Problem 64 - University Physics with Modern Physics

Question

A cylinder with a piston contains 0.150 mol of mitrogen at \(1.80 \times 10^{5} \mathrm{Pa}\) and 300 \(\mathrm{K}\) . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a \(p V\) -diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

Step-by-Step Solution

Answer

Minimum pressure is about \(1.25 \times 10^5 \mathrm{Pa}\).

Step 1: Understand the problem

The problem involves a cylinder with a piston containing nitrogen gas, which undergoes various thermodynamic processes: isobaric compression, adiabatic expansion, and isochoric heating. Our task is to analyze these processes using the ideal gas law and other thermodynamic relations.

Step 2: Isobaric Compression Analysis

In isobaric processes, the pressure remains constant. Since the volume is halved, we use the ideal gas law: \( PV = nRT \). Initially, \( P_i = 1.80 \times 10^5 \, \text{Pa} \), \( V_i \), \( T_i = 300 \, \text{K} \). After compression, the volume \( V_f = \frac{1}{2}V_i \), pressure remains \( P_f = P_i \), and \( T_f = \frac{P_f V_f}{nR} = 150 \, \text{K} \), by solving \( 1.80 \times 10^5 \times \frac{1}{2}V_i = 0.150 \times R \times T_f \).

Step 3: Adiabatic Expansion Analysis

During adiabatic processes, no heat is exchanged. Using the adiabatic relation \( TV^{\gamma - 1} = \text{constant} \), with \( \gamma = 1.4 \) for diatomic gases like nitrogen, and the initial state \( T_i = 150 \, \text{K} \), \( V_i = \frac{1}{2}V \), we get \( T_f (V)^{0.4} = 150 (\frac{1}{2}V)^{0.4} \). Solving for the final temperature when the volume returns to \( V \): \( T_f = 150 \times (0.5)^{0.4 - 1} \approx 243 \, \text{K} \).

Step 4: Isochoric Heating Analysis

In isochoric processes, the volume stays constant. The pressure and temperature are related by \( \frac{P}{T} = \text{constant} \). Using the adiabatic end \( T_i = 243 \) K and the original conditions \( T_f = 300 \) K: \( \frac{P_i}{243} = \frac{1.80 \times 10^5}{300} \). Solving for \( P_i \): \( P_i \approx 2.14 \times 10^5 \, \text{Pa} \). This final adjustment ensures pressure returns to its original value.

Step 5: Calculate Minimum Pressure

The minimum pressure occurs at the end of the adiabatic expansion process. Using poypy relation \( PV^{\gamma} = \text{constant} \) from the end of the adiabatic process, found with reduced temperature \( T = 243 \) K, with volume again \( V \): original pressure \( P_i = 1.80 \times 10^5 \, \text{Pa} \), \( V_i = \frac{1}{2}V \), determine new \( P_f \): \[ P_f = \frac{1.80 \times 10^5}{2^{1.4}} \approx 1.25 \times 10^5 \, \text{Pa} \].

Key Concepts

Ideal Gas Law

The ideal gas law combines several basic principles of chemistry and physics into one equation:

• Pressure (\(P\)
),
• Volume (\(V\)
),
• The number of moles (\(n\)
),
• Temperature (\(T\)
).

These variables are linked through the formula \(PV = nRT\), where \(R\) is the ideal gas constant. This equation helps describe how an ideal gas behaves under different conditions. For instance, if you increase the temperature of a gas while keeping the volume constant, its pressure will also increase. This is because the gas molecules move faster and hit the walls of the container more frequently.

Similarly, if a gas is compressed by decreasing its volume while maintaining its temperature, the pressure will increase. In real-life scenarios, gases deviate from ideal behavior under high pressure or low temperature conditions. However, many gases, especially under standard conditions, adhere closely to the ideal gas law.

Isobaric Process

An isobaric process refers to a thermodynamic change in which the pressure remains constant while other parameters may vary. This is common in scenarios involving gas behaviors, such as a piston compressing or expanding gas under a constant external pressure.These processes are easy to analyze using the ideal gas law. During an isobaric compression, for instance:

• The volume of the gas decreases.
• To compensate, the temperature eventually drops, based on the relation \( P_i V_i = nRT_i\) and \(P_f V_f = nRT_f\).

A classical isobaric process can be visualized easily in a piston-cylinder setup as the piston moves up or down.
Such a process is represented as a horizontal line on a \(PV\)-diagram because the pressure remains unchanged while the volume decreases or increases.

Adiabatic Process

In an adiabatic process, there is no heat exchange with the surroundings (\(Q = 0\)). This makes such processes intriguing, as they occur without transfer of energy as heat across the system’s boundary.Key attributes of adiabatic processes involve:

• Rapid expansions and compressions (often requiring insulated conditions).
• All changes in energy are internal, causing temperature changes.

The formula \( TV^{\gamma - 1} = \text{constant}\) (where \(\gamma\) is the heat capacity ratio), governs these processes.For example, as gas expands adiabatically, it does work on the surroundings and its temperature decreases. Conversely, when adiabatically compressed, the internal energy increases, causing a temperature rise.

In the context of the original problem, the gas expanded back to its original volume without heat exchange, which necessarily affected the final temperature.

Isochoric Process

An isochoric process keeps the volume constant, meaning no work is done (since work \(W = P\Delta V = 0\)). This makes it unique among thermodynamic processes.During an isochoric change, pressure and temperature are directly proportional, represented by \( \frac{P}{T} = \text{constant}\).Some typical applications include:

• Heating or cooling a gas in a closed container, where only temperature and pressure change.
• Analysis is often straightforward due to the simplicity of the constant-volume condition.

In the exercise, the isochoric process was used to restore the gas to its original pressure by heating it at constant volume, showing a direct pressure-temperature relationship.
It is depicted as a vertical line on a \(PV\)-diagram, since volume does not change.