Problem 8 - University Physics with Modern Physics

Question

You close off the nozzle of a bicycle tire pump and very slowly depress the plunger so that the air inside is compressed to half its original volume. Assume the air behaves like an ideal gas. If you do this so slowly that the temperature of the air inside the pump never changes: (a) Is the work done by the air in the pump positive or negative? (b) Is the heat flow to the air positive or negative? (c) What can you say about the relative magnitudes of the heat flow and the work? Explain.

Step-by-Step Solution

Answer

(a) The work is positive. (b) The heat flow is negative. (c) Magnitudes are equal.

Step 1: Identify the Process

The process described in the problem is called an isothermal compression. Since the temperature of the air doesn't change, we can assume this isothermal process.

Step 2: Understand Isothermal Process for an Ideal Gas

For an ideal gas undergoing an isothermal process, the temperature remains constant. According to the first law of thermodynamics, \( \Delta U = Q - W \), where \( \Delta U = 0 \) because temperature is constant, so \( Q = W \).

Step 3: Determine the Work Done by the Air

In an isothermal compression, the work done on the gas is calculated as \( W = -nRT\ln\left(\frac{V_f}{V_i}\right) \). Since the final volume \( V_f \) is less than the initial volume \( V_i \), \( \ln\left(\frac{V_f}{V_i}\right) \) is negative, making \( W \) positive.

Step 4: Evaluate Heat Flow to the Air

Since \( Q = W \) in an isothermal process and work is being done on the gas (air), energy must flow out of the system for \( Q \) to equal \( W \), implying that \( Q \) is negative.

Step 5: Compare Heat Flow and Work

According to the first law of thermodynamics for an isothermal process, since \( Q = W \), the magnitudes are equal. However, since work is done on the gas, \( Q \) is negative, indicating energy is lost as heat.

Key Concepts

Ideal Gas

An ideal gas is a theoretical concept used in physics and chemistry to simplify the study of gases. Ideal gas molecules are considered to behave in a way where they exert no forces on each other except during elastic collisions. This means they move randomly and are very distant from one another, allowing us to ignore any potential energy arising from molecular interactions. Additionally, the molecular size is negligible compared to the volume of the container.

The behavior of ideal gases is described by the Ideal Gas Law:\[ PV = nRT \]where:

• \( P \) is the pressure of the gas
• \( V \) is the volume of the gas
• \( n \) is the number of moles
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

This law is particularly useful because it establishes a relationship between the various parameters that govern the state of an ideal gas. It's important to note that real gases approximate ideal gas behavior under a range of conditions but may deviate significantly under high pressure and low temperature.

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle that describes the conservation of energy in a system. It tells us that energy can neither be created nor destroyed; it can only be transferred or changed from one form to another. This concept is particularly useful when analyzing thermodynamic processes, such as those involving gases.

The First Law of Thermodynamics can be mathematically expressed as:\[ \Delta U = Q - W \]where:

• \( \Delta U \) is the change in internal energy of the system
• \( Q \) is the heat added to the system
• \( W \) is the work done by the system

In the scenario of an isothermal process, the temperature remains constant, leading to a constancy in internal energy, i.e., \( \Delta U = 0 \). Thus, the equation simplifies to:\[ Q = W \]This means that the amount of work done by/on the system is exactly balanced by the heat added or removed, maintaining energy equilibrium.

Work Done in Thermodynamics

In thermodynamics, work done by or on a system is a key concept that helps us understand energy transfers. Here, we discuss an isothermal process, where the temperature of the gas remains constant. In such processes, the work done on the gas can be expressed with the formula:\[ W = -nRT\ln\left(\frac{V_f}{V_i}\right) \]where:

• \( W \) is the work done on the gas
• \( n \) is the number of moles of the gas
• \( R \) is the ideal gas constant
• \( T \) is the absolute temperature
• \( V_f \) and \( V_i \) are the final and initial volumes, respectively

In the context of a bicycle pump, when air is compressed isothermally, this means work is done on the gas (air). Therefore, since the logarithm of a number less than one is negative, the product of the entire equation results in positive work. During this isothermal compression, the external force (your hand) does work on the gas, leading to a transfer of energy out of the system as heat, consistent with the isothermal process' characteristics.