#set document(title: "3.5 Heat Capacities of an Ideal Gas", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 3.5#h(0.6em)Heat Capacities of an Ideal Gas We learned about specific heat and molar heat capacity in Temperature and Heat; however, we have not considered a process in which heat is added. We do that in this section. First, we examine a process where the system has a constant volume, then contrast it with a system at constant pressure and show how their specific heats are related. Let’s start with looking at , which shows two vessels #emph[A] and #emph[B], each containing the same type of ideal gas at a temperature #emph[T] and a volume #emph[V]. The only difference between the two vessels is that the piston at the top of #emph[A] is fixed, whereas the one at the top of #emph[B] is free to move against a constant external pressure #emph[p]. We now consider what happens when the temperature of the gas in each vessel is slowly increased to #math.equation(block: false, alt: "T plus d T")[$T + d T$] with the addition of heat. #figure(figph[Two containers, labeled Vessel A and Vessel B, are shown. Both are filled with gas and are capped by a piston. In vessel A, the piston is pinned in place. In vessel B, the piston is free to slide, as indicated by a double headed arrow near the piston.], alt: "Two containers, labeled Vessel A and Vessel B, are shown. Both are filled with gas and are capped by a piston. In vessel A, the piston is pinned in place. In vessel B, the piston is free to slide, as indicated by a double headed arrow near the piston.", caption: [Two vessels are identical except that the piston at the top of #emph[A] is fixed, whereas that atop #emph[B] is free to move against a constant external pressure #emph[p].]) Since the piston of vessel #emph[A] is fixed, the volume of the enclosed gas does not change. Consequently, the gas does no work, and we have from the first law #math.equation(block: true, alt: "d E sub int equals d Q minus d W equals d Q .")[$d E_("int") = d Q − d W = d Q .$] We represent the fact that the heat is exchanged at constant volume by writing #math.equation(block: true, alt: "d Q equals n C sub V d T ,")[$d Q = n C_(V) d T ,$] where #math.equation(block: false, alt: "C sub V")[$C_(V)$] is the #strong[molar heat capacity at constant volume] of the gas, and, as before, #emph[n] is the number of moles of the gas. In addition, since #math.equation(block: false, alt: "d E sub int equals d Q")[$d E_("int") = d Q$] for this particular process, #math.equation(block: true, alt: "d E sub int equals n C sub V d T .")[$d E_("int") = n C_(V) d T .$] We obtained this equation assuming the volume of the gas was fixed. However, internal energy is a state function that depends on only the temperature of an ideal gas. Therefore, #math.equation(block: false, alt: "d E sub int equals n C sub V d T")[$d E_("int") = n C_(V) d T$] gives the change in internal energy of an ideal gas for any process involving a temperature change #emph[dT]. When the gas in vessel #emph[B] is heated, it expands against the movable piston and does work #math.equation(block: false, alt: "d W equals p d V .")[$d W = p d V .$] In this case, the heat is added at constant pressure, and we write #math.equation(block: true, alt: "d Q equals n C sub p d T ,")[$d Q = n C_(p) d T ,$] where #math.equation(block: false, alt: "C sub p")[$C_(p)$] is the #strong[molar heat capacity at constant pressure] of the gas. Furthermore, since the ideal gas expands against a constant pressure, #math.equation(block: true, alt: "d open parenthesis p V close parenthesis equals d open parenthesis n R T close parenthesis")[$d ( p V ) = d ( n R T )$] becomes #math.equation(block: true, alt: "p d V equals n R d T .")[$p d V = n R d T .$] Finally, inserting the expressions for #emph[dQ] and #emph[pdV] into the first law, we obtain #math.equation(block: true, alt: "d E sub int equals d Q minus p d V equals open parenthesis n C sub p minus n R close parenthesis d T .")[$d E_("int") = d Q − p d V = ( n C_(p) − n R ) d T .$] We have found #math.equation(block: false, alt: "d E sub int")[$d E_("int")$] for both an isochoric and an isobaric process. Because the internal energy of an ideal gas depends only on the temperature, #math.equation(block: false, alt: "d E sub int")[$d E_("int")$] must be the same for both processes. Thus, #math.equation(block: true, alt: "n C sub V d T equals open parenthesis n C sub p minus n R close parenthesis d T ,")[$n C_(V) d T = ( n C_(p) − n R ) d T ,$] and #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "C sub p equals C sub V plus R .")[$C_(p) = C_(V) + R .$] ] The derivation of was based only on the ideal gas law. Consequently, this relationship is approximately valid for all dilute gases, whether monatomic like He, diatomic like #math.equation(block: false, alt: "O sub 2 ,")[$"O"_(2) ,$] or polyatomic like #math.equation(block: false, alt: "CO sub 2 or NH sub 3 .")