CHAPTER 21 HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS ActivPhysics can help with these problems: Activities 8.5-8.13 -. Section 21-1: The First Law of Section 21-2: Thermodynamic Processes . Thermodynamics Problem Problem 8. An ideal gas expands from the state (PI,Vl) to the 1. In a perfectly insulated container, 1.0 kg of water state (9,Vz), where P2 = 2P1 and V2 = 2V1. The is stirred vigorously until its temperature rises by expansion proceeds along the straight diagonal path 7.0°C. How much work was done on the water? AB shown in Fig. 21-26. Find an expression for the work done by the gas during this process. Solution Since the container is perfectly insulated thermally, no heat enters or leaves the water in it. Thus, Q = 0 in Equation 21-1. The change in the internal energy of the water is determined from its temperature rise, AU = rnc AT (see comments in Section 19-4 on internal energy), so W = -AW = -(1 kg) x (4.184 kJ/kg.K)(7 K) = -29.3 kJ. (The negative sign signifies that work was done on the water.) Problem FIGURE 21-26 Problems 8 and 9. 5. The most efficient large-scale electric power generating systems use high-temperature gas turbines and a so-called combined cycle system that Solution maximizes the conversion of thermal energy into The work done by the gas equals the area under the useful work. One such plant produces electrical straight diagonal path AB in Fig. 21-26. The area of energy at the rate of 360 hlW, while extracting this trapezoid is W = +(PI + P2)(V2 - Vl ) = +(PI + energy-from its natural gas fuel at the rate of 2Pl) (24 - Vl) = 2 PI v]. W can also be obtained from 670 MW. (a) At what rate does it reject waste heat Equation 21-3. On the path AB, P = PI + (V - Vl)x to the environment? (b) Find its efficiency, defined (P2 - PI)/(% - Vl). Then as the percent of the total energy extracted from the fuel that ends up as work. Solution (a) If we assume that the generating system operates in a cycle and choose it as "the system," then dli/dt = 0 and Equation 21-2 implies dQ/dt = dW/dt. Here, dW/dt is the rate that the generator supplies energy to its surroundings (360 h/lW in this problem) and dQ/dt Problem . is the net rate of heat flow into the generator from the 11. A balloon contains 0.30 mot of helium. It rises, surroundings. Since the system is just the generator, while maintaining a constant 300 K temperature, the net heat flow is the difference between the heat to an altitude where its volume has expanded extracted from its fuel and the heat exhausted to the 5 times. How much work is done by the gas in the environment, i.e., dQ/dt = (dQ/dt)i, - (dQ/dt)our = - - balloon during this isothermal expansion? Neglect 670 MW - (dQ/dt),,,. Therefore, (dQ/dt),,, = tension forces in the balloon. 670 MW - 360 MW = 310 MW. (Note: If the system is assumed to be the generator and its fuel, as in Solution Example 21-1, then dW/dt is still 360 MW, but the During an isothermal expansion, the work done by a system's internal energy decreases because energy is given amount of ideal gas is W = nR'TIn(h/V1) = extracted from the fuel, dU/dt = -670 MW, and there (0.3 mo1)(8.314 J/mol.K)(300 K)ln(5) = 1.20 kJ (see is no heat input. Then dQ/dt = -670 MW + 360 MW = -310 MW, representing the rate of heat Equation 2 1-4). rejected to the environment.) (b) The efficiency is (dW/dt)/(dQ/dt)i, = 360 MW/670 MW = 53.7% (see Section 22-2). - Problem Problem 37. A bicycle pump consists of a cylinder 30 cm long 19. A gas with y = 1.4 is at 100 kPa pressure and when the pump handle is all the way out. The occupies 5.00 L. (a) How much work does it take pump