Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

You may have noticed that when you compress the air in a bicycle pump, the body of the pump gets warmer. (a) Assuming the pump and the air in it comprise the system, what is the sign of \(w\) when you compress the air? (b) What is the sign of \(q\) for this process? (c) Based on your answers to parts (a) and (b), can you determine the sign of \(\Delta E\) for compressing the air in the pump? If not, what would you expect for the sign of \(\Delta E\) ? What is your reasoning? [Section 5.2\(]\)

Short Answer

Expert verified
For the process of compressing air in a bicycle pump: (a) the sign of work (w) is negative since work is being done on the system, (b) the sign of heat (q) is negative since heat is being removed from the system, and (c) the sign of ∆E is generally positive due to the increased kinetic energy of the molecules as air is compressed, despite heat transfer losses.

Step by step solution

01

(a) Determine the sign of w when compressing air.

Since we are compressing the air in the bicycle pump, work is being done on the system (air inside the pump). According to the definition, when work is done on the system, the work (w) is negative. Therefore, the sign of w is negative.
02

(b) Determine the sign of q for this process.

As we compress the air in the bicycle pump and the pump gets warm, we observe that heat is being transferred from the system (air) to the surroundings (pump body). Since heat is being removed from the system, the sign of heat (q) is negative.
03

(c) Determine the sign of ΔE and explain your reasoning.

We can use the thermodynamic relation \(ΔE = q + w\) to determine the sign of ΔE. In this case, both q and w are negative, which means \(ΔE\) is negative. However, we should also consider the physical process of compression. When air is compressed, its internal energy, mainly due to the increased kinetic energy of the molecules, should increase. This results in a positive change in internal energy (∆E). In reality, we could suppose that due to heat transfer some part of the internal energy increase is lost, but our sign for ∆E would still remain positive in general. So, to summarize, for the process of compressing air in the pump, \(w < 0\), \(q < 0\), and \(\Delta E > 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Work and Energy
Work and energy are central concepts in thermodynamics and understanding them can help explain the changes we observe in physical systems.
When you compress air in a bicycle pump, work is done on the air, which is defined as the system here. In thermodynamics, the sign convention is essential. If work is done on the system, the energy enters the system, which typically should increase its energy.
However, when following the usual sign conventions, we mark work done on the system as negative. This may seem counterintuitive, but in physics, this negative sign signifies the work input that doesn't naturally flow—it is being forced upon the system.

It's helpful to think of this with the formula for calculating work:
  • When work is positive, energy is leaving the system.
  • When work is negative, energy is entering the system.

In practical terms, compressing the bicycle pump requires effort, and thus work is performed to push the air molecules closer together, contrary to their natural motion of spreading apart. Thus, the air receives energy through this process.
Heat Transfer
Heat transfer describes how thermal energy moves from one place to another. In our example of compressing air in a bicycle pump, the air and pump act as a system where heat transfer dynamics are evident.

During compression:
  • Some kinetic energy of the air molecules is converted into thermal energy.
  • This increase in thermal energy can partially flee the system as heat to the surroundings.

This explains why the pump body becomes warm—some of the heat generated is transferred from the air (the system) to the pump (the surroundings).
According to thermodynamic sign conventions, when heat leaves the system, the variable for heat, denoted as \(q\), becomes negative.

Due to this heat being lost, energy that could have contributed to increasing the air's temperature ends up warming the pump externally.
First Law of Thermodynamics
The First Law of Thermodynamics is about the conservation of energy, which states that energy cannot be created or destroyed, only transformed or transferred. This principle is key to solving the problem in the pump exercise.

In thermodynamics, this law is often expressed as:\[ \Delta E = q + w \]
  • \( \Delta E \) is the change in internal energy.
  • \( q \) is the heat exchanged between the system and surroundings.
  • \( w \) is the work done by or on the system.

Let's apply this principle: - In our exercise, both work done (\(w\)) and heat transferred (\(q\)) are negative. - According to this equation, a negative \( q \) and \( w \) would suggest a decrease in internal energy, yet when you compress the air, the real scenario involves an increase due to the kinetic energy boost of the molecules.

The interplay between work done on the system and heat loss results in different contributions to internal energy changes. Although both \( q \) and \( w \) are negative, the actual energy increase, due to compression, means \( \Delta E \) is greater than zero. This showcases how work input can be more significant than heat loss in modifying a system's energy during processes such as compression.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: (a) For each of these substances, calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Calculate the fuel value in \(\mathrm{kJ} / \mathrm{g}\) for each of these compounds. \((\mathrm{c})\) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of AgCl. (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

The decomposition of slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}(s),\) into lime, \(\mathrm{CaO}(s),\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) at constant pressure requires the addition of \(109 \mathrm{~kJ}\) of heat per mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\). (a) Write a balanced thermochemical equation for the reaction. (b) Draw an enthalpy diagram for the reaction.

Identify the force present and explain whether work is being performed in the following cases: (a) You lift a pencil off the top of a desk. (b) A spring is compressed to half its normal length.

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix C, calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s):\) $$ \begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ \Delta H^{\circ}=-127.2 \mathrm{~kJ} \end{aligned} $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free