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Chapter 8: Problem 13

The heat capacities of a substance have been defined as $$C_{v}=\left(\frac{\partial \hat{U}}{\partial T}\right)_{V}, \quad C_{p}=\left(\frac{\partial \hat{H}}{\partial T}\right)_{P}$$ Use the defining relationship between \(\hat{H}\) and \(\hat{U}\) and the fact that \(\hat{H}\) and \(\hat{U}\) for ideal gases are functions only of temperature to prove that \(C_{p}=C_{v}+R\) for an ideal gas (Eq. \(8.3-12\) ).

Short Answer

Expert verified
The relationship \(C_{p}=C_{v}+R\) for an ideal gas is a result of the defining relationship between internal energy (\(\hat{U}\)) and enthalpy (\(\hat{H}\)), the laws of thermodynamics, and the gas constant \(R\).

Step by step solution

01

Define the concepts

We'll start by defining the relevant concepts involved in the scenario:\n- Heat capacities at constant volume, \(C_{v}\), and constant pressure, \(C_{p}\), are defined as the rate at which a substance's internal energy or enthalpy changes with temperature, respectively. - Internal energy, \(\hat{U}\), and enthalpy, \(\hat{H}\), are state variables for a gas that depend on temperature and volume or pressure.- For an ideal gas, both \(\hat{U}\) and \(\hat{H}\) are only functions of temperature.- The ideal gas constant, R, is a constant of proportionality that relates the energy of a mole of gas to its temperature.
02

Understand the relationship between \(\hat{H}\) and \(\hat{U}\)

\(\hat{H}\) and \(\hat{U}\) are related by the following thermodynamic equation: \(\hat{H} = \hat{U} + PV\)Differentiating this equation with respect to temperature, we get: \(\frac{\partial \hat{H}}{\partial T} = \frac{\partial \hat{U}}{\partial T} + \frac{\partial (PV)}{\partial T}\)This equation tells us that the change in enthalpy with respect to temperature is equal to the change in internal energy plus the change in the pressure/volume product with temperature.
03

Apply the 'only function of temperature' condition

Remember, for an ideal gas, both \(\hat{U}\) and \(\hat{H}\) are only functions of temperature, and therefore, are independent of volume and pressure. This means that partial derivative of the pressure/volume product with respect to temperature at constant volume (\(V\)) and pressure (\(P\)) equals to the pressure/volume product times the derivative of temperature with respect to temperature. In the equation, this term simplifies to \(P\). Substituting \(\frac{\partial \hat{U}}{\partial T}\) and \(\frac{\partial \hat{H}}{\partial T}\) with \(C_{v}\) and \(C_{p}\) respectively, we obtain \(C_{p} = C_{v} + P\).
04

Apply ideal gas law

For an ideal gas, applying the ideal gas law \(PV = nRT\), where \(n\) is the number of moles and is equal to 1 for a mole of a substance, substitute \(P\) with \(RT\). Therefore, the expression becomes \(C_{p} = C_{v} + RT\). But, given that \(R\) is in fact the gas constant, the \(T\)'s cancel out, resulting in the final equation: \(C_{p} = C_{v} + R\) as was required.

