A gas sample is heated in a cylinder, using \(375 \mathrm{~kJ}\) of heat. At the same time a piston compresses the gas, using \(645 \mathrm{~kJ}\) of work. What is the change in internal energy of the gas during this process?

Short Answer

Expert verified
The change in internal energy of the gas is \( -270 \mathrm{~kJ} \)

Step by step solution

01

Understand the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. It can be expressed mathematically as \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added, and \(W\) is the work done by the system.
02

Identify the Heat and Work Values

From the problem, we know that the heat \(Q\) added to the gas is \(375 \mathrm{~kJ}\) and the work \(W\) done on the gas (since the piston is compressing it) is \(645 \mathrm{~kJ}\).
03

Calculate the Change in Internal Energy

Using the formula from Step 1, substitute the values of \(Q\) and \(W\) into the equation: \[\Delta U = Q - W = 375 \mathrm{~kJ} - 645 \mathrm{~kJ}\]. Calculating this gives us the change in internal energy.
04

Find the Result and Consider the Sign

Performing the subtraction we get \[\Delta U = 375 \mathrm{~kJ} - 645 \mathrm{~kJ} = -270 \mathrm{~kJ}\]. The negative sign indicates that the internal energy has decreased, which is consistent with the work being done on the system (compression).

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.

Internal Energy
The concept of internal energy is central to understanding thermodynamics and the behaviour of systems. It encompasses the total energy contained within a system, which is the sum of the kinetic and potential energies of all particles within the system. The internal energy of a system changes when the system undergoes a physical or chemical change.

In the exercise given, the gas within the cylinder has internal energy that can change due to heat transfer or work done on the gas. If we consider the microscopic view, when heat is applied to the gas, molecules move faster, thereby increasing the system's kinetic energy. Conversely, when work is done on the gas, such as by a piston compressing it, the potential energy between molecules can increase due to the decreased volume and increased intermolecular forces. Understanding these changes is pivotal for solving problems related to the internal energy.
Heat Transfer
Heat transfer refers to the process by which thermal energy is exchanged between physical systems or between different parts within a system. It is crucial to the field of thermodynamics as it explains how and why energy transfers occur. Heat can be transferred in three primary ways: conduction, convection, and radiation.

In the context of the described problem, heat is transferred to the gas when it is heated, which increases the internal energy of the gas. This addition of heat is quantified as positive because it represents energy entering the system. Always recall that in thermodynamics, we follow the convention of considering 'heat absorbed by the system' as positive and 'heat released by the system' as negative. This is important for correctly applying the First Law of Thermodynamics.
Work Done on Gas
Work done on a gas represents the energy transferred to or from the gas when an external force, such as a moving piston, acts upon it. If the gas is compressed, as we have in our exercise example, the volume of the gas decreases, and work is considered to be done on the gas.

It's important to note that, according to the sign convention in thermodynamics, the work done by the system is negative if the system does the work (like when a gas expands and does work on the surroundings), and it's positive if work is done on the system (like when the gas is compressed). The more work that is done, the more the internal energy is affected, in this case, reduced, since the gas is being compressed.
Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships and conversions between heat and other forms of energy. The First Law of Thermodynamics, highlighted in the exercise, describes one of the fundamental principles of this science: the conservation of energy. It expresses that energy cannot be created or destroyed in an isolated system, only transformed.

The exercise presented showcases the practical application of the First Law by calculating the change in internal energy of the gas. As students tackle such problems, they should remember that the First Law provides a balance sheet for energy exchanges, understanding that changes in internal energy are the result of both heat transfer and work interactions with the surroundings. Grasping this concept, students will be better equipped to solve a variety of thermodynamic problems with confidence.

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

When \(25.0 \mathrm{~g}\) of a metal at a temperature of \(90.0^{\circ} \mathrm{C}\) is added to \(50.0 \mathrm{~g}\) of water at \(25.0^{\circ} \mathrm{C}\), the water temperature rises to \(29.8^{\circ} \mathrm{C}\). The specific heat capacity of water is \(4.184 \mathrm{~J}-\left({ }^{\circ} \mathrm{C}\right)^{-1} \mathrm{~g}^{-1}\). What is the specific heat capacity of the metal?

Calculate the lattice enthalpy of solid potassium bromide, \(\mathrm{KBr}(\mathrm{s}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{Br}^{-}(\mathrm{g})\), from the following information: $$ \begin{aligned} &\Delta H_{i}{ }^{\circ}\left(\mathrm{KBr}, \text { s) }=-394 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right. \\ &\Delta H_{i}{ }^{\circ}(\mathrm{K}, \mathrm{g})=+89.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ First ionization energy of \(\mathrm{K}(\mathrm{g})=+425.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) \(\Delta H_{\text {vap }}{ }^{\circ}\left(\mathrm{Br}_{2}, \mathrm{l}\right)=+30.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Br-Br bond dissociation cuthalpy \(=+192.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Electron attachment to \(\mathrm{Br}(\mathrm{g})\) : $$ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{Be}^{-}(\mathrm{g}), \quad \Delta H^{*}=-331.0 \mathrm{~kJ} $$

Write the thermochemical equations that give the values of the standard enthalpies of formation for (a) \(\mathrm{CH}_{2} \mathrm{COOH}\) (I); (b) \(\mathrm{SO}_{2}\) (g); (c) \(\mathrm{CO}_{2}\) (g).

The internal energy of a system increased by \(400 \mathrm{~J}\) when it absorbed \(600 \mathrm{~J}\) of heat. (a) Was work done by or on the system? (b) How much work was done?

A gas in a cylinder was placed in a heater and gained \(7000 \mathrm{~kJ}\) of heat. If the cylinder increased in volume from \(700 \mathrm{~mL}\) to \(1450 \mathrm{~mL}\) against an atmospheric pressure of 750 Torr during this process, what is the change in internal energy of the gas in the cylinder?

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