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Consider the following reaction in a vessel with a movable piston. $$ \mathrm{X}(g)+\mathrm{Y}(g) \longrightarrow \mathrm{Z}(l) $$ As the reaction occurs, the system loses \(1185 \mathrm{~J}\) of heat. The piston moves down and the surroundings do \(623 \mathrm{~J}\) of work on the system. What is \(\Delta E ?\)

Short Answer

Expert verified
Answer: The change in internal energy (ΔE) of the system is -562 J.

Step by step solution

01

Recall the first law of thermodynamics formula

The first law of thermodynamics states the conservation of energy in a system: $$ \Delta E = q + w $$ Where: \(ΔE\) is the change in internal energy of the system, \(q\) is the heat transferred to the system, and \(w\) is the work done on the system.
02

Determine the heat transferred and work done

From the given information in the exercise, we know that: Heat lost: q = -1185 J (Since the system loses heat, we use a negative sign.) Work done on the system: w = 623 J
03

Calculate the change in internal energy

Use the first law of thermodynamics formula and plug in the values of q and w: $$ \Delta E = (-1185 \mathrm{J}) + (623 \mathrm{J}) $$
04

Solve for ΔE

Perform the arithmetic operation to find the change in internal energy: $$ \Delta E = -1185 \mathrm{J} + 623 \mathrm{J} = -562 \mathrm{J} $$ The internal energy change of the system is -562 J. Since this value is negative, it indicates that the system's internal energy has decreased during the reaction.

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Key Concepts

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

Change in Internal Energy
The change in internal energy, denoted as \(\Delta E\), represents the difference in a system's energy before and after a process occurs. While dealing with internal energy, it's crucial to understand that it encompasses all forms of energy within a system, including kinetic and potential energies of the molecules.

In the given exercise, when a chemical reaction takes place resulting in a gas transforming into a liquid, the internal energy of the system changes. This change is quantified by measuring the heat lost or gained and the work done on or by the system, applying the first law of thermodynamics equation \(\Delta E = q + w\). As the exercise shows, a negative value of \(\Delta E\) indicates that the system's internal energy has decreased, likely because the system has lost more energy as heat than the work it has received.
Heat Transfer
Heat transfer refers to the movement of thermal energy from one body or system to another due to a temperature difference. It is a fundamental concept in thermodynamics and can occur through processes such as conduction, convection, and radiation.

In thermodynamic terms, heat transferred to the system is denoted as \(q\) and is considered positive when the system gains heat and negative when the system loses heat, as evidenced by the exercise where \(q = -1185\ J\), indicating the system has released heat.
Work Done
In thermodynamics, work done refers to the energy transferred when an external force is applied to a system, resulting in displacement. Work is symbolized as \(w\), and in the context of a movable piston in a reaction vessel, work is done when the volume of a system changes.

The direction of work is key: if the system does work on the surroundings (like when a gas expands), \(w\) is negative. Conversely, when work is done on the system (such as when a piston compresses a gas), \(w\) is positive. The exercise clarifies this concept with the given work done on the system as \(w = 623\ J\), showing that the surroundings have done work on the system by compressing it, thus adding energy to the system.
Thermodynamic Processes
Thermodynamic processes describe how a system goes from one state to another, involving changes like pressure, volume, temperature, and also include exchanges of work and heat with the surroundings. These processes can be adiabatic (no heat transfer), isothermal (constant temperature), isobaric (constant pressure), or isochoric (constant volume).

The discussed exercise showcases a process where a system consisting of gases experiences a change because of a reaction, which alters the system's volume as the piston moves. This indicates a process where both work and heat are involved, and their relationship with internal energy change is mapped according to the first law of thermodynamics.
Conservation of Energy
The conservation of energy is a fundamental principle of physics stating that the total energy of an isolated system remains constant—it can neither be created nor destroyed, only converted from one form to another. This principle is the cornerstone of the first law of thermodynamics, which relates to the internal energy change for any system.

The exercise exemplifies this principle by accounting for all energy changes within the system. The energy lost as heat and the energy gained from work done on the system must result in a corresponding change in the system's internal energy. Thus, the equation \(\Delta E = q + w\) balances these changes, preserving the total energy of the system.

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Most popular questions from this chapter

Isooctane is a primary component of gasoline and gives gasoline its octane rating. Burning \(1.00 \mathrm{~mL}\) of isooctane \((d=0.688 \mathrm{~g} / \mathrm{mL})\) releases \(33.0 \mathrm{~kJ}\) of heat. When \(10.00 \mathrm{~mL}\) of is ooctane is burned in a bomb calorime- ter, the temperature in the bomb rises from \(23.2^{\circ} \mathrm{C}\) to \(66.5^{\circ} \mathrm{C}\). What is the heat capacity of the bomb calorimeter?

Glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s),\left(\Delta H_{\mathrm{f}}^{\circ}=-1275.2 \mathrm{~kJ} / \mathrm{mol}\right)\) is converted to ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\), and carbon dioxide in the fermentation of grape juice. What quantity of heat is liberated when \(750.0 \mathrm{~mL}\) of wine containing \(12.0 \%\) ethyl alcohol by volume \(\left(d=0.789 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is produced by the fermentation of grape juice?

Stainless steel accessories in cars are usually plated with chromium to give them a shiny surface and to prevent rusting. When \(5.00 \mathrm{~g}\) of chromium at \(23.00^{\circ} \mathrm{C}\) absorbs \(62.5 \mathrm{~J}\) of heat, the temperature increases to \(50.8^{\circ} \mathrm{C}\). What is the specific heat of chromium?

In the late eighteenth century Priestley prepared ammonia by reacting \(\mathrm{HNO}_{3}(g)\) with hydrogen gas. The thermodynamic equation for the reaction is $$ \mathrm{HNO}_{3}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-637 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) when one mole of hydrogen gas reacts. (b) What is \(\Delta H\) when \(10.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g)\) is made to react with an excess of steam to form \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{2}\) gases?

When \(35.0 \mathrm{~mL}\) of \(1.43 \mathrm{M} \mathrm{NaOH}\) at \(22.0^{\circ} \mathrm{C}\) is neutralized by \(35.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) also at \(22.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, the temperature of the final solution rises to \(31.29^{\circ} \mathrm{C}\). Assume that the specific heat of all solutions is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), that the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and that volumes are additive. (a) Calculate \(q\) for the reaction. (b) Calculate \(q\) for the neutralization of one mole of \(\mathrm{NaOH}\).

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