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Chapter 6: Problem 49

A gas is cooled and loses \(82 \mathrm{~J}\) of heat. The gas contracts as it cools, and work done on the system equal to \(29 \mathrm{~J}\) is exchanged with the surroundings. What are \(q, w\), and \(\Delta U ?\)

Short Answer

Expert verified
\(q = -82 \, \mathrm{J}\), \(w = 29 \, \mathrm{J}\), \(\Delta U = -53 \, \mathrm{J}\).

Step by step solution

01

Understanding the Given Information

We are given two important pieces of information about the energy changes in the system: the gas loses 82 J of heat (meaning \(q = -82 \, \mathrm{J}\)) and work is done on the system as it contracts, to the amount of 29 J (meaning \(w = 29 \, \mathrm{J}\)). The signs are important: heat loss is negative, while work done on the system is positive.
02

Applying the First Law of Thermodynamics

We use the first law of thermodynamics, which states that the change in internal energy \(\Delta U\) of a system is equal to the heat exchanged with the surroundings \(q\) plus the work done on or by the system \(w\). This is represented by the equation: \( \Delta U = q + w \).
03

Substitution into the Equation

Substitute the given values into the equation. We have \(q = -82 \, \mathrm{J}\) and \(w = 29 \, \mathrm{J}\). Therefore, \(\Delta U = -82 \, \mathrm{J} + 29 \, \mathrm{J}\).
04

Solve for \(\Delta U\)

Calculate \(\Delta U\) by performing the addition: \(-82 + 29 = -53\). Therefore, \(\Delta U = -53 \, \mathrm{J}\).
05

Conclusion

The values are, \(q = -82 \, \mathrm{J}\), \(w = 29 \, \mathrm{J}\), and \(\Delta U = -53 \, \mathrm{J}\). The negative \(\Delta U\) indicates the system's internal energy decreases.

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Key Concepts

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

Internal Energy Change
Internal energy change, represented by \( \Delta U \), is a core concept of the First Law of Thermodynamics. This change refers to the difference between the energy of a system at the end and the beginning of a process. In the context of thermodynamics, when a system, like a gas, undergoes changes such as heating or cooling, its internal energy changes accordingly.

The formula for the internal energy change is given by the equation of the first law of thermodynamics:
  • \( \Delta U = q + w \)
Where \( q \) is the heat exchange and \( w \) is the work done on or by the system.

In simpler terms, \( \Delta U \) measures how much energy is stored in the system after heat and work interactions. In our example, the internal energy decrease is \(-53 \, \mathrm{J}\), meaning the system loses more energy than it gains.
Heat Exchange
Heat exchange, denoted by \( q \), describes the transfer of thermal energy between a system and its surroundings. It is essential to consider both the magnitude and the direction of heat flow to understand any thermodynamic process.

  • If heat is supplied to the system, \( q \) is positive (endothermic process).
  • If heat is lost from the system, \( q \) is negative (exothermic process).

In this exercise, the gas loses \( 82 \mathrm{~J} \) of heat as it cools, signifying an exothermic process, so \( q = -82 \, \mathrm{J} \). The negative sign reflects the energy flowing out of the system, which is key in applying the First Law of Thermodynamics correctly.
Work Done on System
Work done on or by a system, represented by \( w \), is another crucial part of thermodynamic transitions. It occurs when there is a mechanical interaction, such as expansion or compression of gases.

According to conventions,
  • Work done on the system is positive, as energy is being added to it.
  • Work done by the system is negative, indicating energy leaving the system due to its action on the surroundings.
In the case of our system, \( 29 \mathrm{~J} \) of work is done on the gas, implying \( w = 29 \, \mathrm{J} \). This positive value points toward energy entering the system, contributing to the calculation of \( \Delta U \) using the first law of thermodynamics. It is vital for students to grasp how work affects energy change in the context of thermodynamic equations.

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Most popular questions from this chapter

When \(15.3 \mathrm{~g}\) of sodium nitrate, \(\mathrm{NaNO}_{3}\), was dissolved in water in a constant-pressure calorimeter, the temperature fell from \(25.00^{\circ} \mathrm{C}\) to \(21.56^{\circ} \mathrm{C}\). If the heat capacity of the solution and the calorimeter is \(1071 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) what is the enthalpy change when \(1 \mathrm{~mol}\) of sodium nitrate dissolves in water? The solution process is $$ \mathrm{NaNO}_{3}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) ; \Delta H=? $$

Nitric acid, a source of many nitrogen compounds, is produced from nitrogen dioxide. An old process for making nitrogen dioxide employed nitrogen and oxygen. $$ \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ The reaction absorbs \(66.2 \mathrm{~kJ}\) of heat per \(2 \mathrm{~mol} \mathrm{NO}_{2}\) produced. Is the reaction endothermic or exothermic? What is the value of \(q\) ?

A piece of iron was heated to \(95.4^{\circ} \mathrm{C}\) and dropped into a constant-pressure calorimeter containing \(284 \mathrm{~g}\) of water at \(32.2^{\circ} \mathrm{C}\). The final temperature of the water and iron was \(51.9^{\circ} \mathrm{C}\). Assuming that the calorimeter itself absorbs a negligible amount of heat, what was the mass (in grams) of the piece of iron? The specific heat of iron is \(0.449 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\), and the specific heat of water is \(4.18 \mathrm{~J} /\) \(\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\)

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), is a colorless liquid whose solutions are used as a bleach and an antiseptic. \(\mathrm{H}_{2} \mathrm{O}_{2}\) can be prepared in a process whose overall change is $$ \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(l) $$ Calculate the enthalpy change using the following data: $$ \begin{array}{l} \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g) ; \Delta H=-98.0 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) ; \Delta H=-571.6 \mathrm{~kJ} \end{array} $$

The head of a "strike anywhere" match contains tetraphosphorus trisulfide, \(\mathrm{P}_{4} \mathrm{~S}_{3} .\) In an experiment, a student burned this compound in an excess of oxygen and found that it evolved \(3651 \mathrm{~kJ}\) of heat per mole of \(\mathrm{P}_{4} \mathrm{~S}_{3}\) at a constant pressure of 1 atm. She wrote the following thermochemical equation: $$ \begin{array}{c} \mathrm{P}_{4} \mathrm{~S}_{3}(s)+8 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+3 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-3651 \mathrm{~kJ} \end{array} $$ Calculate the standard enthalpy of formation of \(\mathrm{P}_{4} \mathrm{~S}_{3}, \mathrm{us}-\) ing this student's result and the following standard enthalpies of formation: \(\mathrm{P}_{4} \mathrm{O}_{10}(s),-3009.9 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{SO}_{2}(g),-296.8\) \(\mathrm{kJ} / \mathrm{mol}\). How does this value compare with the value given in Appendix C?
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