can solve for \(P_2\): \(P_2 = 200\,\text{kPa} \times 1.02 \times \frac{315.15}{295.15} = 209.36\,\text{kPa}\)

Final Answer

The pressure of the air in the tube at the temperature of \(42^{\circ}C\) will be approximately: (A) \(209\,\text{kPa}\)

Key concepts

Pressure and Temperature Relationship

The relationship between pressure and temperature in gases is crucial in understanding how they behave when heated or cooled. This principle is rooted in the Ideal Gas Law, which states that for a fixed amount of gas at constant volume, the pressure is directly proportional to its temperature. Hence, as the temperature of the gas increases, so does its pressure, assuming the gas does not escape or change phases.

In the scenario where gas is contained, such as in a car tire, if the temperature rises significantly due to surrounding environmental changes, the pressure within the tire will increase. This is because the molecules move more vigorously at higher temperatures, colliding with the walls of the container more forcefully and frequently, thereby increasing pressure.

It's essential to monitor these changes because excessive pressure can lead to unexpected issues, like tire blowouts. Understanding this relationship helps in designing better safety measures and systems that can efficiently adjust to temperature fluctuations.

Volume Expansion

The concept of volume expansion deals with how substances, particularly gases, expand when temperature changes. As temperature increases, the kinetic energy of molecules likewise increases, causing them to occupy more space, leading to volume expansion.

In the practice problem mentioned above, the tire's volume increased by 2%. This expansion is calculated based on the initial volume, demonstrating how even a small temperature change can significantly affect volume and, consequently, pressure.

Volume expansion is crucial to consider because it can impact pressure calculations, as seen when determining the final pressure of the tire. Since the volume and final temperature both affect the result, considering expansion is key to an accurate understanding of such thermodynamic processes.

Temperature Conversion

When dealing with thermodynamic calculations, especially those involving gases, converting temperatures to a common scale is essential for consistency. The Kelvin scale is the preferred unit of temperature in these scenarios because it starts at absolute zero, the point where molecular motion ceases, providing a true measure of energy changes.

In the provided exercise, converting Celsius to Kelvin ensures that the relationships described by the Ideal Gas Law are correctly applied.

• An increase in temperature from 22°C to 42°C was converted to Kelvin, resulting in 295.15 K and 315.15 K respectively.

This conversion eliminates negative values and proportional discrepancies that could arise using Celsius or Fahrenheit scales.

Understanding temperature conversion helps to ensure that calculations are accurate and reliable, which is fundamental when predicting how gases will react to temperature changes.