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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

Determine whether the statements given below are true or false. Consider an endothermic process taking place in a beaker at room temperature. (a) Heat flows from the surroundings to the system. (b) The beaker is cold to the touch. (c) The pressure of the system decreases. (d) The value of \(q\) for the system is positive.

Given $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=3351.4 \mathrm{~kJ} $$ (a) What is the heat of formation of aluminum oxide? (b) What is \(\Delta H^{\circ}\) for the formation of \(12.50 \mathrm{~g}\) of aluminum oxide?

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?

Microwave ovens convert radiation to energy. A microwave oven uses radiation with a wavelength of \(12.5 \mathrm{~cm}\). Assuming that all the energy from the radiation is converted to heat without loss, how many moles of photons are required to raise the temperature of a cup of water \((350.0 \mathrm{~g}\), specific heat \(=4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) ) from \(23.0^{\circ} \mathrm{C}\) to \(99.0^{\circ} \mathrm{C} ?\)

Calcium carbide, \(\mathrm{CaC}_{2}\), is the raw material for the production of acetylene (used in welding torches). Calcium carbide is produced by reacting calcium oxide with carbon, producing carbon monoxide as a byproduct. When one mole of calcium carbide is formed, \(464.8 \mathrm{~kJ}\) is absorbed. (a) Write a thermochemical equation for this reaction. (b) Is the reaction exothermic or endothermic? (c) Draw an energy diagram showing the path of this reaction. (Figure \(8.4\) is an example of such an energy diagram.) (d) What is \(\Delta H\) when \(1.00 \mathrm{~g}\) of \(\mathrm{CaC}_{2}(\mathrm{~g})\) is formed? (e) How many grams of carbon are used up when \(20.00 \mathrm{~kJ}\) of heat is absorbed?

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