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Calculate (a) \(g\) when a system does \(54 \mathrm{~J}\) of work and its energy decreases by \(72 \mathrm{~J}\). (b) \(\Delta E\) for a gas that releases \(38 \mathrm{~J}\) of heat and has \(102 \mathrm{~J}\) of work done on it.

Short Answer

Expert verified
Answer: (a) \(g = -126 \mathrm{~J}\); (b) \(\Delta E = 64 \mathrm{~J}\).

Step by step solution

01

Write down the mechanical energy conservation equation

The equation is: \(\Delta E = W_g + g\).
02

Substitute the given values

Substitute \(\Delta E = -72 \mathrm{~J}\) and \(W_g = 54 \mathrm{~J}\) into the equation: \(-72 = 54 + g\).
03

Solve for \(g\)

Rearrange the equation and solve for \(g\): \(g = -72 - 54\), so \(g = -126 \mathrm{~J}\). Thus, the value of \(g\) when the system does \(54 \mathrm{~J}\) of work and its energy decreases by \(72 \mathrm{~J}\) is \(-126 \mathrm{~J}\). For (b):
04

Write down the first law of thermodynamics equation

The equation is: \(\Delta E = Q + W\).
05

Substitute the given values

Substitute \(Q = -38 \mathrm{~J}\) (since heat is released) and \(W = 102 \mathrm{~J}\) (since work is done on the gas) into the equation: \(\Delta E = -38 + 102\).
06

Solve for \(\Delta E\)

Calculate the change in energy: \(\Delta E = -38 + 102\), so \(\Delta E = 64 \mathrm{~J}\). Thus, the change in energy for a gas that releases \(38 \mathrm{~J}\) of heat and has \(102 \mathrm{~J}\) of work done on it is \(64 \mathrm{~J}\).

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Key Concepts

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

Conservation of Energy
The conservation of energy is a fundamental principle in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of thermodynamics, this law plays a crucial role in understanding how energy is transferred within a system.
When a system undergoes a process, such as expanding or contracting, energy may change form, for example from potential to kinetic energy or vice versa. However, the total energy of a closed system must remain constant. This concept is closely related to the first law of thermodynamics, which asserts 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 on its surroundings.
For students, this concept is important to recognize that whenever work is done or heat is transferred, the system's internal energy will adjust accordingly to maintain energy conservation. For example, in a scenario where a system does work on its surroundings, such as in part (a) of the exercise, the system's internal energy decreases, which is consistent with the law of conservation of energy.
Thermodynamic Work
Thermodynamic work, often represented by the letter 'W' in equations, refers to the energy transferred from a system to its surroundings, or vice versa, due to a force acting through a distance. It's important to differentiate this work from other types of work, like electrical or mechanical, because it specifically involves thermodynamic processes, such as the compression or expansion of a gas.

Understanding Work in Thermodynamics

When a gas expands against a resisting pressure, for example, it does work on the environment. Conversely, when the environment does work on the gas - compressing it - we consider this work as done on the system. In mathematical terms, work is often a product of pressure and volume change.
  • If the volume of a system increases, work is done by the system.
  • If the volume of a system decreases, work is done on the system.
The concept of work is critical for solving problems like those in the textbook exercise, where understanding the direction of energy flow is key to using the correct signs when calculating work and changes in energy.
Internal Energy Change
Internal energy change, denoted as \(\Delta E\), signifies the difference in a system's energy before and after a process occurs. In thermodynamics, internal energy encompasses all the forms of energy within a system, which includes kinetic energy due to the motion of molecules, potential energy from molecular interactions, and more.
An important takeaway is that internal energy change is the net result of heat exchange and work done in a thermodynamic process. It's reflected in the first law of thermodynamics formula: \(\Delta E = Q + W\), where \(Q\) is the heat added to the system and \(W\) is the work done. A positive \(\Delta E\) means the system gained energy, and a negative \(\Delta E\) indicates energy loss.

Calculating Internal Energy Change

In the context of part (b) of the textbook exercise, heat is released from the system (which would decrease internal energy), but work is done on the system (which increases internal energy), leading to a positive \(\Delta E\) because the work input is greater. This illustrates how different energy interactions can combine to affect the overall energy of the system.

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Most popular questions from this chapter

Nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\), is a powerful explosive used in rock blasting when roads are created. When ignited, it produces water, nitrogen, carbon dioxide, and oxygen. Detonation of one mole of nitroglycerine liberates \(5725 \mathrm{~kJ}\) of heat. (a) Write a balanced thermochemical equation for the reaction for the detonation of four moles of nitroglycerine. (b) What is \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l) ?\)

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?

Determine whether the statements given below are true or false. Consider specific heat. (a) Specific heat represents the amount of heat required to raise the temperature of one gram of a substance by \(1^{\circ} \mathrm{C}\). (b) Specific heat is the amount of heat flowing into the system. (c) When 20 J of heat is added to equal masses of different materials at \(25^{\circ} \mathrm{C}\), the final temperature for all these materials will be the same. (d) Heat is measured in \({ }^{\circ} \mathrm{C}\).

Mercury has a specific heat of \(0.140 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). Assume that a thermometer has 20 (2 significant figures) grams of mercury. How much heat is absorbed by the mercury when the temperature in the thermometer increases from \(98.6^{\circ} \mathrm{F}\) to \(103.2^{\circ} \mathrm{F} ?\) (Assume no heat loss to the glass of the thermometer.)

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 \(\Delta E ?\)

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