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Key Concepts

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

Heat Capacity
Heat capacity is an important concept in thermodynamics. It represents how much heat energy is needed to change the temperature of a substance by a certain amount. There are two common types of heat capacity to consider:
  • Heat Capacity at Constant Volume (\(C_{v} \))
    • At constant volume, heat capacity (\(C_{v}\)) measures how much heat energy is required to raise the temperature of a system without allowing it to expand. This is particularly useful for closed systems where volume remains unchanged.
  • Heat Capacity at Constant Pressure (\(C_{p} \))
    • This measures the heat capacity when the system is allowed to expand or contract, still maintaining a constant pressure. It's useful in many real-world scenarios, such as cooking or heating buildings where pressure is often constant.
It's crucial to understand that these heat capacities are vital when determining how a substance behaves under different heating conditions. They are defined through thermodynamic equations that relate to the substance's internal energy or enthalpy.
Internal Energy
Internal energy (\(\hat{U}\)) is a fundamental concept in thermodynamics. It encompasses all of the energy contained within a system, stemming from both the motion and interaction of molecules. For ideal gases, internal energy is exclusively a function of temperature, not volume or pressure.
Understanding internal energy helps predict how much energy is in a system and how it changes, providing insight into the system's behavior when subjected to different temperatures. The relationship between heat capacity at constant volume (\(C_{v}\)) and internal energy is described as the rate of change of internal energy with temperature:\[C_{v} = \left(\frac{\partial \hat{U}}{\partial T}\right)_{V}\]This shows that for an ideal gas, a rise in temperature increases internal energy correspondingly, and \(C_{v}\) quantifies this change.
Grasping the concept of internal energy is crucial for thermodynamics as it affects calculations in energy exchange processes.
Thermodynamic Equations
Thermodynamic equations are the foundation of how we understand energy transformations in systems. One key equation for ideal gases is the relationship between enthalpy (\(\hat{H}\)) and internal energy (\(\hat{U}\)):\[\hat{H} = \hat{U} + PV\]This equation shows how enthalpy changes (\(\hat{H}\)) are related to changes in internal energy (\(\hat{U}\)) and the product of pressure and volume (\(PV\)).
When differentiating this equation with respect to temperature, we find:\[\frac{\partial \hat{H}}{\partial T} = \frac{\partial \hat{U}}{\partial T} + \frac{\partial (PV)}{\partial T}\]For an ideal gas, this relationship simplifies as both \(\hat{H}\) and \(\hat{U}\) depend solely on temperature. This equation helps in understanding how the different variables in a system interact and influence each other, particularly in processes where temperature variation occurs.
These equations serve not only as mathematical expressions but also as tools that provide deep insights into the behavior of gases under various conditions.
Enthalpy
Enthalpy (\(\hat{H}\)) is a significant concept because it combines the internal energy of a system with the product of its pressure and volume. Essentially, it is the total heat content of a system. For ideal gases, enthalpy is based solely on temperature, much like internal energy.
The heat capacity at constant pressure (\(C_{p}\)) is linked to enthalpy through:\[C_{p} = \left(\frac{\partial \hat{H}}{\partial T}\right)_{P}\]This represents the rate of change of enthalpy with temperature at constant pressure. It shows how much heat is required to raise the temperature of a substance while keeping the pressure steady.
For ideal gases, using the equation \(\hat{H} = \hat{U} + PV\) and applying the ideal gas law (\(PV = nRT\)), we can derive a vital relationship: \(C_{p} = C_{v} + R\). This highlights how the difference in heat capacities for ideal gases is the ideal gas constant \(R\), thus linking \(C_{p}\) and \(C_{v}\) directly.
Understanding enthalpy is crucial for processes that involve heat exchange, such as chemical reactions and phase changes.

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Most popular questions from this chapter

In gas adsorption a vapor is transferred from a gas mixture to the surface of a solid. (See Section \(6.7 .\) ) An approximate but useful way of analyzing adsorption is to treat it simply as condensation of vapor on a solid surface. Suppose a nitrogen stream at \(35^{\circ} \mathrm{C}\) and 1 atm containing carbon tetrachloride with a \(15 \%\) relative saturation is fed at a rate of \(10.0 \mathrm{mol} / \mathrm{min}\) to a \(6-\mathrm{kg}\) bed of activated carbon. The temperature and pressure of the gas do not change appreciably from the inlet to the outlet of the bed, and there is no \(\mathrm{CCl}_{4}\) in the gas leaving the adsorber. The carbon can adsorb 40\% of its own mass of carbon tetrachloride before becoming saturated, at which point it must be either regenerated (remove the carbon tetrachloride) or replaced with a fresh bed of activated carbon. Neglect the effect of temperature on the heat of vaporization of \(\mathrm{CCl}_{4}\) when solving the following problems: (a) Estimate the rate ( \(\mathrm{kJ} / \mathrm{min}\) ) at which heat must be removed from the adsorber to keep the process isothermal, and the time (min) it will take to saturate the bed. (b) The surface-to-volume ratio of spherical particles is \((3 / r)\left(\mathrm{cm}^{2} \text { outer surface }\right) /\left(\mathrm{cm}^{3} \text { volume }\right)\) First, derive that formula. Second, use it to explain how decreasing the average diameter of the particles in the carbon bed might make the adsorption process more efficient. Third, since most of the area on which adsorption takes place is provided by pores penetrating the particle, explain why the surface-to-volume ratio, as calculated by the above expression, might be relatively unimportant.