contains air (7 = 1.4) at 20°C. If the pump to compress the gas adiabatically to 2.50 L? .. outlet is blocked and the handle pushed until the (b) What is its final pressure? internal length of the pump cylinder is 17 cm, by Solution how much does the air temperature rise? Assume that no heat is lost. The work done by an ideal gas undergoing an adiabatic process is W12 = (PIK - P2V2)/(y - 1) (see . Equation 21-14). Since the compression is specified by Solution given values of PI, Vl, and V2, we first find the final If no heat is lost (or gained) by the gas, the pressure from the adiabatic gas law. (b) P2 = Pl (Vl+ compression is adiabatic and Equation 21-13b gives V2l7 = (100 kPa)(5 L12.5 L)1.4 = 264 kPa. (a) Then TV7-' = ~~~o7-l.Therefore, the temperature rise is the work done on the gas (which is -W12) is -W12 = T - To = AT = To[(Vo/V)7-I - 1). Since Vo/V = (P2V2-PlVl)/(y-l) = [(264 kPa)(2.5 L)-(100 kPa)x (30 cm/17 cm), AT = [(30/17)O.~- 1](293 K) = (5 L)]/0.4 = 399 J. 74.7 CO. Problem 25:By how much must the volume of a gas with 43. A mixture of monatomic and diatomic gases has specific heat ratio y = 1.52. What fraction of the 7 = 1.4 be changed in an adiabatic process if the kelvin temperature is to double? molecules are monatomic? Solution Solution V/G = (T~/T)'/(Y-') = (0.5)'/~.~= 0.177 The internal energy of a mixture of two ideal gases is (Equation 21-13b). U = flNE~ + fiNE2, where fi is the fraction of the total number of molecules, N, of type 1, and El is the average energy of a molecule of type 1, etc. Problem Classically, E = g(;kT), where g is the number of 31. An ideal gas with y = 1.67 starts at point A in degrees of freedom. The molar specific heat at constant Fig. 21-29, where its volume and pressure are volume is Cv = (i)(d~/dT)= (N~/N)dldTx 1.00 m3 and 250 kPa, respectively. It then (flNgl;kT + f2Ng24kT) = @(fig1 + f2g2). Suppose undergoes an adiabatic expansion that triples its that the temperature range is such that 91 = 3 for the volume, ending at point B. It's then heated at monatomic gas, and 92 = 5 for the diatomic gas, as constant volume to point C, then compressed discussed in Section 21-3. Then Cv = R(1.5fl + isothermally back to A. Find (a) the pressure at 2.5 f2) = R(2.5 - fl), where f2 = 1 - fl since the sum B, (b) the pressure at C, and (c) the net work of the fractions of the mixture is one. Now, Cv can done on the gas. also be specified by the ratio 7 = Cp/Cv = 1 + R/CV, or Cv = R/(y - I), so in this problem, 2.5 - fl = 110.52, or5= 57.7%. Problem 45. A gas mixture contains monatomic argon and diatomic oxygen. An adiabatic expansion that doubles its volume results in the pressure dropping to one-third of its original value. What fraction of the molecules are argon? FIGURE 21-29 Problem 31. Solution From the pressures and volumes in the described Solution adiabatic expansion, PoV2 = (iPo)(2Vo)7,we can calculate that y = In 3/ In 2 = 1.58. Then the result of (a) From the adiabatic law for an ideal gas Problem 43 gives 2.5 - fA, = 110.58, Or fAr = 79.0%. (Equation 21-13a), PB = PA(v~/VB)7= (250 kPa) x = 39.9 kPa. (b) Point C lies on an isotherm with A, so the ideal gas law (Equation 20-2) yields PC = PAVA/VC= (250 kPa)(i) = 83.3 kPa. (c) Wnet = WAB + WBC + WCA. WAB is for an adiabatic process (Equation 21-14) and equals (PAVA - PBVB)/(~- 1) = [(250 kPa) (1 m3) - (39.9 kPa)(3 m3)]/0.67 = 194 kJ; WBC is for an isovolumic process and equals zero; WCA is for an isothermal Process (Equation 21-4) and equals ~WAIn(v~/vc) = PAVA ln(v~/V~)= 250 kJ In(?) = -275 kJ. Thus, Wnet = -80.2 kJ. The work done on the gas is the negative of this. .