[$"CO"_(2) attach(#h(0.2em) "or NH", b: 3) "."$] In the preceding chapter, we found the molar heat capacity of an ideal gas under constant volume to be #math.equation(block: true, alt: "C sub V equals the fraction d over 2 R ,")[$C_(V) = frac(d, 2) R ,$] where #emph[d] is the number of degrees of freedom of a molecule in the system. shows the molar heat capacities of some dilute ideal gases at room temperature lists the experimental values for some gases. The heat capacities of real gases are somewhat higher than those predicted by the expressions of #math.equation(block: false, alt: "C sub V")[$C_(V)$] and #math.equation(block: false, alt: "C sub p")[$C_(p)$] given in . This indicates that vibrational motion in polyatomic molecules is significant, even at room temperature. Nevertheless, the difference in the molar heat capacities, #math.equation(block: false, alt: "C sub p minus C sub V ,")[$C_(p) − C_(V) ,$] is very close to #emph[R], even for the polyatomic gases. #figure(table( columns: 3, align: left, inset: 6pt, [Molar Heat Capacities of Dilute Ideal Gases at Room Temperature], [], [], [Type of Molecule], [#math.equation(block: false, alt: "C sub p")[$C_(p)$] #linebreak() (J/mol K)], [#math.equation(block: false, alt: "C sub V")[$C_(V)$] #linebreak() (J/mol K)], [Monatomic], [#math.equation(block: false, alt: "the fraction 5 over 2 R equals 20.79")[$frac(5, 2) R = 20.79$]], [#math.equation(block: false, alt: "the fraction 3 over 2 R equals 12.47")[$frac(3, 2) R = 12.47$]], [Diatomic], [#math.equation(block: false, alt: "the fraction 7 over 2 R equals 29.10")[$frac(7, 2) R = 29.10$]], [#math.equation(block: false, alt: "the fraction 5 over 2 R equals 20.79")[$frac(5, 2) R = 20.79$]], [Polyatomic], [#math.equation(block: false, alt: "4 R equals 33.26")[$4 R = 33.26$]], [#math.equation(block: false, alt: "3 R equals 24.94")[$3 R = 24.94$]], )) === Summary - For an ideal gas, the molar capacity at constant pressure #math.equation(block: false, alt: "C sub p")[$C_(p)$] is given by #math.equation(block: false, alt: "C sub p equals C sub V plus R equals d R / 2 plus R")[$C_(p) = C_(V) + R = d R "/" 2 + R$], where d is the number of degrees of freedom of each molecule/entity in the system. - A real gas has a specific heat close to but a little bit higher than that of the corresponding ideal gas with #math.equation(block: false, alt: "C sub p ≃ C sub V plus R .")[$C_(p) ≃ C_(V) + R .$] === Conceptual Questions How can an object transfer heat if the object does not possess a discrete quantity of heat? Most materials expand when heated. One notable exception is water between #math.equation(block: false, alt: "0 ° C")[$0 #h(0.2em) "°" "C"$] and #math.equation(block: false, alt: "4 ° C,")[$4 #h(0.2em) "°" "C,"$] which actually decreases in volume with the increase in temperature. Which is greater for water in this temperature region, #math.equation(block: false, alt: "C sub p")[$C_(p)$] or #math.equation(block: false, alt: "C sub V")[$C_(V)$] ? #solutionbox[ Typically #math.equation(block: false, alt: "C sub p")[$C_(p)$] is greater than #math.equation(block: false, alt: "C sub V")[$C_(V)$] because when expansion occurs under constant pressure, it does work on the surroundings. Therefore, heat can go into internal energy and work. Under constant volume, all heat goes into internal energy. In this example, water contracts upon heating, so if we add heat at constant pressure, work is done on the water by surroundings and therefore, #math.equation(block: false, alt: "C sub p")[$C_(p)$] is less than #math.equation(block: false, alt: "C sub V")[$C_(V)$]. ] Why are there two specific heats for gases #math.equation(block: false, alt: "C sub p")[$C_(p)$] and #math.equation(block: false, alt: "C sub V")[$C_(V)$], yet only one given for solid? === Problems The temperature of an ideal monatomic gas rises by 8.0 K. What is the change in the internal energy of 1 mol of the gas at constant volume? #solutionbox[ 100 J ] For a temperature increase of #math.equation(block: false, alt: "10 ° C")[$10 #h(0.2em) "°" "C"$] at constant volume, what is the heat absorbed by (a) 3.0 mol of a dilute monatomic gas; (b) 0.50 mol of a dilute diatomic gas; and (c) 15 mol of a dilute polyatomic gas? If the gases of the preceding problem are initially at 300 K, what are their internal energies after they absorb the heat? #solutionbox[ a. 11,600 J; b. 3220 J; c. 116,000 J ] Consider 0.40 mol of dilute carbon dioxide at a pressure of 0.50 atm and a volume of 50 L. What is the internal energy of the gas? When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by #math.equation(block: false, alt: "10 ° C")[$10 #h(0.2em) "°" "C"$]. What is the work done on the gas? #solutionbox[ 850 J ] One mole of a dilute diatomic gas occupying a volume of 10.00 L expands against a constant pressure of 2.000 atm when it is slowly heated. If 400.0 J of heat are added in the process, what is its final volume?