Problem 66 - University Physics with Modern Physics

Question

CALC A cylinder with a piston contains 0.150 mol of Nnitrogen at \(1.80 \times 10^{5}\) 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 occurs during adiabatic expansion.

Step 1: Understanding the Processes

We are dealing with nitrogen in a piston undergoing several processes: isobaric compression, adiabatic expansion, and isochoric heating. In an isobaric process, pressure remains constant, in an adiabatic process, there is no heat exchange with surroundings, and in an isochoric process, volume remains constant.

Step 2: Initial Conditions and Calculating Initial Volume

Given: number of moles, \(n = 0.150\) mol; initial pressure, \(P_1 = 1.80 \times 10^{5}\) Pa; initial temperature, \(T_1 = 300\, \mathrm{K}\). Using the ideal gas law \(PV = nRT\) to calculate the initial volume, \(V_1 = \frac{nRT_1}{P_1}\), where \(R = 8.314\, \text{J/mol K}\).

Step 3: Isobaric Compression

In isobaric compression, the volume is compressed to half its original volume. Therefore, the final volume \(V_2 = \frac{V_1}{2}\) while pressure remains \(P_1 = 1.80 \times 10^5\) Pa.

Step 4: Adiabatic Expansion Calculation

When the gas expands adiabatically back to its original volume, we use the adiabatic expansion equations: \(P_2V_2^\gamma = P_3V_1^\gamma\) and \(T_2V_2^{\gamma-1} = T_3V_1^{\gamma-1}\), where \(\gamma = \frac{C_p}{C_v}\) (for nitrogen, \(\gamma \approx 1.4\)). Solve for \(T_3\), the temperature at the end of the adiabatic expansion.

Step 5: Isochoric Heating Back to Initial Pressure

In the isochoric process, the volume remains constant at \(V_1\). Use the ideal gas law to solve for temperature holding \(V_1\) constant and adjusting pressure back to \(P_1\). Therefore, \(T_4 = \frac{P_1V_1}{nR}\) at original pressure \(P_1\).

Step 6: Computing Minimum Pressure

The minimum pressure occurs during the adiabatic expansion due to combined effects of increased volume and decreased temperature. We use the relation for adiabatic processes: \(P_3 = \frac{nRT_3}{V_1}\), solving for \(P_3\), where \(V = V_1\) at the end of the adiabatic process.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation used to understand the behavior of gases under various conditions. It combines several gas laws into one formula: \[ PV = nRT \] where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles of the gas,
• \( R \) is the universal gas constant, which is about \( 8.314 \, \mathrm{J/mol \cdot K} \),
• \( T \) is the temperature in Kelvin.

This formula helps you to calculate one of the variables if the other three are known. In the given problem, it assists in determining the initial volume of nitrogen gas in the cylinder. You simply rearrange the formula to solve for volume: \[ V = \frac{nRT}{P} \] This law assumes that the gas molecules do not interact and occupy negligible space, treating the gas as "ideal." In practical applications, the ideal gas law provides a reasonable approximation under many conditions.

Adiabatic Process

An adiabatic process involves changes in a gas where no heat is exchanged with the surroundings. In this scenario, the system is insulated so that energy changes occur only in the form of work done by or on the gas. An important property for adiabatic processes is that they follow the relation:\[ P V^\gamma = \text{constant} \] where \( \gamma \) is the heat capacity ratio \( \frac{C_p}{C_v} \). For nitrogen, \( \gamma \approx 1.4 \). During an adiabatic expansion, like in this problem, the gas does work as it expands, leading to a fall in both pressure and temperature. To calculate the final temperature of the gas after this process, we use:\[ T_2 V_2^{\gamma-1} = T_3 V_1^{\gamma-1} \] This allows you to find the temperature \( T_3 \) at the end of the adiabatic expansion, which is crucial for further steps.

Isobaric Process

An isobaric process is characterized by constant pressure. Under these conditions, the volume of the gas changes, while the pressure \( P \) remains constant throughout the process. In the exercise described, the gas undergoes an isobaric compression, which reduces the volume of the gas to half its initial value. Utilizing the ideal gas law in such a process helps relate the initial and final states of the gas:- Since \( P_1 = P_2 \), we have:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] This relation helps to understand how temperature changes when volume changes, given constant pressure.For the compression, this tells us that decreasing the volume at constant pressure requires a proportionate decrease in temperature.

Isochoric Process

In an isochoric process, the volume of the gas remains constant. Volume being constant means that any heat added to the gas results in a change in pressure and temperature.In the given problem, the final step involves heating the gas isochorically to its original pressure. Because the volume does not change in an isochoric process, the relation between pressure and temperature is straightforward:\[ \frac{P}{T} = \text{constant} \] This implies that if you increase the temperature, the pressure will rise as well, provided that volume stays the same. By using the ideal gas law, one can find the final temperature by holding volume constant and adjusting the system back to initial pressure \( P_1 \):\[ T_4 = \frac{P_1V_1}{nR} \] This calculation is vital to ensure that the system returns to its original state, completing the thermodynamic cycle.