Question

A gas is confined to one side of a rigid, insulated container divided by a partition. The other side is initially evacuated. The following data are known for the initial state of the gas: \(p_{1}=5\) bar, \(T_{1}=500 \mathrm{~K}\), and \(V_{1}=0.2 \mathrm{~m}^{3}\). When the partition is removed, the gas expands to fill the entire container, which has a total volume of \(0.5 \mathrm{~m}^{3}\). Assuming ideal gas behavior, determine the final pressure, in bar.

Short answer

The final pressure is 2 bar.

Step-by-step solution

- Define the Ideal Gas Law

The Ideal Gas Law is used to relate the pressure, volume, and temperature of an ideal gas. The formula is given by: \[ PV = nRT \] Where: - \(P\) is the pressure.- \(V\) is the volume.- \(n\) is the number of moles of gas.- \(R\) is the ideal gas constant.- \(T\) is the temperature.

- Calculate the Initial Number of Moles

Use the initial state information to find the number of moles of gas. Rearrange the Ideal Gas Law to solve for \(n\): \[ n = \frac{P_1 V_1}{RT_1} \] Substitute \(P_1 = 5\) bar (or \(5 \times 10^5\) Pa), \(V_1 = 0.2\) m\(^3\), \(R = 8.314\) J/(mol K), and \(T_1 = 500\) K: \[ n = \frac{5 \times 10^5 \times 0.2}{8.314 \times 500} \approx 24.04 \text{ moles} \]

- Determine Final Conditions and Apply the Ideal Gas Law

After the partition is removed, the gas expands, and the final volume \(V_2\) becomes 0.5 m\(^3\). Since the container is rigid and insulated, the temperature remains the same (\(T_2 = T_1 = 500\) K). Use the Ideal Gas Law to find the final pressure \(P_2\): \[ P_2 = \frac{nRT_2}{V_2} \] Substitute the values \(n = 24.04\) moles, \(R = 8.314\) J/(mol K), \(T_2 = 500\) K, and \(V_2 = 0.5\) m\(^3\): \[ P_2 = \frac{24.04 \times 8.314 \times 500}{0.5} \approx 200,000 \text{ Pa} = 2 \text{ bar} \]

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics, often used to describe the behavior of gases under various conditions. The law is expressed as \[ PV = nRT \]Where

• \(P\) is the pressure of the gas.
• \(V\) is the volume the gas occupies.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant, which is approximately 8.314 J/(mol K).
• \(T\) is the temperature in Kelvin.

The Ideal Gas Law helps us understand how these properties are related. For example, if we know any three properties, we can calculate the fourth.

Thermodynamics

Thermodynamics is the study of energy, heat, and work, and how they relate to each other. In the context of the Ideal Gas Law, thermodynamics describes how a gas behaves under different conditions of temperature, volume, and pressure. Thermodynamic processes can be characterized as:

• **Isothermal:** Processes occurring at constant temperature.
• **Adiabatic:** Processes where no heat is transferred to or from the system.
• **Isochoric:** Processes at constant volume.
• **Isobaric:** Processes at constant pressure.

Understanding these processes helps us predict how gases will react when subjected to different changes, making it crucial for solving problems involving gas behavior.

Gas Expansion

Gas expansion occurs when a gas increases in volume. This can involve the gas spreading out to fill a larger container, as in the current exercise. When the partition is removed, the gas expands from its initial volume \(V_1\) of 0.2 m\(^3\) to the final volume \(V_2\) of 0.5 m\(^3\). Throughout this process, the temperature remains constant, and the gas follows the principles outlined in the Ideal Gas Law. This example is an isothermal expansion since the temperature does not change. The final pressure can be determined by applying the changes in volume while calculating the final pressure.

Pressure Calculation

Calculating gas pressure using the Ideal Gas Law involves rearranging the formula. After finding the number of moles, you can calculate the pressure. In this exercise, the initial conditions given are:

• \(p_1 = 5\) bar
• \(T_1 = 500\) K
• \(V_1 = 0.2\) m\(^3\)

With these values, we use

• \[ n = \frac{P_1 V_1}{RT_1} \]

Next, using the final volume and same temperature:

• \[ P_2 = \frac{nRT_2}{V_2} \]

Inputting values:

• \(n = 24.04\) moles, \(R = 8.314\), \(T_2 = T_1 = 500\) K, \(V_2 = 0.5\) m\(^3\)

We get \[ P_2 = 2 \text{ bar} \]

Volume and Temperature Relations

The relationship between volume and temperature in gases is crucial for understanding their behavior. For an ideal gas, under constant pressure, the volume is directly proportional to its temperature, as shown by **Charles's Law**. Conversely, when the volume is kept constant, the pressure is directly proportional to the temperature, described by **Gay-Lussac's Law**. In this exercise, the temperature remains constant (isothermal expansion), which means the product of pressure and volume remains constant according to the Ideal Gas Law (\(PV = \text{constant}\)). When the volume increases, the pressure must decrease to maintain this constant product. Understanding these relationships helps us solve problems related to changes in gas properties.