Among the best-known building blocks in nanotechnology applications are nanoparticles of noble metals. For example, colloidal suspensions of silver or gold nanoparticles (10-200 nm) exhibit vivid colors because of intense optical absorption in the visible spectrum, making them useful in colorimetric sensors. In the illustration shown below, a suspension of gold nanoparticles of a fairly uniform size in water exhibits peak absorption near a wavelength of \(525 \mathrm{nm}\) (near the blue region of the visible spectrum of light). When one views the solution in ambient (white) light, the solution appears wine-red because the blue part of the spectrum is largely absorbed. When the nanoparticles aggregate to form large particles, an optical absorption peak near \(600-700 \mathrm{nm}\) (near the red region of the visible spectrum) is observed. The breadth of the peak reflects a fairly broad particle size distribution. The solution appears bluish because the unabsorbed light reaching the eye is dominated by the short (blue-violet) wavelength region of the spectrum. since the optical properties of metallic nanoparticles are a strong function of their size, achieving a narrow particle size distribution is an important step in the development of nanoparticle applications. A promising way to do so is laser photolysis, in which a suspension of particles of several different sizes is irradiated with a high-intensity laser pulse. By carefully selecting the wavelength and energy of the pulse to match an absorption peak of one of the particle sizes (e.g., irradiating the red solution in the diagram with a \(525 \mathrm{nm}\) laser pulse), particles of or near that size can be selectively vaporized. (a) A spherical silver nanoparticle of diameter \(D\) at \(25^{\circ} \mathrm{C}\) is to be heated to its normal boiling point and vaporized with a pulsed laser. Considering the particle a closed system at constant pressure, write the energy balance for this process, look up the physical properties of silver that are required in the energy balance, and perform all the required substitutions and integrations to derive an expression for the energy \(Q_{\text {abs }}(\mathrm{J})\) that must be absorbed by the particle as a function of \(D(\mathrm{nm})\) (b) The total energy absorbed by a single particle \(\left(Q_{\text {abs }}\right)\) can also be calculated from the following relation: $$ Q_{\mathrm{abs}}=F A_{\mathrm{p}} \sigma_{\mathrm{abs}} $$ where \(F\left(\mathrm{J} / \mathrm{m}^{2}\right)\) is the energy in a single laser pulse per unit spot area (area of the laser beam) and \(A_{\mathrm{p}}\left(\mathrm{m}^{2}\right)\) is the total surface area of the nanoparticle. The effectiveness factor, \(\sigma_{\mathrm{ahs}},\) accounts for the efficiency of absorption by the nanoparticle at the wavelength of the laser pulse and is dependent on the particle size, shape, and material. For a spherical silver nanoparticle irradiated by a laser pulse with a peak wavelength of \(532 \mathrm{nm}\) and spot diameter of \(7 \mathrm{mm}\) with \(D\) ranging from 40 to \(200 \mathrm{nm}\), the following empirical equation can be used for \(\sigma_{\mathrm{abs}}\) $$ \sigma_{\mathrm{abs}}=\frac{1}{4}\left[0.05045+2.2876 \exp \left(-\left(\frac{D-137.6}{41.675}\right)^{2}\right)\right] $$ where \(\sigma_{\text {abs }}\) and the leading \(\frac{1}{4}\) are dimensionless and \(D\) has units of nm. Use the results of Part (a) to determine the minimum values of F required for complete vaporization of single nanoparticles with diameters of \(40.0 \mathrm{nm}, 80.0 \mathrm{nm},\) and \(120.0 \mathrm{nm}\). If the pulse frequency of the laser is \(10 \mathrm{Hz}\) (i.e., 10 pulses per second), what is the minimum laser power \(P(\mathrm{W})\) required for each of those values of D? (Hint: Set up a dimensional equation relating \(P\) to \(F\).) (c) Suppose you have a suspension of a mixture of \(D=40 \mathrm{nm}\) and \(D=120 \mathrm{nm}\) spherical silver nanoparticles and a \(10 \mathrm{Hz} / 532 \mathrm{nm}\) pulsed laser source with a \(7 \mathrm{nm}\) diameter spot and adjustable power. Describe how you would use the laser to produce a suspension of particles of only a single size and state what that size would be.

