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Chapter 5: Problem 23

(a) State the first law of thermodynamics. (b) What is meant by the internal energy of a system? (c) By what means can the internal energy of a closed system increase?

Short Answer

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(a) The first law of thermodynamics states that energy cannot be created nor destroyed, but can be converted from one form to another. It can be expressed as: \[\Delta U = Q - W\] (b) The internal energy of a system is the total energy, including kinetic and potential energy, of all the particles within a system. It is a state function and an extensive property. (c) The internal energy of a closed system can increase through heat transfer (\(Q\)) and work done on the system (\(-W\)).

Step by step solution

01

(a) State the first law of thermodynamics.

The first law of thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created nor destroyed, but it can be converted from one form to another. Mathematically, this law can be expressed as: \[\Delta U = Q - W\] where \(\Delta U\) represents the change in internal energy of a system, \(Q\) is the heat added to the system, and \(W\) is the work done by the system.
02

(b) What is meant by the internal energy of a system?

The internal energy of a system refers to the total energy, including kinetic and potential energy, of all the particles (atoms, molecules, ions, etc.) within a system. It is a state function, meaning that it depends solely on the current state of the system and can be used to determine the net energy change during a process. Internal energy is an extensive property, implying that it scales with the size or amount of substance in the system. To compute it, we usually need to consider changes in the internal energy rather than the internal energy itself.
03

(c) By what means can the internal energy of a closed system increase?

The internal energy of a closed system can increase through two primary means: 1. Heat transfer (\(Q\)): If heat is added to the system, meaning that the surrounding transfers more energy to the system as heat, the internal energy of the system increases. 2. Work done on the system (\(-W\)): When work is done on the system by the surroundings, the internal energy of the system increases. Note that it is written as "-W" because work done on the system has a negative sign, as per the convention. In both cases, the result is an increase in the system's internal energy, as described by the first law of thermodynamics.

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Most popular questions from this chapter

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36\) \(\mathrm{mol}\) of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

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Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l)\) : $$ \begin{array}{r} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-726.5 \mathrm{~kJ} \end{array} $$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l)\), would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

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