(a) What is meant by the term system in thermodynamics? (b) What is a closed system? (c) What do we call the part of the universe that is not part of the system?

Short Answer

Expert verified
(a) In thermodynamics, a "system" refers to a specific portion of the universe being observed or studied, surrounded by a boundary separating it from its surroundings. (b) A "closed system" is a type of thermodynamic system that can exchange energy with its surroundings but does not exchange matter. (c) The part of the universe not part of the system being studied is called the "surroundings."

Step by step solution

01

(a) Definition of System in Thermodynamics

In thermodynamics, a "system" is a specific portion of the universe that is being observed or studied. It can be any specific region or object, like a container of gas, a metal, or a chemical reaction. The system is surrounded by a boundary that separates it from its surroundings (the rest of the universe). The boundary could be physical or imaginary. The main focus of the thermodynamic analysis is on the internal processes and properties of the system.
02

(b) Definition of Closed System

A "closed system" is a type of thermodynamic system that can exchange energy (in the form of work and heat) with its surroundings but does not exchange matter. In other words, no mass or particles can move across the boundary, and the total amount of matter within the closed system remains constant during the process.
03

(c) Name for the Part of the Universe Not Part of the System

The part of the universe that is not part of the system being studied is called the "surroundings." In thermodynamic analysis, the surroundings include everything that is outside the system, encompassing the rest of the universe. Both the system and its surroundings together constitute the "universe" in a thermodynamic context.

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!

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

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 \(\mathrm{CaC}_{2}(s):\) $$ \begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ \Delta H^{\circ}=-127.2 \mathrm{~kJ} \end{aligned} $$

Given the data $$ \begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g) & & \Delta H=+180.7 \mathrm{~kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) & & \Delta H=-113.1 \mathrm{~kJ} \\ 2 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H &=-163.2 \mathrm{~kJ} \end{aligned} $$ use Hess's law to calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) $$

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J} ;(\mathbf{b})\) a system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system; (c) the system absorbs \(7.25 \mathrm{~kJ}\) of heat from the surroundings while its volume remains constant (assume that only \(P-V\) work can be done).

Consider the following reaction: $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) \quad \Delta H=-1204 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(3.55 \mathrm{~g}\) of \(\mathrm{Mg}(s)\) reacts at constant pressure. (c) How many grams of \(\mathrm{MgO}\) are produced during an enthalpy change of \(-234 \mathrm{~kJ}\) ? (d) How many kilojoules of heat are absorbed when \(40.3 \mathrm{~g}\) of \(\mathrm{MgO}(s)\) is decomposed into \(\mathrm{Mg}(s)\) and \(\mathrm{O}_{2}(g)\) at constant pressure?

The heat of combustion of fructose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) is \(-2812 \mathrm{~kJ} / \mathrm{mol}\). If a fresh golden delicious apple weighing 4.23 oz \((120 \mathrm{~g})\) contains \(16.0 \mathrm{~g}\) of fructose, what caloric content does the fructose contribute to the apple?

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