Problem 31 - Chemistry and Chemical Reactivity

Question

A volume of 1.50 L of argon gas is confined in a cylinder with a movable piston under a constant pressure of \(1.22 \times 10^{5}\) Pa. When \(1.25 \mathrm{kJ}\) of energy in the form of heat is transferred from the surroundings to the gas, the internal energy of the gas increases by 1.11 kJ. What is the final volume of argon gas in the cylinder?

Step-by-Step Solution

Answer

The final volume of argon gas is 2.65 L.

Step 1: Identify the Given Values and Requirements

We have the initial volume of the gas, \( V_i = 1.50 \) L, and the constant pressure \( P = 1.22 \times 10^5 \) Pa. Energy added to the gas is \( q = 1.25 \) kJ and the increase in internal energy is \( \Delta U = 1.11 \) kJ. We need to find the final volume \( V_f \) of the argon gas.

Step 2: Apply the First Law of Thermodynamics

The First Law of Thermodynamics states: \( \Delta U = q - W \), where \( \Delta U \) is the change in internal energy, \( q \) is the heat added, and \( W \) is the work done by the system. Here, \( \Delta U = 1.11 \) kJ and \( q = 1.25 \) kJ.

Step 3: Calculate the Work Done by the Gas

Since \( W = q - \Delta U \), we plug in the values: \( W = 1.25 \text{ kJ} - 1.11 \text{ kJ} = 0.14 \text{ kJ} \). Convert this to joules: \( W = 140 \text{ J} \).

Step 4: Use the Work Formula for Gas Expansion

For isobaric processes (constant pressure), work done \( W \) is given by \( W = P \Delta V = P(V_f - V_i) \). Reorganize to find \( \Delta V: \Delta V = \frac{W}{P} \).

Step 5: Calculate the Change in Volume \( \Delta V \)

Plug in the values: \( \Delta V = \frac{140 \text{ J}}{1.22 \times 10^5 \text{ Pa}} = 1.15 \times 10^{-3} \text{ m}^3 \). Convert units: \( 1 \text{ m}^3 = 1000 \text{ L} \), so \( \Delta V = 1.15 \text{ L} \).

Step 6: Determine the Final Volume \( V_f \)

Using the relationship \( V_f = V_i + \Delta V \), we find \( V_f = 1.50 \text{ L} + 1.15 \text{ L} = 2.65 \text{ L} \).

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a foundational principle in thermodynamics. It is often written as \( \Delta U = q - W \). This law states that the change in internal energy \( \Delta U \) of a system is equal to the heat \( q \) added to the system minus the work \( W \) done by the system. This concept is very important as it shows the relationship between heat transfer and work done within a system.
Understanding this law helps in analyzing how energy is conserved within closed systems. Energy cannot be created or destroyed, only transformed or transferred from one form to another. For students, it's essential to grasp that when heat is added to a gas, it can either increase the gas’s internal energy or do work by expanding against the walls of its container.
In our exercise, the gas receives 1.25 kJ of heat from the surroundings, resulting in an internal energy increase of 1.11 kJ. The remaining energy is used to perform work. This is a direct application of the First Law of Thermodynamics.

Isobaric Process

An isobaric process is a thermodynamic process that takes place at a constant pressure. In an isobaric process, while the pressure remains unchanged, other properties like volume and temperature can vary.
In the context of our original exercise, the process involving the argon gas can be classified as isobaric because the pressure remains constant at \(1.22 \times 10^5\) Pa. This constancy of pressure makes the calculations more straightforward and helps in using specific formulas such as: \( W = P \Delta V \) (where \( W \) is the work done, \( P \) is the constant pressure, and \( \Delta V \) is the change in volume).
This scenario simplifies many real-world applications, like in engines or syringes, where maintaining consistent pressure is vital.

Internal Energy

Internal energy is the sum of all kinetic and potential energies of the particles in a system. It is a state function, meaning that it depends only on the current state of the system, not on how the system reached that state. Any change in internal energy \( \Delta U \) is due to heat added and work done as per the First Law of Thermodynamics.
For our exercise, the internal energy of the argon gas increases by 1.11 kJ when heat is added. This increase indicates that not all the added heat is used for doing work. Some of the energy remains inside, increasing the kinetic energy of gas molecules and thus, its temperature.
In practical terms, internal energy provides a measure of a substance's ability to perform work or transfer heat. Knowing the internal energy change can help determine the system's behavior under specific conditions.

Gas Laws

Gas laws describe how gases behave in different conditions of pressure, volume, and temperature. They are mathematical relationships that predict the behavior of an ideal gas.
In our problem, the constancy of pressure suggests the use of the isobaric condition. Here, we work with equations that derive from the Ideal Gas Law \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the amount of substance (in moles), \( R \) is the ideal gas constant, and \( T \) is temperature.
The focus in this exercise is mainly on how volume changes when energy (heat) is added under constant pressure. This correlates to gas expansion where heat energy enables the gas particles to move more freely, causing an increase in volume. Understanding how these variables interact can be critically important in fields like chemistry, engineering, and environmental science.