Problem 36 - Chemistry An Atoms First Approach

Question

A piston performs work of \(210 . \mathrm{L}\). \(\mathrm{atm}\) on the surroundings, while the cylinder in which it is placed expands from \(10. \mathrm{L}\) to \(25 \mathrm{L}\). At the same time, \(45 \mathrm{J}\) of heat is transferred from the surroundings to the system. Against what pressure was the piston working?

Step-by-Step Solution

Answer

The pressure against which the piston was working is 14 atm.

Step 1: Convert the work done from L.atm to J

We are given the work done by the piston as 210 L.atm. To work with SI units across the problem, we need to convert this value to Joules. We can use the following conversion factor: 1 L.atm = 101.325 J So, the work done in Joules is: \(W = 210 \ L \cdot atm \times \frac{101.325 \ J}{1 \ L \cdot atm} = 21278.25 \ J\)

Step 2: Apply the first law of thermodynamics

The first law of thermodynamics states that the change in internal energy ΔU of a system is equal to the heat transferred to the system (q) minus the work done by the system (W). This can be written as: \(\Delta U = q - W\) We are given the heat transferred from the surroundings to the system as 45 J, so q = 45 J. We calculated the work done by the system in Joules as 21278.25 J. Now we can find the change in internal energy: \(\Delta U = 45 \ J - 21278.25 \ J = -21233.25 \ J\)

Step 3: Calculate the work done using the pressure and volume changes

The work done by a piston is related to the pressure against which it works and the change in volume of the cylinder. The formula for the work done by the piston is given by: \(W = -P \cdot \Delta V\) Here, ΔV is the change in volume of the cylinder, which is the final volume minus the initial volume: \(\Delta V = V_f - V_i = 25 \ L - 10 \ L = 15 \ L\) We know that the work done by the piston (W) is 21278.25 J. With the volume change, we can now find the pressure against which the piston was working: \(21278.25 \ J = -P \cdot (15 \ L)\)

Step 4: Calculate the pressure

Now, we can solve for P: \(P = -\frac{21278.25 \ J}{15 \ L}\) \(P = -1418.55 \frac{J}{L}\) The negative sign indicates that the pressure was applied in the opposite direction of the expansion, which is expected for a piston working against a pressure. To report the pressure as a positive value, we can take the absolute value: \(P = 1418.55 \frac{J}{L}\)

Step 5: Convert the pressure back to atm

Finally, we need to convert the pressure back to atm. We know that 1 atm is equivalent to 101.325 J/L. So, we can use the conversion factor: \(P_\text{atm} = 1418.55 \frac{J}{L} \times \frac{1 \ atm}{101.325 \ J/L} = 14 \ atm\) The pressure against which the piston was working is 14 atm.

Key Concepts

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. It tells us how energy is transferred within systems and how it affects matter. The core laws of thermodynamics lay the foundation for understanding concepts such as energy conservation, heat transfer, and the efficiency of engines.

Specifically, the first law of thermodynamics, also known as the law of energy conservation, asserts that energy cannot be created or destroyed in an isolated system. Instead, energy can only change forms or be transferred from one part of the system to another. This law is crucial in solving chemistry problems that involve heat and work, as it allows us to account for all energy changes within a reaction or process.

Work and Energy in Chemistry

In the context of chemistry, work is often related to the expansion or compression of gases. Energy, in a broad sense, includes both the kinetic energy of particles and the potential energy stored in chemical bonds. Whenever a chemical reaction occurs, energy is either absorbed or released, demonstrating the interplay between work and energy.

When we talk about a system doing work on its surroundings, such as a gas expanding against a piston, we are discussing how the system uses its energy to cause a displacement. In these scenarios, physicists and chemists use the concept of work to quantify the energy transferred through mechanical means.

Pressure-Volume Work

Pressure-volume work, commonly denoted as P-V work, refers to the work done by or on a system as its volume changes under pressure. This is the most common type of work in chemistry, especially with gases where changes in volume can occur against atmospheric pressure.

The work done during the expansion or compression of the gas is calculated by multiplying the pressure against which the gas is working by the change in volume. It's important to note that work is considered positive when the system does work on its surroundings and negative when work is done on the system. This is consistent with the first law of thermodynamics and provides a way to calculate energy changes during chemical reactions.

Internal Energy Change

Internal energy change, symbolized as \(\Delta U\), is a key concept when discussing energy exchanges in thermodynamics. This change reflects the sum of all energy transfers, including heat and work, into or out of a system. When a system undergoes any process, its internal energy can increase or decrease.

In our exercise, when the heat is transferred to the system and work is performed by it, the internal energy change can be calculated using the first law of thermodynamics formula: \(\Delta U = q - W\). Here, \(q\) represents the heat added to the system, and \(W\) signifies the work done by the system. This equation helps us understand how the energy balance within a system is maintained or altered through various processes.