Problem 51 - University Physics with Modern Physics

Question

In a certain process, \(2.15 \times 10^{5} \mathrm{J}\) of heat is liberated by a system, and at the same time the system contracts under a constant external pressure of \(9.50 \times 10^{5} \mathrm{Pa}\) . The internal energy of the system is the same at the beginning and end of the process. Find the change in volume of the system. (The system is not an ideal gas.)

Step-by-Step Solution

Answer

The change in volume of the system is \(-0.0226 \, \mathrm{m^3}\).

Step 1: Understand the Given Problem

We know that the heat released by the system is \( Q = 2.15 \times 10^{5} \, \mathrm{J} \) and the external pressure is \( P = 9.50 \times 10^{5} \, \mathrm{Pa} \). Also, we are given that the internal energy change is zero, meaning \( \Delta U = 0 \). We are asked to find the change in volume.

Step 2: Apply the First Law of Thermodynamics

According to the first law of thermodynamics, \( \Delta U = Q - W \), where \( W \) is the work done by the system. Given that \( \Delta U = 0 \), we have: \[ 0 = Q - W \] This means \( W = Q \). Since heat is liberated, \( Q = -2.15 \times 10^{5} \, \mathrm{J} \). Therefore, \( W = -2.15 \times 10^{5} \, \mathrm{J} \).

Step 3: Calculate Work Done by the System

The work done by the system on its surroundings under constant pressure is given by \( W = P \Delta V \), where \( \Delta V \) is the change in volume. Since we have \( W = -2.15 \times 10^{5} \, \mathrm{J} \), we substitute to get:\[ -2.15 \times 10^{5} = 9.50 \times 10^{5} \times \Delta V \]

Step 4: Solve for Change in Volume

Isolate \( \Delta V \) in the equation:\[ \Delta V = \frac{-2.15 \times 10^{5}}{9.50 \times 10^{5}} \]Calculation:\[ \Delta V = -0.226 \times 10^{-1} \, \mathrm{m^3} = -0.0226 \, \mathrm{m^3} \] The change in volume is \(-0.0226 \, \mathrm{m^3}\). This negative sign indicates contraction.

Key Concepts

Internal Energy

Internal energy represents the total energy contained within a system. It encompasses all kinetic and potential energy of the particles within that system.
This concept is crucial in thermodynamics as it is a state function, meaning it depends solely on the current state of the system and not on how it arrived there.

• Internal energy is denoted by the symbol "U."
• If internal energy does not change over a process, we denote this by \( \Delta U = 0 \).

In the given exercise, the internal energy does not change. This implies that any energy entering or leaving the system as heat or work is balanced out. A crucial takeaway is that understanding changes in internal energy involves accounting for heat transfer and work done in processes.

Work Done

Work done in a thermodynamic process is the energy transferred by the system to its surroundings due to a force acting over a distance.
It can be visualized as the mechanism through which the system exchanges energy with its environment.

• In our scenario, work done is calculated using \( W = P \Delta V \), where \( P \) is the constant external pressure.
• Positive work indicates energy is added to the system, while negative work means energy is extracted.

The negative sign in work \( W = -2.15 \times 10^{5} \, \mathrm{J} \) implies that energy has been given away, typical of a system contracting. Recognizing whether work is positive or negative helps us comprehend the energetic dynamics in physical processes.

Heat Transfer

Heat transfer refers to the mechanism by which thermal energy is exchanged between systems or their surroundings. It plays a vital role in energy interactions in thermodynamics.
In simple terms, heat flows from hotter areas to cooler ones until thermal equilibrium is reached.

• The symbol "Q" denotes heat, quantified by the amount of energy transferred as heat.
• In the exercise, heat liberated means \( Q \) is negative since the system loses heat.

Understanding heat transfer helps us better explain the energy balance in a process. In this exercise, the liberated heat signifies the energy dispersed from the system during its contraction, impacting calculations of work and volume change.

Volume Change

Volume change describes the alteration in a system's volume due to internal pressure dynamics or external factors. It is intimately linked with work done in processes involving gases or liquids.
In thermodynamics, volume change is essential to gauge how much space a system occupies before and after a process.

• The change in volume \( \Delta V \) can be deduced from work formulas when pressure is known.
• The negative result \( \Delta V = -0.0226 \, \mathrm{m^3} \) illustrates the system's contraction.

This contraction confirms energy transferred during work done matches the heat energy lost. Volume changes hallmark the transition states of systems, reflecting physical transformation in enclosed settings.