Physics Solution Manual for 1100 and 2101

15 we will solve equation 1515 for th the temperature

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Unformatted text preview: quation 15.8) and ∆T = Tf − Ti into Equation (1) yields 2 Chapter 15 Problems n= = Q = C V ∆T 787 Q 2Q = R ( Tf − Ti ) 3R ( Tf − Ti ) () 3 2 2 ( 8500 J ) = 11 mol 3 8.31 J/ ( mol ⋅ K ) ( 279 K − 217 K ) 37. SSM REASONING When the temperature of a gas changes as a result of heat Q being added, the change ∆T in temperature is related to the amount of heat according to Q = Cn ∆T ( Equation 15.6 ) , where C is the molar specific heat capacity, and n is the number of moles. The heat QV added under conditions of constant volume is QV = CV n ∆TV , where CV is the specific heat capacity at constant volume and is given by CV = 3 R ( Equation 15.8 ) and R is the universal gas constant. The heat QP added under 2 conditions of constant pressure is QP = CP n ∆TP , where CP is the specific heat capacity at constant pressure and is given by CP = 5 R ( Equation 15.7 ) . It is given that QV = QP, and 2 this fact will allow us to find the change in temperature of the gas whose pressure remains constant. SOLUTION Setting QV = QP, gives CV n ∆TV = CP n ∆TP 1 24 1 24 43 43 QP QV Algebraically eliminating n and solving for ∆TP, we obtain C ∆TP = V C P 3R 2 ∆TV = 5 ( 75 K ) = 45 K R 2 38. REASONING Under constant-pressure conditions, the heat QP required to raise the temperature of an ideal gas is given by QP = CP n ∆T (Equation 15.6), where CP is the molar specific heat capacity at constant pressure, n is the number of moles of the gas, and ∆T is the change in the temperature. The molar specific heat capacity CP at constant pressure is greater than the molar specific heat capacity at constant volume CV, as we see from CP = CV + R (Equation 15.10), where R = 8.31 J/(mol·K). Equation 15.6 also applies to the constant-volume process, so we have that QV = CV n ∆T , where QV = 3500 J is the heat required for the constant-volume process. We will use these relations, and the fact that the 788 THERMODYNAMICS change in temperature and the number of moles of the gas are the same for both processes, to find the heat QP required for the constant-pressure process. SOLUTION We begin by substituting CP = CV + R (Equation 15.10) into QP = CP n ∆T (Equation 15.6) to obtain QP = CP n ∆T = ( CV + R ) n ∆T Solving QV = CV n ∆T (Equation 15.6) for CV yields CV = (1) QV n ∆T . Substituting this result into Equation (1), we find that Q QP = ( CV + R ) n ∆T = V + R n ∆T = QV + Rn ∆T n ∆T (2) Therefore, the heat required for the constant-pressure process is QP = 3500 J + (1.6 mol ) [8.31 J ( mol ⋅ K )] ( 75 K ) = 4500 J 39. REASONING AND SOLUTION The heat added at constant pressure is Q = Cpn∆T = (5R/2) n∆T The work done during the process is W = P∆V. The ideal gas law requires that ∆V = nR∆T/P, so W = nR∆T. The required ratio is then Q/W = 5/2 40. REASONING According to Equations 15.6 and 15.7, the heat supplied to a monatomic ideal gas at constant pressure is Q = CP n ∆T , with CP = 5 R . Thus, Q = 5 nR ∆T . The 2 2 percentage of this heat used to increase the internal energy by an amount ∆U is ∆U Percentage = Q ∆U ×100 % = 5 nR ∆T 2 ×100 % (1) But according to the first law of thermodynamics, ∆U = Q − W . The work W is W = P ∆V , and for an ideal gas P ∆V = nR ∆T . Therefore, the work W becomes W = P ∆V = nR ∆T 3 and the change in the internal energy is ∆U = Q − W = 5 nR ∆T − nR∆T = 2 nR∆T . 2 Chapter 15 Problems 789 Combining this expression for ∆U with Equation (1) above yields a numerical value for the percentage of heat being supplied to the gas that is used to increase its internal energy. SOLUTION a. The percentage is ∆U Percentage = 5 2 nR ∆T 3 nR ∆T 2 5 nR ∆T 2 × 100 % = 3 ×100 % = ×100 % = 60.0 % 5 b. The remainder of the heat, or 40.0 % , is used for the work of expansion. 41. REASONING The change in the internal energy of the gas can be found using the first law of thermodynamics, since the heat added to the gas is known and the work can be calculated by using Equation 15.2, W = P ∆V. The molar specific heat capacity at constant pressure can be evaluated by using Equation 15.6 and the ideal gas law. SOLUTION a. The change in the internal energy is ∆U = Q − W = Q − P ∆V ( )( ) = 31.4 J − 1.40 × 104 Pa 8.00 × 10−4 m3 − 3.00 × 10−4 m3 = 24.4 J b. According to Equation 15.6, the molar specific heat capacity at constant pressure is CP = Q/(n ∆T). The term n ∆T can be expressed in terms of the pressure and change in volume by using the ideal gas law: P ∆V = n R ∆T or n ∆T = P ∆V/R Substituting this relation for n ∆T into CP = Q/(n ∆T), we obtain CP = Q 31.4 J = = 37.3 J/(mol ⋅ K) 4 P ∆V 1.40 × 10 Pa 5.00 × 10−4 m3 R R ( )( ) 42. REASONING The heat added is given by Equation 15.6 as Q = CV n ∆T, where CV is the molar specific heat capacity at constant volume, n is the number of moles, and ∆T is the change in temperature. But the heat is supplied by the heater at a rate of ten watts...
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