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Chapter 5: Problem 24

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

Short Answer

Expert verified
(a) The first law of thermodynamics in terms of heat and work can be written as: \(鈭哢 = q - w\), where 鈭哢 represents the change in internal energy, q is the heat transfer into the system, and w is the work done by the system on its surroundings. (b) The quantities q and w will be negative under the following conditions: - q is negative when the system loses heat during an exothermic process (i.e., it transfers energy to its surroundings). - w is negative when the system absorbs work from its surroundings, leading to an increase in its internal energy (e.g., during compression of a gas at constant temperature).

Step by step solution

01

First Law of Thermodynamics Equation

The first law of thermodynamics is also known as the conservation of energy principle. It states that the change in the internal energy of a system is equal to the heat transfer into the system minus the work done by the system on its surroundings. Mathematically, we can write this as: \[ 鈭哢 = q - w \] Here, - 鈭哢 represents the change in internal energy of the system - q is the heat transfer into the system - w is the work done by the system on its surroundings
02

When is q Negative?

Heat transfer into the system, q, will be negative when heat is transferred from the system to its surroundings. This occurs during an exothermic process, where the system loses energy in the form of heat. In such a case, we denote the heat transfer out of the system as a negative value to indicate that the system is losing energy. For example, when a hot metal rod cools down, it transfers heat to its surroundings, and thus, q would be negative in this scenario.
03

When is w Negative?

Work done by the system, w, will be negative when the system absorbs work from its surroundings. This happens when the surroundings do work on the system, causing an increase in the system's internal energy. For example, when a gas is compressed at constant temperature, the surroundings (like a piston) do work by pushing against the gas, thereby increasing its internal energy. In this scenario, the work done by the system (w) would be negative because it absorbs work from its surroundings. In summary: - q is negative when the system loses heat, and it transfers energy to its surroundings (exothermic process). - w is negative when the system absorbs work from its surroundings, leading to an increase in its internal energy.

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Most popular questions from this chapter

Consider the twodiagramsbelow. (a) Based on \((t)\), write an equation showing how \(\Delta H_{\mathrm{A}}\) is related to \(\Delta H_{\mathrm{B}}\) and \(\Delta H_{\mathrm{C}}\). How do both diagram (i) and your equation relate to the fact that enthalpy is a state function? (b) Based on (ii), write an equation relating \(\Delta H_{Z}\) to the other enthalpy changes in the diagram. (c) How do these diagrams relate to Hess's law? [Section 5.6]

(a) Calculate the standard enthalpy of formation of gaseous diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) using the following thermochemical information: \(\begin{array}{clrl}4 \mathrm{~B}(\mathrm{~s})+3 \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{~B}_{2} \mathrm{O}_{3}(s) & \Delta H^{\circ}=-2509.1 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}= & -571.7 \mathrm{~kJ} \\\ \mathrm{~B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-2147.5 \mathrm{~kJ}\end{array}\) (b) Pentaborane \(\left(\mathrm{B}_{5} \mathrm{H}_{9}\right)\) is another boron hydride. What experiment or experiments would you need to perform to yield the data necessary to calculate the heat of formation of \(\mathrm{B}_{5} \mathrm{H}_{9}(l) ?\) Explain by writing out and summing any applicable chemical reactions.

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is heated by adding 850 J of heat. It expands, doing \(382 \mathrm{~J}\) of work on the atmosphere. (b) A \(50-g\) sample of water is cooled from \(30^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\), thereby losing approximately \(3140 \mathrm{~J}\) of heat. (c) A chemical reaction releases \(6.47 \mathrm{~kJ}\) of heat and does no work on the surroundings.

Many cigarette lighters contain liquid butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\). Using standard enthalpies of formation, calculate the quantity of heat produced when \(5.00 \mathrm{~g}\) of butane is completely combusted in air under standard conditions.

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) For a given process at constant pressure, \(\Delta H\) is negative. Is the process endothermic or exothermic?
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