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Chapter 19: Problem 15

A thermodynamic system undergoes processes in which \(|Q|=100 \mathrm{~J}\) and \(|W|=300 \mathrm{~J}\). Find \(Q\) and \(W,\) including whether the quantity is positive or negative, if the change in internal energy is (a) \(+400 \mathrm{~J},(\mathrm{~b})+200 \mathrm{~J},(\mathrm{c})-200 \mathrm{~J},\) and \((\mathrm{d})-400 \mathrm{~J} .\)

Short Answer

Expert verified
Solution for the exercise includes 4 different scenarios involving the same values of \(Q = 100 J\) and \(W = 300 J\) but different values for the change in internal energy \(螖U\). The four solutions are (a) \(Q = +100 J, W = -300 J\), (b) \(Q = +100 J, W = 0 J\), (c) \(Q = -100 J, W = +300 J\), and (d) \(Q = -100 J, W = +300 J\).

Step by step solution

01

Determine the Positive or Negative for Heat and Work (Case a)

Given that \(螖U = +400J\). Knowing that if a system has an increase in internal energy, it has absorbed heat, therefore, \(Q\) is positive as \(Q=+100J\). Also, work is done on the system (as \(螖U = Q - W\) and \(Q < 螖U\), therefore, we can conclude that \(W\) is negative as \(W = -300J\).
02

Determine the Positive or Negative for Heat and Work (Case b)

Given that \(螖U = +200J\). The system has absorbed less heat than in case (a), So, \(Q\) is still positive: \(Q=+100J\). Considering that \(螖U = Q - W\) and \(Q = 螖U\), hence, no work is done, i.e., \(W = 0J\).
03

Determine the Positive or Negative for Heat and Work (Case c)

Given that \(螖U = -200J\). The internal energy of the system decreases which means the system has done work, therefore, \(Q\) is negative or \(Q = -100J\). The work has been done by the system (as, \( 螖U = Q - W \) and \( |螖U| > |Q| \)), hence, \(W\) is positive i.e., \(W = +300J\).
04

Determine the Positive or Negative for Heat and Work (Case d)

Given \(螖U = -400J\). The system has done more work than in case (c), So \(Q\) remains negative: \(Q = -100J\), but the work increases. As, \(螖U = Q - W\), \(W > Q\) implies that the work done by the system is \(W = +300J\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Energy
Internal energy is a central concept in thermodynamics, representing the total energy contained within a system. It includes:
  • The kinetic energy of particles moving and vibrating
  • The potential energy associated with the intermolecular forces
Changes in internal energy occur when a system exchanges energy with its surroundings.
When a problem states a change in internal energy, denoted as \( \Delta U \), it describes the energy increase or decrease.
A positive \( \Delta U \) means energy has been absorbed, while a negative \( \Delta U \) indicates energy loss.
Understanding how internal energy is affected by heat and work helps in analyzing the system鈥檚 behavior and energy dynamics.
Heat Transfer
Heat transfer is the movement of thermal energy between a system and its surroundings due to a temperature difference.
This energy flow can occur in various forms, including conduction, convection, and radiation.
In thermodynamics, heat is denoted by \( Q \), and its sign indicates direction:
  • Positive \( Q \): The system absorbs heat.
  • Negative \( Q \): The system releases heat.
The value of heat transfer can greatly affect the internal energy.
Understanding heat transfer helps to determine how a system responds to temperature changes and external influences.
Work Done
Work done refers to the energy transferred to or from a system by mechanical or external forces. In thermodynamics, it is denoted by \( W \), and can manifest through:
  • Compression or expansion of gases
  • Movement of components within the system
The direction of work transfer is marked by its sign:
  • Positive \( W \): Work done by the system, losing energy.
  • Negative \( W \): Work done on the system, gaining energy.
Work, alongside heat, helps determine the changes in a system鈥檚 internal energy.
Grasping the impact of work done is vital for assessing how systems interact with external forces and conditions.
Laws of Thermodynamics
The laws of thermodynamics explain the principles governing energy transfer within systems.
  • The First Law of Thermodynamics, also known as the Law of Energy Conservation, states: \( \Delta U = Q - W \). This law emphasizes that energy can neither be created nor destroyed, only transferred or transformed.
    Using this law helps clarify how much energy a system gains or loses.
  • The Second Law of Thermodynamics introduces the concept of entropy. It suggests that energy transfers are not 100% efficient, and systems naturally progress towards disorder.
    This law explains why some energy is lost as heat during processes.
Knowing these laws builds a foundation to analyze various thermodynamic systems, enabling predictions about energy flow and transformations.

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Most popular questions from this chapter

An experimenter adds \(970 \mathrm{~J}\) of heat to \(1.75 \mathrm{~mol}\) of an ideal gas to heat it from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C}\) at constant pressure. The gas does \(+223 \mathrm{~J}\) of work during the expansion. (a) Calculate the change in internal energy of the gas. (b) Calculate \(\gamma\) for the gas.

A player bounces a basketball on the floor, compressing it to \(80.0 \%\) of its original volume. The air (assume it is essentially \(\mathrm{N}_{2}\) gas) inside the ball is originally at \(20.0^{\circ} \mathrm{C}\) and 2.00 atm. The ball's inside diameter is \(23.9 \mathrm{~cm}\). (a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal. (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

\(\cdot\) A cylinder contains \(0.0100 \mathrm{~mol}\) of helium at \(T=27.0^{\circ} \mathrm{C}\) (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C} ?\) Draw a \(p V\) -diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

Comparing Thermodynamic Processes. In a cylinder, \(1.20 \mathrm{~mol}\) of an ideal monatomic gas, initially at \(3.60 \times 10^{5} \mathrm{~Pa}\) and \(300 \mathrm{~K},\) expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a \(p V\) -diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

A quantity of \(2.00 \mathrm{~mol}\) of a monatomic ideal gas undergoes a compression during which the volume decreases from \(0.0800 \mathrm{~m}^{3}\) to \(0.0500 \mathrm{~m}^{3}\) while the pressure stays constant at a value of \(1.80 \times 10^{4} \mathrm{~Pa}\). (a) What is the work \(W ?\) Is work done by the gas or on the gas? (b) What is the heat flow \(Q\) ? Does heat enter or leave the gas? (c) What is the internal energy change for the gas? Does the internal energy of the gas increase or decrease?
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