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Chapter 7: Problem 55

What is the change in internal energy of a system if the system (a) absorbs \(58 \mathrm{J}\) of heat and does \(58 \mathrm{J}\) of work; (b) absorbs 125 J of heat and does 687 J of work; (c) evolves 280 cal of heat and has 1.25 kJ of work done on it?

Short Answer

Expert verified
The change in internal energy of the system are (a) 0J, (b) -562J, and (c) 2421.52J.

Step by step solution

01

Calculate Internal Energy Change for Part (a)

Use the first law of Thermodynamics formula which is \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, Q is the heat absorbed by the system and W is the work done by the system. Substituting the given values, we get \( \Delta U = 58J - 58J = 0J \).
02

Calculate Internal Energy Change for Part (b)

Similarly for part (b), using the first law of Thermodynamics formula, \( \Delta U = Q - W \). Substitute the given values, \( \Delta U = 125J - 687J = -562J \). The negative sign indicates that energy has left the system.
03

Convert Calories to Joules for Part (c)

Given 1 cal = 4.184 J. So, 280 cal would be \( 280cal * 4.184 J/cal = 1171.52J \).
04

Calculate Internal Energy Change for Part (c)

Now, compute the change in internal energy for part (c). We need to remember that the work is done on the system this time, so it's considered negative. Converting work done from kJ to J (since 1 kJ = 1000 J), we get \( 1.25kJ * 1000 J/kJ = 1250J \). Hence, interior energy change is calculated as \( \Delta U = Q - W = 1171.52J - (-1250J) = 2421.52J \).

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Key Concepts

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

Internal Energy
When we talk about internal energy, we're referring to the total energy contained within a system. This includes the kinetic energy of molecules moving about and the potential energy associated with them. Internal energy is crucial in understanding how energy transfers affect a system.

One essential point is that the internal energy of a system can change due to exchanges of heat and work. If you ever find yourself wondering why a system feels hotter or colder, or why it seems to move with more or less vigor, it might be because its internal energy has shifted.
  • Increase in internal energy means the system becomes more energetic.
  • Decrease in internal energy generally implies the system loses some energy to its surroundings.
By observing internal energy, we can better predict and describe the behavior of physical and chemical systems under various conditions.
Heat Absorption
Heat absorption is a critical concept when studying energy changes within a system. It is one way energy can be transferred into the system, causing a warming effect or fueling a reaction.

When a system absorbs heat:
  • The temperature of the system may increase, although in some cases, heat might cause changes in state without altering temperature.
  • Potential energy within the system could increase, depending on the system's characteristics.
The sign convention in the first law of thermodynamics states that absorbed heat is considered positive (as it's entering the system). If a system absorbs more heat than it loses, you can expect its internal energy to rise, which might affect the system's overall activity.
Work Done on System
The concept of work done on a system is another path by which energy can influence a system's internal energy. Whenever an external force performs work on a system, it essentially adds energy to it.

In thermodynamics:
  • If work is done on a system, this work is treated as negative in calculations because it adds energy to the system.
  • If the system itself does work on its surroundings, this work is positive as it expends energy from the system.
For a practical example, consider when you compress a gas inside a cylinder: work is done on the gas, which increases its internal energy. Hence, a system that has work done on it tends to experience a rise in its energy level, possibly influencing its temperature, volume, or phase.

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Most popular questions from this chapter

A 0.205 g pellet of potassium hydroxide, \(\mathrm{KOH}\), is added to \(55.9 \mathrm{g}\) water in a Styrofoam coffee cup. The water temperature rises from 23.5 to \(24.4^{\circ} \mathrm{C}\). [Assume that the specific heat of dilute \(\mathrm{KOH}(aq)\) is the same as that of water.] (a) What is the approximate heat of solution of \(\mathrm{KOH}\) expressed as kilojoules per mole of \(\mathrm{KOH}?\) (b) How could the precision of this measurement be improved without modifying the apparatus?

In the Are You Wondering \(7-1\) box, the temperature variation of enthalpy is discussed, and the equation \(q_{P}=\) heat capacity \(\times\) temperature change \(=C_{P} \times \Delta T\) was introduced to show how enthalpy changes with temperature for a constant-pressure process. Strictly speaking, the heat capacity of a substance at constant pressure is the slope of the line representing the variation of enthalpy (H) with temperature, that is $$C_{P}=\frac{d H}{d T} \quad(\text { at constant pressure })$$ where \(C_{P}\) is the heat capacity of the substance in question. Heat capacity is an extensive quantity and heat capacities are usually quoted as molar heat capacities \(C_{P, \mathrm{m}},\) the heat capacity of one mole of substance; an intensive property. The heat capacity at constant pressure is used to estimate the change in enthalpy due to a change in temperature. For infinitesimal changes in temperature, $$d H=C_{p} d T \quad(\text { at constant pressure })$$ To evaluate the change in enthalpy for a particular temperature change, from \(T_{1}\) to \(T_{2}\), we write $$\int_{H\left(T_{1}\right)}^{H\left(T_{2}\right)} d H=H\left(T_{2}\right)-H\left(T_{1}\right)=\int_{T_{1}}^{T_{2}} C_{P} d T$$ If we assume that \(C_{P}\) is independent of temperature, then we recover equation (7.5) $$\Delta H=C_{P} \times \Delta T$$ On the other hand, we often find that the heat capacity is a function of temperature; a convenient empirical expression is $$C_{P, \mathrm{m}}=a+b T+\frac{c}{T^{2}}$$ What is the change in molar enthalpy of \(\mathrm{N}_{2}\) when it is heated from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) The molar heat capacity of nitrogen is given by$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$

In each of the following processes, is any work done when the reaction is carried out at constant pressure in a vessel open to the atmosphere? If so, is work done by the reacting system or on it? (a) Neutralization of \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) by \(\mathrm{HCl}(\mathrm{aq}) ;\) (b) conversion of gaseous nitrogen dioxide to gaseous dinitrogen tetroxide; (c) decomposition of calcium carbonate to calcium oxide and carbon dioxide gas.

An alternative approach to bomb calorimetry is to establish the heat capacity of the calorimeter, exclusive of the water it contains. The heat absorbed by the water and by the rest of the calorimeter must be calculated separately and then added together. A bomb calorimeter assembly containing \(983.5 \mathrm{g}\) water is calibrated by the combustion of \(1.354 \mathrm{g}\) anthracene. The temperature of the calorimeter rises from 24.87 to \(35.63^{\circ} \mathrm{C} .\) When \(1.053 \mathrm{g}\) citric acid is burned in the same assembly, but with 968.6 g water, the temperature increases from 25.01 to \(27.19^{\circ} \mathrm{C}\). The heat of combustion of anthracene, \(\mathrm{C}_{14} \mathrm{H}_{10}(\mathrm{s}),\) is \(-7067 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{C}_{14} \mathrm{H}_{10} \cdot\) What is the heat of combustion of citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) expressed in \(\mathrm{kJ} / \mathrm{mol} ?\)

The internal energy of a fixed quantity of an ideal gas depends only on its temperature. A sample of an ideal gas is allowed to expand at a constant temperature (isothermal expansion). (a) Does the gas do work? (b) Does the gas exchange heat with its surroundings? (c) What happens to the temperature of the gas? (d) What is \(\Delta U\) for the gas?
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