Physics Solution Manual for 1100 and 2101

# This change in pressure is given by the ideal gas law

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Unformatted text preview: pressure step is W = P∆V. But the ideal gas law applies, so W = P∆V = nR ∆T = (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 800.0 K – 400.0 K ) = 3320 J The first law of thermodynamics indicates that the heat is Q = ∆U + W = 3 nR ∆T + nR ∆T 2 = 5 2 (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 800.0 K – 400.0 K ) = 8310 J Step B → C The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is ∆U = 3 nR ∆T 2 = 3 2 (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 400.0 K – 800.0 K ) = – 4990 J The volume is constant in this step, so the work done by the gas is W = 0 J . Chapter 15 Problems 781 The first law of thermodynamics indicates that the heat is Q = ∆U + W = ∆U = – 4990 J Step C → D The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is ∆U = 3 nR ∆T 2 = 3 2 (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 200.0 K – 400.0 K ) = – 2490 J The work for this constant pressure step is W = P∆V. But the ideal gas law applies, so W = P∆V = nR ∆T = (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 200.0 K – 400.0 K ) = – 1660 J The first law of thermodynamics indicates that the heat is Q = ∆U + W = 3 nR ∆T + nR ∆T 2 = 5 2 (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 200.0 K – 400.0 K ) = – 4150 J Step D → A The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is ∆U = 3 nR ∆T 2 = 3 2 (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 400.0 K – 200.0 K ) = 2490 J The volume is constant in this step, so the work done by the gas is W = 0 J The first law of thermodynamics indicates that the heat is Q = ∆U + W = ∆U = 2490 J 30. REASONING The cylinder containing the gas is perfectly insulated, so no heat can enter or leave. Therefore, the compression of the gas is adiabatic, and the initial and final pressures (Pi, Pf) and volumes (Vi, Vf) of the gas are related by PViγ = Pf Vfγ (Equation 15.5), where i γ = 5 because the gas is monatomic and ideal. The final Kelvin temperature Tf of the gas is 3 twice the initial temperature Ti, so we have that Tf = 2Ti. We will use the ideal gas law PV = nRT (Equation 14.1) to determine the final pressure Pf of the gas from its volume and Kelvin temperature. In Equation 14.1, n is the number of moles of the gas and R = 8.31 J/(mol·K) is the universal gas constant. 782 THERMODYNAMICS SOLUTION In order to make use of the ideal gas law PV = nRT (Equation 14.1), it is convenient to rewrite PViγ = Pf Vfγ (Equation 15.5) so that the product PV is raised to the i power γ : Pi1−γ Piγ Viγ = Pf1−γ Pfγ Vfγ γ γ Pi1−γ ( PVi ) = Pf1−γ ( Pf Vf ) i or (1) In Equation (1), we have used the fact that P(a + b) = PaPb. Substituting PVi = nRTi and i Pf Vf = nRTf (Equation 14.1) into Equation (1) yields γ γ Pi1−γ ( nR Ti ) = Pf1−γ ( nR Tf ) Pi1−γ Tiγ = Pf1−γ Tf γ or (2) Solving Equation (2) for Pf1−γ and then raising both sides to the power 1 (1−γ ) , we obtain Pf1−γ = Pi1−γ Tiγ Tf γ γ (1−γ ) γ T = Pi1−γ i Tf T Pf = Pi i Tf or Equation (3) makes use of the fact that ( P a ) 1a = Pa a (3) = P1 = P . Substituting Tf = 2Ti and γ = 5 into Equation (3), we find that 3 Ti Pf = (1.50 × 105 Pa ) 2 Ti ( 5 3) 1−( 5 3) = (1.50 × 105 1 Pa ) 2 −( 5 2 ) = 8.49 × 105 Pa 31. REASONING AND SOLUTION a. Since the curved line between A and C is an isotherm, the initial and final temperatures are the same. Since the internal energy of an ideal monatomic gas is U = (3/2)nRT, the initial and final energies are also the same, and the change in the internal energy is ∆U = 0. The first law of thermodynamics, then, indicates that for the process A→B→C, we have ∆U = 0 = Q − W or Q =W The heat is equal to the work. Determining the work from the area beneath the straight line segments AB and BC, we find that ( ) Q = W = – ( 4.00 × 105 Pa ) 0.400 m3 – 0.200 m 3 = – 8.00 × 10 4 J Chapter 15 Problems 783 b. The minus sign is included because the gas is compressed, so that work is done on the gas. Since the answer for Q is negative, we conclude that heat flows out of the gas . 32. REASONING We seek the difference Wdiff in work between the work Wad done by the gas during the adiabatic process and the work Walt done by the gas during the alternative process: Wdiff = Wad − Walt (1) Considering the alternative process first, we bear in mind that, according to W = P (Vf − Vi ) (Equation 15.2), a gas only does work if its volume increases. As the pressure of the gas decreases to Pf from Pi during the first step, its volume remains constant, and, therefore, the gas does no work. Consequently, the total work Walt done by the gas during the entire alternative process occurs during the second step, when the gas expands to a volume Vf from a volume Vi at a constant pressure Pf: Walt = Pf (Vf − Vi ) (2) Only the initial volume Vi is given, so the final volume Vf in Equation (2) must be determined. Because the gas is ideal and monatomic, and can undergo an adiabatic process that takes it from the initial volume Vi to the final volume Vf , we will use PViγ = Pf Vfγ i (Equation 15.5), with γ = 5 , to determine th...
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