571_PartUniversity Physics Solution

571_PartUniversity Physics Solution - 19-6 Chapter 19 (b)...

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19-6 Chapter 19 (b) Using Equation (19.14), 645 J 119.9 K (0.185 mol)(29.07 J mol K) p Q T nC Δ= = = and 900 K. T = The pV -diagram is sketched in Figure 19.23b. EVALUATE: At constant pressure some of the heat energy added to the gas leaves the gas as expansion work and the internal energy change is less than if the same amount of heat energy is added at constant volume. T Δ is proportional to . U Δ Figure 19.23 19.24. IDENTIFY and SET UP: Use information about the pressure and volume in the ideal gas law to determine the sign of , T Δ and from that the sign of Q . EXECUTE: For constant p , p Qn C T Since the gas is ideal, p Vn R T = and for constant p , . p RT Δ p p C pV C nR R ⎛⎞ Δ = ⎜⎟ ⎝⎠ Since the gas expands, 0 V Δ> and therefore 0. Q > 0 Q > means heat goes into gas. EVALUATE: Heat flows into the gas, W is positive and the internal energy increases. It must be that . QW > 19.25. IDENTIFY: . UQW Δ=− For an ideal gas, , V UCT Δ and at constant pressure, . WpVn =Δ= Δ SET UP: 3 2 V CR = for a monatomic gas. EXECUTE: 333 222 () . UnRT pV W Δ= Δ= Then 5 2 5 2 ,so . QU WWW Q = Δ+ = = EVALUATE: For diatomic or polyatomic gases, V C is a different multiple of R and the fraction of Q that is used for expansion work is different. 19.26. IDENTIFY: For an ideal gas, , V Δ= Δ and at constant pressure, . p Δ SET UP: 3 2 V = for a monatomic gas. EXECUTE: 3 43 3 3 3 33 22 2 (4.00 10 Pa)(8.00 10 m 2.00 10 m ) 360 J. UnRT pV −− × × × = EVALUATE: 2 3 240 J. Wn U =Δ=Δ= 55 23 6 0 0 J . p CTnRT U = Δ = Δ = 600 J of heat energy flows into the gas. 240 J leaves as expansion work and 360 J remains in the gas as an increase in internal energy. 19.27. IDENTIFY: For a constant volume process, . V CT = Δ For a constant pressure process, . p For any process of an ideal gas, . V Un SET UP: From Table 19.1, for 2 N , 20.76 J/mol K V C = and 29.07 J/mol K. p C = Heat is added, so Q is positive and 1557 J. Q =+ EXECUTE: (a) 1557 J 25.0 K (3.00 mol)(20.76 J/mol K) V Q T nC = = + (b) 1557 J 17.9 K (3.00 mol)(29.07 J/mol K) p Q T nC = = + (c) V for either process, so U Δ is larger when T Δ is larger. The final internal energy is larger for the constant volume process in (a). EVALUATE: For constant volume 0 W = and all the energy added as heat stays in the gas as internal energy. For the constant pressure process the gas expands and 0. W > Part of the energy added as heat leaves the gas as expansion work done by the gas.
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The First Law of Thermodynamics 19-7 19.28. IDENTIFY: Apply p Vn R T = to calculate T . For this constant pressure process, . Wp V = Δ . p Qn CT Use UQW Δ=− to relate Q , W and . U Δ SET UP: 5 2.50 atm 2.53 10 Pa. For a monatomic ideal gas, 12.47 J/mol K V C = and 20.78 J/mol K. p C =⋅ EXECUTE: (a) 52 2 1 1 (2.53 10 Pa)(3.20 10 m ) 325 K. (3.00 mol)(8.314 J/mol K) pV T nR ×× == = 2 2 2 (2.53 10 Pa)(4.50 10 m ) 456 K. (3.00 mol)(8.314 J/mol K) pV T nR = (b) 3 2 3 3 (2.53 10 Pa)(4.50 10 m 3.20 10 m ) 3.29 10 J V −− =Δ= × × × = × (c) 3 (3.00 mol)(20.78 J/mol K)(456 K 325 K) 8.17 10 J p = = × (d) 3 4.88 10 J Δ=− = × EVALUATE: We could also calculate U Δ as 3 (3.00 mol)(12.47 J/mol K)(456 K 325 K) 4.90 10 J, V Un Δ= Δ= = × which agrees with the value we calculated in part (d).
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571_PartUniversity Physics Solution - 19-6 Chapter 19 (b)...

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