Consider an ideal gas with constant cp =R and in an initial state at 300K and 100 kPa. (a) Compute the molar heat transfer, Q, required to raise the temperature of the gas to 500K at constant pressure. (answer: 4,157 J/mol) (b) Do the same as in (a), except assume the temperature is increased to 500K at constant volume. (answer: 2,494 J/mol) (c) Show that in general if an ideal gas with constant cp (and c) undergoes a transition from T₁ to T₂ in which the pressure varies linearly with volume, then: q= -1/[(₁x + ‚ + c,)(T₂ - T;) + R (RT₂ - BT;)] = ½ [(cv +cp)(T₂− T₁)+(P‚v₂− P₂v₁)] = 21 [ (Gr + C₂) (T₁ - T. )+ R ( ^ T, - + T.;)] V₂ (d) Using the result of (c), graph Q vs. V₂ using the given initial state of 300K, 100 kPa and a final temperature of 500K. (e) For the processes considered in (d), identify the final state for which Q = 0 and quantitatively illustrate the process in a P vs. V graph, complete with isotherms for T₁ and T₂. (You should see that the linear process actually cuts above the 500K isotherm before reaching 500K again at the final state.) (answer: V₂ : 0.0130352 m³/mol, P₂: 318,925 Pa) (f) Show that for the linear process described in (e), the molar heat transfer up to any intermediate state P, V, is given by: Q.,.,. = 1 [SR(V, - V‚) + D} = B (4V² — SVV, +V²)] Depending on how you group terms, you may end up with an expression that appears different from that above. Regardless, your result should reveal that Q=0 when V = V₁ and V = V₂. The key is to arrive at a heat transfer than can be computed based on the initial and final states and the intermediate molar volume V¡ . Recall that c, = R. (g) Quantitatively graph the result of part (f), allowing V to vary all the way from the initial molar volume V, to the final molar volume V₂ . Since V₁ > V₂, it might also be helpful to reverse the x axis, or think about whether heat is being added or removed as moving from right to left; however, the parabolic symmetry of the graph somewhat trivializes this exercise. (h) One way to think about the graph in (g) is that you really graphed the sum of all the incremental heat transfers, 8Q, up to a given point in the process; i.e., №u, w, – ſ", ôē . Determine the state along that path where 8Q=0. (answer: 0.0189894 m³/mol, 209,463 Pa, 478.388K)

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Consider an ideal gas with constant cp =/R_and in an initial state at 300K and 100 kPa.
(a) Compute the molar heat transfer, Q, required to raise the temperature of the gas to 500K at
constant pressure. (answer: 4,157 J/mol)
(b) Do the same as in (a), except assume the temperature is increased to 500K at constant
volume. (answer: 2,494 J/mol)
(c) Show that in general if an ideal gas with constant cp (and c) undergoes a transition from
T to T₂ in which the pressure varies linearly with volume, then:
q=; (c +c)(T_ -
-¹/[(
= ±[(cv +cp)(T₂− T₁)+(P,‚v₂ − P₂v₁)]
21 [ (cv + c₂)(T₂ - T.) + R ( ²^ T, - "^_T₂)
V₂
) + R ( 1 T₁ - T)]
P₁
(d) Using the result of (c), graph Q vs. V₂ using the given initial state of 300K, 100 kPa and a
final temperature of 500K.
(e) For the processes considered in (d), identify the final state for which Q=0 and
quantitatively illustrate the process in a P vs. V graph, complete with isotherms for T₁ and
T₂. (You should see that the linear process actually cuts above the 500K isotherm before
reaching 500K again at the final state.) (answer: V₂: 0.0130352 m³/mol, P₂: 318,925 Pa)
(f) Show that for the linear process described in (e), the molar heat transfer up to any
intermediate state P, V, is given by:
Ñ‚,-…, =
- 1 [SR(V-V.) + D =D (4V² - SVV,+V:)]
Depending on how you group terms, you may end up with an expression that appears
different from that above. Regardless, your result should reveal that → = 0 when V₁ = V₁
and V = V₂. The key is to arrive at a heat transfer than can be computed based on the initial
and final states and the intermediate molar volume V¡ . Recall that c = R.
(g) Quantitatively graph the result of part (f), allowing V, to vary all the way from the initial
molar volume V, to the final molar volume V₂ . Since V₁ > V₂, it might also be helpful to
reverse the x axis, or think about whether heat is being added or removed as moving from
right to left; however, the parabolic symmetry of the graph somewhat trivializes this exercise.
(h) One way to think about the graph in (g) is that you really graphed the sum of all the
incremental heat transfers, dē, up to a given point in the process; i.e., ē,→ = ſ" dē .
Determine the state along that path where dQ=0.
(answer: 0.0189894 m³/mol, 209,463 Pa, 478.388K)
Transcribed Image Text:Consider an ideal gas with constant cp =/R_and in an initial state at 300K and 100 kPa. (a) Compute the molar heat transfer, Q, required to raise the temperature of the gas to 500K at constant pressure. (answer: 4,157 J/mol) (b) Do the same as in (a), except assume the temperature is increased to 500K at constant volume. (answer: 2,494 J/mol) (c) Show that in general if an ideal gas with constant cp (and c) undergoes a transition from T to T₂ in which the pressure varies linearly with volume, then: q=; (c +c)(T_ - -¹/[( = ±[(cv +cp)(T₂− T₁)+(P,‚v₂ − P₂v₁)] 21 [ (cv + c₂)(T₂ - T.) + R ( ²^ T, - "^_T₂) V₂ ) + R ( 1 T₁ - T)] P₁ (d) Using the result of (c), graph Q vs. V₂ using the given initial state of 300K, 100 kPa and a final temperature of 500K. (e) For the processes considered in (d), identify the final state for which Q=0 and quantitatively illustrate the process in a P vs. V graph, complete with isotherms for T₁ and T₂. (You should see that the linear process actually cuts above the 500K isotherm before reaching 500K again at the final state.) (answer: V₂: 0.0130352 m³/mol, P₂: 318,925 Pa) (f) Show that for the linear process described in (e), the molar heat transfer up to any intermediate state P, V, is given by: Ñ‚,-…, = - 1 [SR(V-V.) + D =D (4V² - SVV,+V:)] Depending on how you group terms, you may end up with an expression that appears different from that above. Regardless, your result should reveal that → = 0 when V₁ = V₁ and V = V₂. The key is to arrive at a heat transfer than can be computed based on the initial and final states and the intermediate molar volume V¡ . Recall that c = R. (g) Quantitatively graph the result of part (f), allowing V, to vary all the way from the initial molar volume V, to the final molar volume V₂ . Since V₁ > V₂, it might also be helpful to reverse the x axis, or think about whether heat is being added or removed as moving from right to left; however, the parabolic symmetry of the graph somewhat trivializes this exercise. (h) One way to think about the graph in (g) is that you really graphed the sum of all the incremental heat transfers, dē, up to a given point in the process; i.e., ē,→ = ſ" dē . Determine the state along that path where dQ=0. (answer: 0.0189894 m³/mol, 209,463 Pa, 478.388K)
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