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800 Chapter 19 / The First Law of Thermodynamics SOLUTION: We use the equation = The relevant chemical equation for the formation of one mole of NH3(g) from its elements is NH3(g) and = - = -31.21 + 10-3 - (5.895 The integral of is 1000 - = (-21.91+14.07-1.913) - = -9.75 (1000 K) = = = The pressure dependence of H (which we will study in Chapter 22) is usually r much smaller than its temperature dependence. Problems 19-1. Suppose that a 10-kg mass of iron at 20°C is dropped from a height of 100 meters. What is the kinetic energy of the mass just before it hits the ground? What is its speed? What would be the final temperature of the mass if all its kinetic energy at impact is transformed into internal energy? Take the molar heat capacity of iron to be P = 25.1 and the gravitational acceleration constant to be 9.80 m.s-2. 19-2. Consider an ideal gas that occupies 2.50 at a pressure of 3.00 bar. If the gas is compressed isothermally at a constant external pressure, that the final volume is 0.500 calculate the smallest value Pext can have. Calculate the work involved using this value of 19-3. A one-mole sample of occupies 2.00 at a temperature of 300 K. If the gas is compressed isothermally at a constant external pressure, that the final volume isProblems 801 0.750 calculate the smallest value Pext can have, assuming that satisfies the van der Waals equation of state under these conditions. Calculate the work involved using this value of 19-4. Calculate the work involved when one mole of an ideal gas is compressed reversibly from 1.00 bar to 5.00 bar at a constant temperature of 300 K. 19-5. Calculate the work involved when one mole of an ideal gas is expanded reversibly from 20.0 to 40.0 at a constant temperature of 300 K. 19-6. Calculate the minimum amount of work required to compress 5.00 moles of an ideal gas isothermally at 300 K from a volume of 100 to 40.0 19-7. Consider an ideal gas that occupies 2.25 L at 1.33 bar. Calculate the work required to compress the gas isothermally to a volume of 1.50 L at a constant pressure of 2.00 bar followed by another isothermal compression to 0.800 L at a constant pressure of 2.50 bar (Figure 19.4). Compare the result with the work of compressing the gas isothermally and reversibly from 2.25 L to 0.800 L. 19-8. Show that for an isothermal reversible expansion from a molar volume V1 to a final molar volume V2, 2' the work is given by RT In for the Redlich-Kwong equation. 19-9. Use the result of Problem 19-8 to calculate the work involved in the isothermal reversible expansion of one mole of (g) from a volume of 1.00 to 5.00 at 300 K. (See Table 16.4 for the values of A and B.) 19-10. Repeat the calculation in Problem 19-9 for a van der Waals gas. 19-11. Derive an expression for the reversible isothermal work of an expansion of a gas that obeys the Peng-Robinson equation of state. 19-12. One mole of a monatomic ideal gas initially at a pressure of 2.00 bar and a temperature of 273 K is taken to a final pressure of 4.00 bar by the reversible path defined by P/V = constant. Calculate the values of AU, q, and for this process. Take to be equal to 12.5 19-13. The isothermal compressibility of a substance is given by (1) For an ideal gas, B = /P, but for a liquid, B is fairly constant over a moderate pressure range. If B is constant, show that (2)802 Chapter 19 / The First Law of Thermodynamics where V 0 is the volume at a pressure Use this result to show that the reversible isothermal work of compressing a liquid from a volume (at a pressure to a volume V (at a pressure P) is given by (3) (You need to use the fact The fact that liquids are incompressible is reflected by B being small, that 1 for moderate pressures. Show that (4) Calculate the work required to compress one mole of toluene reversibly and isothermally from 10 bar to 100 bar at 20°C. Take the value of B to be 8.95 X 10-5 and the molar volume to be 0.106 at 20°C. 19-14. In the previous problem, you derived an expression for the reversible, isothermal work done when a liquid is compressed. Given that B is typically show that 1 for pressures up to about 100 bar. This result, of course, reflects the fact that liquids are not very compressible. We can exploit this result by substituting dV = - -BVdP from the defining equation of into = - PdV and then treating V as a constant. Show that this approximation gives Equation 4 of Problem 19-13. 