Thermodynamics Homework: Energy Conversion & Ideal Gas Processes, Assignments of Chemistry

Download Thermodynamics Homework: Energy Conversion & Ideal Gas Processes and more Assignments Chemistry in PDF only on Docsity! CHEM 110A 1ST HOMEWORK SET :SOLUTIONS 1. The potential energy of the water, V = mgh , is converted into thermal energy when thewater reaches the bottom of the falls. The thermal energy is seen as a change in temperature, q = mC∆T . Thus, V = mgh = mC∆T h = C∆T g = 4.2JK −1g−1 × 0.10K 9.8ms−2 ≈ 0.04kgm 2s−2g−1 ms−2 = 40m 2. Step A: isochoric; dV = 0, therefore w = 0 and q = ∆U = Cv∆T = -(3R/2)•273 = - 3.40 kJ. Step B: isobaric; dP = 0, ∆U = Cv∆T = +3.40 kJ, qp = CpD∆T = (5R/2)•273 = + 5.67 kJ and from 1st law: w = q – ∆U = 2.27 kJ. Step C: ∆U = 0 since there was no overall change in temperature. T does vary along the path, however, so w in not given by the work isothermal equation i.e., w = -nRTln(V2/V1). The work is obtained by evaluating the area under the curve PdV∫ . From the geometry of a straight-line path w = - area = - (2•11.2 +(1/2)•2•11.2) L•atm = -33.6 L•atm = - 3.40 kJ. Thus Since ∆U = 0, we have from the 1st law: q = w = - 3.40 kJ. For the complete cycle: ∆U = - 3.40 kJ + 3.40 kJ + 0 = 0; q = - 3.40 kJ + 5.67 kJ – 3.40 kJ = - 1.13 kJ; and w = 0 + 2.27 kJ – 3.40 kJ = -1.13 kJ (since q = w). 3. a) P = nRT V 1+ Bn V     , where B = b − a RT w = − PdV = − nRT V 1+ Bn V     dV∫∫ w = −nRT dV V∫ − Bn dV V2∫     = −nRT ln V2 V1 − nB 1 V2 − 1 V1             b) B = b − a RT = 115.40 cm3mol −1 − 18.240 × 10 6 cm 6atmmol −2 82.06cm3 atmK −1mol−1 × 300K B = (115.40 − 740.9)cm3mol −1 = −625.5cm3 mol−1 w = −8.314JK −1 × 300K ln 5.00 1.00 − 625.5cm3 1 5.00 × 103 cm 3 − 1 1.00 × 103 cm3             w = −2.494 × 103 J × 1.609 + 0.500( ) = −4.01 kJ Note that in this case the correction for nonideality, makes a contribution of only about 5% to the work. 4. From the chain rule we have: ∂V ∂T     P ∂T ∂P     V ∂P ∂V     T = −1 so, ∂P ∂T     V = − ∂V ∂T     P ∂P ∂V     T = − ∂V ∂T     P ∂P ∂V     T Now α = 1 V ∂V ∂T     P and κ = − 1 V ∂V ∂P     T So ∂P ∂T       V = α κ = 1.120 × 10 −3 K −1 0.768 × 10−4 atm −1 = 14.58 atmK −1 ∆P = ∂P ∂T       v dT =∫ ∂P ∂T       v ∆T = 14.58 atmK −1 × 10K = 145.8 atm 5. a) At State B: P = 20.00 atm, V = 0.05 L, and T = 313 K; n = PV/RT = 0.389 mol. At State B and C: From graph V = 0.50 L (also for State B from V = nRT/P = 0.5 L) b) w = w(A->C) + w(C->B) = - P∆V + 0 = - 1.00 atm•(0.50–10.00) L = + 9.50 L•atm = + 0.963 kJ