Ever wonder why espresso costs much more per cup than regular drip coffee? Part of the reason is the expensive equipment needed to brew a proper espresso. A high-powered burr grinder first shears the coffee beans to a fine powder without producing too much heat. (Heating the coffee in the grinding stage prematurely releases the volatile oils that give espresso its rich flavor and aroma.) The ground coffee is put into a cylindrical container called a gruppa and tamped down firmly to provide an even flow of water through it. An electrically heated boiler inside the espresso machine maintains water in a reservoir at 1.4 bar and \(109^{\circ} \mathrm{C}\). An electric pump takes cold water at \(15^{\circ} \mathrm{C}\) and 1 bar, raises its pressure to slightly above 9 bar, and feeds it into a heating coil that passes through the reservoir. Heat transferred from the reservoir through the coil wall raises the water temperature to \(96^{\circ} \mathrm{C}\). The heated water flows into the top of the gruppa at \(96^{\circ} \mathrm{C}\) and 9 bar, passes slowly through the tightly packed ground beans, and dissolves the oils and some of the solids in the beans to become espresso, which decompresses to 1 atm as it exits the machine. The water temperature and uniform flow through the bed of packed coffee in the gruppa lead to the more intense flavor of espresso relative to normal drip coffee. Water drawn directly from the reservoir is expanded to atmospheric pressure where it forms steam, which is used to heat and froth milk for lattes and cappuccinos. (a) Sketch this process, using blocks to represent the pump, reservoir, and gruppa. Label all heat and work flows in the process, including electrical energy. (b) To make a 14-oz latte, you would steam 12 ounces of cold milk (3^'C) until it reaches 71 ^ C and pour it over 2 ounces of espresso. Assume that the steam cools but none of it condenses as it bubbles through the milk. For each latte made, the heating element that maintains the reservoir temperature must supply enough energy to heat the espresso water plus enough to heat the milk, plus additional energy. Assuming $$ \left(C_{p}\right)_{\operatorname{milk}}=3.93 \frac{\mathrm{J}}{\mathrm{g} \cdot^{\circ} \mathrm{C}}, \quad \mathrm{SG}_{\mathrm{milk}}=1.03 $$ calculate the quantity of electrical energy that must be provided to the heating element to accomplish those two functions. Why would more energy than what you calculate be required? (There are several reasons.) (c) Coffee beans contain a considerable amount of trapped carbon dioxide, not all of which is released when the beans are ground. When the hot pressurized water percolates through the ground beans, some of the carbon dioxide is absorbed in the liquid. When the liquid is then dispensed at atmospheric pressure, fine \(\mathrm{CO}_{2}\) bubbles come out of solution. In addition, one of the chemical compounds formed when the coffee beans are roasted and extracted into the espresso is melanoidin, a surfactant. Surfactant molecules are asymmetrical, with one end being hydrophilic (drawn to water) and the other end hydrophobic (repelled by water). When the bubbles (thin water films containing \(\mathrm{CO}_{2}\) ) pass through the espresso liquid, the hydrophilic ends of the melanoidin molecules attach to the bubbles and the dissolved bean oils in turn attach to the hydrophobic ends. The result is that the bubbles emerge coated with the oils to form the crema , the familiar reddish brown stable foam at the surface of good espresso. Speculate on why you don't see crema in normal drip coffee. (Hint: Henry's law should show up in your explanation.) Note: All soaps and shampoos contain at least one surfactant species. (A common one is sodium lauryl sulfate.) Its presence explains why if you have greasy hands, washing with plain water may leave the grease untouched but washing with soap removes the grease. (d) Explain in your own words (i) how espresso is made, (ii) why espresso has a more intense flavor than regular drip coffee, (iii) what the crema in espresso is, how it forms, and why it doesn't appear in regular drip coffee, and (iv) why washing with plain water does not remove grease but washing with soap does. (Note: Many people automatically assume that all chemical engineers are extraordinarily intelligent. If you can explain those four things, you can help perpetuate that belief.)

A liquid stream containing 50.0 mole \(\%\) benzene and the balance toluene at \(25^{\circ} \mathrm{C}\) is fed to a continuous single-stage evaporator at a rate of \(1320 \mathrm{mol} / \mathrm{s}\). The liquid and vapor streams leaving the evaporator are both at \(95.0^{\circ} \mathrm{C}\). The liquid contains 42.5 mole \(\%\) benzene and the vapor contains 73.5 mole\% benzene. (a) Calculate the heating requirement for this process in \(\mathrm{kW}\). (b) Using Raoult's law (Section 6.4b) to describe the equilibrium between the vapor and liquid outlet streams, determine whether or not the given benzene analyses are consistent with each other. If they are, calculate the pressure (torr) at which the evaporator must be operating; if they are not, give several possible explanations for the inconsistency.

The brakes on an automobile act by forcing brake pads, which have a metal support and a lining, to press against a disk (rotor) attached to the wheel. Friction between the pads and the disk causes the car to slow or stop. Each wheel has an iron brake disk with a mass of \(15 \mathrm{lb}_{\mathrm{m}}\) and two brake pads, each having a mass of \(11 \mathrm{b}_{\mathrm{m}}\). (a) Suppose an automobile is moving at 55 miles per hour when the driver suddenly applies the brakes and brings the car to a rapid halt. Take the heat capacity of the disk and brake pads to be \(0.12 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) and assume that the car stops so rapidly that heat transfer from the disk and pads has been insignificant. Estimate the final temperature of the disk and pads if the car is (i) a Toyota Camry, which has a mass of about \(3200 \mathrm{Ib}_{\mathrm{m}},\) or (ii) a Cadillac Escalade, which has a mass of about \(5.900 \mathrm{lb}_{\mathrm{m}}.\) (b) Why are the linings on brake pads no longer made of asbestos? Your answer should provide information on specific issues or concerns caused by the use of asbestos.
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