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What do the terms system and surroundings mean? What is the difference between an isolated system and a closed system? What is the universe in terms of thermodynamics?

Short Answer

Expert verified
The 'system' is the part of the universe being studied, while 'surroundings' are everything outside the system. An isolated system exchanges neither energy nor matter with its surroundings, whereas a closed system can exchange energy but not matter. The 'universe' in thermodynamics refers to the system plus surroundings with no external interaction.

Step by step solution

01

Defining System

In thermodynamics, a 'system' refers to the part of the universe that is under consideration or study, typically defined by physical boundaries. It is separated from the rest by a boundary that may be real or imaginary, fixed or movable.
02

Defining Surroundings

The 'surroundings' or 'environment' consists of everything outside the system. It is the part of the universe that is not contained within the system's boundaries, and it can exchange heat, work, or matter with the system.
03

Differentiating an Isolated and Closed System

An isolated system cannot exchange energy (heat or work) or matter with its surroundings, essentially making it closed off from the rest of the universe. In contrast, a closed system can exchange energy but not matter with its surroundings, having impermeable boundaries that prevent matter exchange.
04

Understanding the Universe in Thermodynamics

In thermodynamics, the 'universe' is considered as the combination of the system and its surroundings. It encompasses everything, implying that there is nothing outside of the universe to exchange energy or matter with.

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Key Concepts

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

Isolated System
An isolated system in thermodynamics is one that stands alone, quarantined from the rest of the universe. Imagine a perfectly insulated container where no heat, light, or sound can escape or enter. In such a system, no transfer of energy or matter occurs with the surroundings. This is an idealized concept because in reality, perfect isolation is tough to achieve. Even the most insulated systems we can create, like a thermos, will eventually experience some form of energy transfer over time.

Understanding the nature of isolated systems is crucial in theoretical scenarios. It allows scientists to simplify calculations by assuming that the system's total energy remains constant since it is unaffected by external factors. This concept is pivotal when discussing the first law of thermodynamics, also known as the law of energy conservation. In an equation, it's represented as \( \Delta E_{system} = Q - W = 0 \) for an isolated system, where \( Q \) refers to heat and \( W \) denotes work.
Closed System
On the flip side, a closed system is slightly more connected with the rest of the world. Although no atoms or molecules can leave or enter, energy is free to cross the boundary. It's like a pot with a lid on a stove: while the lid prevents the soup's ingredients from spilling out, heat from the stove can still pass into the pot and cook the soup.

A home's heating system is a real-world example of a closed system. Heat can be added through the furnace or taken away through air conditioning, but the air inside the home (assuming no windows or doors are opened) remains the same. This system is often analyzed using the first law of thermodynamics in the form \( \Delta E_{system} = Q - W \), where changes in the system's energy are due to heat transfer \( Q \) and work done \( W \), but mass remains constant. Understanding closed systems helps us design more efficient engines, heating and cooling systems, and even model Earth's atmosphere.
Thermodynamics Universe
When thermodynamicists speak of the universe, they're not just referring to the vast expanse of stars and galaxies. It’s the all-encompassing term that includes the system being studied and everything around it—its surroundings. This concept is a bit abstract but is crucial for understanding that within this 'universe,' the laws of thermodynamics operate under the assumption that energy and matter cannot magically disappear or appear; they can only be transformed or relocated.

If you mix both the system and surroundings, you get the 'thermodynamics universe.' For instance, if a cup of hot tea (the system) is sitting in a room (surroundings), the 'thermodynamics universe' includes both the tea and the room. According to the conservation of energy, the energy lost by the tea as it cools must be equal to the energy gained by the surrounding air in the room. There's a practical outcome of this: we have energy budgets for isolated areas, like when calculating the energy needs for heating a home or running a refrigerator. In these cases, understanding the thermodynamic universe allows us to predict and control the direction and flow of energy.

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Most popular questions from this chapter

Toluene, \(\mathrm{C}_{7} \mathrm{H}_{8}\), is used in the manufacture of explosives such as TNT (trinitrotoluene). A \(1.500 \mathrm{~g}\) sample of liquid toluene was placed in a bomb calorimeter along with excess oxygen. When the combustion of the toluene was initiated, the temperature of the calorimeter rose from \(25.000^{\circ} \mathrm{C}\) to \(26.413^{\circ} \mathrm{C}\). The products of the combustion were \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l),\) and the heat capacity of the calorimeter was \(45.06 \mathrm{~kJ}^{\circ} \mathrm{C}^{-1}\) (a) Write the balanced chemical equation for the reaction in the calorimeter. (b) How many joules were liberated by the reaction? (c) How many joules would be liberated under similar conditions if 1.000 mol of toluene was burned?

How much heat, in joules and in calories, must be removed from \(1.75 \mathrm{~mol}\) of water to lower its temperature from \(25.0^{\circ} \mathrm{C}\) to \(15.0^{\circ} \mathrm{C}^{?}\)

Given the following thermochemical equations, $$ 3 \mathrm{Mg}(s)+2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+3 \mathrm{H}_{2}(g) $$ \(\Delta H^{\circ}=-371 \mathrm{~kJ}\) $$ \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g) \quad \Delta H^{\circ}=-46 \mathrm{~kJ} $$ calculate \(\Delta H^{\circ}\) (in kilojoules) for the following reaction: $$ 3 \mathrm{Mg}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}(s) $$

Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 \(\left(\mathrm{CFCl}_{3}\right)\) and Freon-12 \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right),\) which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the "ozone hole" that develops at the end of the Antarctic winter. In other parts of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earth's inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, \(\mathrm{O}_{3}\). Possible reactions are $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{ClO}(g) & \Delta H^{\circ}=-126 \mathrm{~kJ} \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) & \Delta H^{\circ}=-268 \mathrm{~kJ} \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) & \Delta H^{\circ}=? \end{array} $$ The \(\mathrm{O}\) atoms in the second equation come from the breaking apart of \(\mathrm{O}_{2}\) molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of \(\Delta H^{\circ}\) (in kilojoules) for the last equation, the net reaction for the removal of \(\mathrm{O}_{3}\) from the atmosphere.

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