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Chapter 6: Problem 50

An ideal gas expands isothermally (at constant temperature). The internal energy of an ideal gas remains constant during an isothermal change. If \(q\) is \(-76 \mathrm{~J}\), what are \(\Delta U\) and \(w ?\)

Short Answer

Expert verified
\(\Delta U = 0\) J; \(w = 76\) J.

Step by step solution

01

Understanding Isothermal Process

In an isothermal process, the temperature remains constant. For an ideal gas, the change in internal energy (\(\Delta U\)) is zero because internal energy depends only on temperature. Therefore, \(\Delta U = 0\) J.
02

Applying the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat exchanged (\(q\)) plus the work done (\(w\)) on or by the system: \[\Delta U = q + w\].
03

Substituting Values

Plug the known values into the formula: \(\Delta U = 0\) J, and \(q = -76\) J. Thus, \[0 = -76 + w\].
04

Solving for Work Done (w)

Rearrange the equation to solve for \(w\): \[w = 76 \text{ J}\]. This means the work done by the system is 76 J.

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. This model helps simplify calculations and can closely approximate behavior of real gases under many conditions.
  • **Behavior of Ideal Gases**: Ideal gases follow the ideal gas law, PV = nRT, where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature in Kelvin.
  • **Characteristics**: In ideal gases, molecules neither attract nor repel each other. The only interactions are collisions between molecules or with the walls of their container.
  • **Importance in Thermodynamics**: While no real gas behaves exactly like an ideal gas, this assumption simplifies thermodynamic calculations and provides insights into more complex behavior.
Understanding the ideal gas helps in comprehending processes like the isothermal change, where these gases can expand or compress without changing temperature.
First Law of Thermodynamics
The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. It states that energy cannot be created or destroyed, only transformed or transferred.
The mathematical expression for this law is:
\[ \Delta U = q + w \]Here,
  • \( \Delta U \) is the change in internal energy of the system.
  • \( q \) represents the heat added to or removed from the system.
  • \( w \) stands for the work done on or by the system.
**Application in Isothermal Processes**:
For isothermal processes in ideal gases, the internal energy change \( \Delta U \) is zero. Thus, the first law simplifies to balancing the heat exchange and the work done.This principle helps us understand how energy changes between different forms, like heat and mechanical work, to keep the total energy constant during processes.
Internal Energy
Internal energy is the total energy contained within a system. It considers both the kinetic energy from the molecules' movements and potential energy from molecular interactions.
For an ideal gas, only kinetic energy contributes, since potential energy (from molecular forces) is negligible.
In an isothermal process, particularly for an ideal gas:
  • **Constant Temperature**: As temperature stays the same, so does the kinetic energy of gas molecules.
  • **Resulting \( \Delta U = 0 \)**: Since internal energy depends solely on temperature for ideal gases, eliminating any temperature change results in no change in internal energy.
Understanding internal energy and its constancy during isothermal processes is crucial for predicting behavior of gases in various thermodynamic scenarios.
Work Done
Work done in thermodynamics refers to the energy transfer that occurs when a force causes displacement. In the context of gases, work is often related to volume changes at constant pressure.
During an isothermal process in an ideal gas, when heat provided is utilized entirely for doing work, keeping temperature constant:
  • **Work Done Formula**: In the given isothermal process, using \( q = -76 \text{ J} \) and substitution in \( \Delta U = q + w \), solving gives \( w = 76 \text{ J} \).
  • **Direction of Work**: Positive work implies that work is done by the system to the surroundings, often indicating expansion.
  • **Understanding in Context**: As the gas expands isothermally, the energy required for expansion matches the magnitude of heat loss (or gain), thereby performing work.
This concept explains the intimate relationship between heat exchanged and work performed during thermodynamic processes.

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Most popular questions from this chapter

The decomposition of ozone, \(\mathrm{O}_{3},\) to oxygen, \(\mathrm{O}_{2},\) is an exothermic reaction. What is the sign of \(q\) ? If you were to touch a flask in which ozone is decomposing to oxygen, would you expect the flask to feel warm or cool?

Calculate the grams of oxygen gas required to produce \(7.60 \mathrm{~kJ}\) of heat when hydrogen gas burns at constant pressure to give liquid water, given the following: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-484 \mathrm{~kJ} $$ Liquid water has a heat of vaporization of \(44.0 \mathrm{~kJ}\) per mole at \(25^{\circ} \mathrm{C}\)

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), is a common fuel gas. Use the following to calculate the grams of propane you would need to provide \(369 \mathrm{~kJ}\) of heat. $$ \begin{aligned} \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) ; \\ \Delta H &=-2043 \mathrm{~kJ} \end{aligned} $$

Sulfur dioxide gas reacts with oxygen, \(\mathrm{O}_{2}(g)\), to produce \(\mathrm{SO}_{3}(g) .\) This reaction releases \(99.0 \mathrm{~kJ}\) of heat (at constant pressure) for each mole of sulfur dioxide that reacts. Write the thermochemical equation for the reaction of \(2 \mathrm{~mol}\) of sulfur dioxide, and then also for the decomposition of \(3 \mathrm{~mol}\) of sulfur trioxide gas into oxygen gas and sulfur dioxide gas. Do you need any other information to answer either question?

Hydrogen cyanide is used in the manufacture of clear plastics such as Lucite and Plexiglas. It is prepared from ammonia and natural gas \(\left(\mathrm{CH}_{4}\right)\). \(2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) The reaction evolves \(469 \mathrm{~kJ}\) of heat per \(\mathrm{mol}\) of \(\mathrm{HCN}\) formed. Is the reaction endothermic or exothermic? What is the value of \(q\) when \(2 \mathrm{~mol} \mathrm{HCN}\) forms? 6.55 What is \(\Delta U\) when \(1.00 \mathrm{~mol}\) of liquid water vaporizes at \(100^{\circ} \mathrm{C} ?\) The heat of vaporization, \(\Delta H_{v a p}^{\circ},\) of water at \(100^{\circ} \mathrm{C}\) is \(40.66 \mathrm{~kJ} / \mathrm{mol}\)
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