Question

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.110 m\(^3\) of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 m\(^3\). If the temperature remains constant, what is the final value of the pressure?

Short answer

The final pressure is 0.100 atm.

Step-by-step solution

Identify the Known Variables

First, note down the given information from the problem. The initial volume of the gas is \( V_1 = 0.110 \, \text{m}^3 \), and the initial pressure is \( P_1 = 0.355 \, \text{atm} \). The final volume of the gas once the piston is pulled out is \( V_2 = 0.390 \, \text{m}^3 \). The temperature is constant, so we are assuming an isothermal process.

Use Boyle’s Law

Since the temperature is constant, we use Boyle’s Law, which states that \( P_1 \times V_1 = P_2 \times V_2 \). This can be rearranged to find the final pressure: \( P_2 = \frac{P_1 \times V_1}{V_2} \).

Substitute the Known Values

Substitute the known values into the equation from Step 2: \( P_2 = \frac{0.355 \, \text{atm} \times 0.110 \, \text{m}^3}{0.390 \, \text{m}^3} \).

Calculate the Final Pressure

Perform the calculation: \( P_2 = \frac{0.03905 \, \text{atm} \cdot \text{m}^3}{0.390 \, \text{m}^3} = 0.1001282 \, \text{atm} \). Rounding the answer, the final pressure is \( 0.100 \, \text{atm} \).

Key concepts

Isothermal Process

An isothermal process is a fundamental concept in thermodynamics where the temperature of the system remains constant. This is crucial because even though the pressure and volume might change, the energy related to temperature remains stable. To ensure this, heat must be exchanged with the environment. In our exercise scenario, the air inside the cylindrical tank undergoes an expansion by pulling the piston out. Since the temperature stays the same, energy that might have been gained or lost due to volume change is compensated by the exchange of heat with the surroundings. This keeps the internal energy of the gas constant, which is a key characteristic of an isothermal process.

Pressure-Volume Relationship

The pressure-volume relationship, described by Boyle's Law, is pivotal in understanding gas behaviors. Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.In mathematical terms, it's expressed as \( P \times V = \, \text{constant} \), or \( P_1 \times V_1 = P_2 \times V_2 \). This equation illustrates that when you increase the volume (like pulling a piston out), the pressure must decrease, provided the temperature does not change.In our given problem, we had an increase in volume from 0.110 m³ to 0.390 m³. As a result, Boyle's Law helps us calculate and understand the resultant decrease in pressure from 0.355 atm to approximately 0.100 atm.

Ideal Gas Law

The ideal gas law is an umbrella equation in thermodynamics that brings together pressure, volume, and temperature for an ideal gas. It is represented as \( PV = nRT \), where:

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

In the context of an isothermal process, changes in pressure and volume do not alter the product of \( n \), \( R \), and \( T \), as temperature remains constant. Although our problem specifically centers on Boyle's Law, it is a component of the broader ideal gas law, which helps us in partially understanding varied gas behaviors under different conditions. Understanding how these equations interconnect provides clarity into gas dynamics within closed systems.