19-15. Show that for a reversible adiabatic expansion of an ideal gas. 19-16. Show that for a reversible, adiabatic expansion of a monatomic gas that obeys the equation of state - b) = RT. Extend this result to the case of a diatomic gas. 19-17. Show that for a reversible adiabatic expansion of an ideal gas. 19-18. Show thatProblems 803 for an adiabatic expansion of an ideal gas. Show that this formula reduces to Equation 19.23 for a monatomic gas. 19-19. Calculate the work involved when one mole of a monatomic ideal gas at 298 K expands reversibly and adiabatically from a pressure of 10.00 bar to a pressure of 5.00 bar. 19-20. A quantity of N2(g) at 298 K is compressed reversibly and adiabatically from a volume of 20.0 to 5.00 Assuming ideal behavior, calculate the final temperature of the N2(g). 19-21. A quantity of CH4(g) at 298 K is compressed reversibly and adiabatically from 50.0 bar to 200 bar. Assuming ideal behavior, calculate the final temperature of the CH4(g). Take C, = 3R. 19-22. One mole of ethane at 25°C and one atm is heated to 1200°C at constant pressure. Assuming ideal behavior, calculate the values of q, AU, and given that the molar heat capacity of ethane is given by - over the above temperature range. Repeat the calculation for a constant-volume process. 19-23. The value of at 25°C and one bar is +290.8 kJ for the reaction Assuming ideal behavior, calculate the value of for this reaction. 19-24. Liquid sodium is being considered as an engine coolant. How many grams of sodium are needed to absorb 1.0 MJ of heat if the temperature of the sodium is not to increase by more than 10°C. Take Cp = for and for H2O(1). 19-25. A 25.0-g sample of copper at 363 K is placed in 100.0 g of water at 293 K. The copper and water quickly come to the same temperature by the process of heat transfer from copper to water. Calculate the final temperature of the water. The molar heat capacity of copper is 24.5 19-26. A 10.0-kg sample of liquid water is used to cool an engine. Calculate the heat removed (in joules) from the engine when the temperature of the water is raised from 293 K to 373 K. Take H2O(I). 19-27. In this problem, we will derive a general relation between Cp and Cv. Start with U U(P,T) and write (1) We could also consider V and T to be the independent variables of U and write (2)804 Chapter 19 / The First Law of Thermodynamics Now take and substitute its expression for dV into Equation 2 to obtain Compare this result with Equation 1 to obtain (3) and (4) Last, substitute PV into the left side of Equation (4) and use the definitions of Cp and Cv to obtain Show that - as it is for an ideal gas. 19-28. Following Problem 19-27, show that 19-29. Starting with H=U+PV, show that Interpret this result physically. 19-30. Given that for an ideal gas, prove that = 0 for an ideal gas. 19-31. Given that = 0 for an ideal gas, prove that = 0 for an ideal gas. 19-32. Show that if as is true for an ideal gas. 19-33. Differentiate H=U+PV with respect to V at constant temperature to show that = ideal gas. 19-34. Given the following data for sodium, plot H(T) - H(0) against T for sodium: melting point, K; boiling point, 1156 K; = 19-35. The values for the following equations are 3 Fe3O4(s) Use these data to calculate the value of for the reaction described byProblems 805 19-36. Given the following data, HF(g) = calculate the value of for the reaction described by 19-37. The standard molar heats of combustion of the isomers m-xylene and p-xylene are -4553.9 and -4556.8 respectively. Use these data, together with Hess's Law, to calculate the value of for the reaction described by m-xylene p-xylene 19-38. Given that = -2826.71 kJ for the combustion of 1.00 mol of fructose at 298.15 K, and the H° data in Table 19.2, calculate the value of H° for fructose at 298.15 K. 19-39. Use the data in Table 19.2 to calculate the value of H° for the combustion reactions described by the equations: a. b. Compare the heat of combustion per gram of the fuels and 19-40. Using Table 19.2, calculate the heat required to vaporize 1.00 mol of CC14(1) at 298 K. 19-41. Using the H° data in Table 19.2, calculate the values of for the following: b. In each case, state whether the reaction is endothermic or exothermic. 19-42. Use the following data to calculate the value of of water at 298 K and compare vap your answer to the one you obtain from Table 19.2: at 373 K = 40.7 = 33.6806 Chapter 19 / The First Law of Thermodynamics 19-43. Use the following data and the data in Table 19.2 to calculate the standard reaction enthalpy of the water-gas reaction at 1273 K. Assume that the gases behave ideally under these conditions. C(s) = - = = = -0.6366 + (7.049 X - + 19-44. The standard molar enthalpy of formation of at 298 K is -393.509 Use the following data to calculate the value of H° at 1000 K. Assume the gases behave ideally under these conditions. = = = -0.6366 + (7.049 X - + 19-45. The value of the standard molar reaction enthalpy for is -802.2 at 298 K. Using the heat-capacity data in Problems 19-43 and 19-44 in addition to = 2.099 + (7.272 X + (1.34 - to derive a general equation for the value of H° at any temperature between 300 K and r 1500 K. Plot versus T. Assume that the gases behave ideally under these 19-46. In all the calculations thus far, we have assumed the reaction takes place at constant temperature, SO that any energy evolved as heat is absorbed by the surroundings. Suppose, however, that the reaction takes place under adiabatic conditions, SO that all the energy released as heat stays within the system. In this case, the temperature of the system will increase, and the final temperature is called the adiabatic flame temperature. One relatively simple way to estimate this temperature is to suppose the reaction occurs at the initial temperature of the reactants and then determine to what temperature the products can be raised by the quantity H°. Calculate the adiabatic flame temperature if one mole of (g) is burned in two moles of at an initial temperature of 298 K. Use the results of the previous problem.Problems 807 19-47. Explain why the adiabatic flame temperature defined in the previous problem is also called the maximum flame temperature. 19-48. How much energy as heat is required to raise the temperature of 2.00 moles of from 298 K to 1273 K at 1.00 bar? Take = - 19-49. When one mole of an ideal gas is compressed adiabatically to one-half of its original volume, the temperature of the gas increases from 273 K to 433 K. Assuming that is independent of temperature, calculate the value of V for this gas. 19-50. Use the van der Waals equation to calculate the minimum work required to expand one mole of isothermally from a volume of 0.100 to a volume of 100 at 273 K. Compare your result with that which you calculate assuming ideal behavior. 19-51. Show that the work involved in a reversible, adiabatic pressure change of one mole of an ideal gas is given by where is the initial temperature and and P2 are the initial and final pressures, respec- tively. 19-52. In this problem, we will discuss a famous experiment called the Joule-Thomson exper- iment. In the first half of the 19th century, Joule tried to measure the temperature change when a gas is expanded into a vacuum. The experimental setup was not sensitive enough, however, and he found that there was no temperature change, within the limits of his error. Soon afterward, Joule and Thomson devised a much more sensitive method for measuring the temperature change upon expansion. In their experiments (see Figure 19.11), a constant applied pressure causes a quantity of gas to flow slowly from one chamber to another through a porous plug of silk or cotton. If a volume, of gas is pushed through the porous plug, the work done on the gas is 1 The pressure on the other side of the plug Initial state n moles at P2 Porous plug n moles P1 at P2 Final state FIGURE 19.11 A schematic description of the Joule-Thomson experiment.808 Chapter 19 / The First Law of Thermodynamics is maintained at P2, if a volume enters the right-side chamber, then the net work is given by The apparatus is constructed that the entire process is adiabatic, SO q = 0. Use the First Law of Thermodynamics to show that or that = 0 for a Joule-Thomson expansion. Starting with show that Interpret physically the derivative on the left side of this equation. This quantity is called the Joule-Thomson coefficient and is denoted by In Problem 19-54 you will show that it equals zero for an ideal gas. Nonzero values of directly reflect intermolecular interactions. Most gases cool upon expansion [a positive value of and a Joule- Thomson expansion is used to liquefy gases. 19-53. The Joule-Thomson coefficient (Problem 19-52) depends upon the temperature and pressure, but assuming an average constant value of 0.15 for N2(g), calculate the drop in temperature if N2(g) undergoes a drop in pressure of 200 bar. 19-54. Show that the Joule-Thomson coefficient (Problem 19-52) can be written as Show that 19-55. Use the rigid rotator-harmonic oscillator model and the data in Table 18.2 to plot Cp ,(T) for CO(g) from 300 K to 1000 K. Compare your result with the expression given in Problem 19-43. 19-56. Use the rigid rotator-harmonic oscillator model and the data in Table 18.4 to plot for CH4(g) from 300 K to 1000 K. Compare your result with the expression given in Problem 19-45. 19-57. Why do you think the equations for the dependence of temperature on volume for a reversible adiabatic process (see Equation 19.22 and Example 19.6) depend upon whether the gas is a monatomic gas or a polyatomic gas?