MATE 510 Homework 1 solutions(2)

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Drexel University *

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510

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Mechanical Engineering

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Apr 3, 2024

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MATE 510 homework 1 (5 questions) is due at 12:00PM on 1/22/2024. Please submit your homework in BBlearn. Steps of calculation need to be shown clearly in your answers. Homework solutions will be posted on 1/22/2024 before 1:00PM. 1. When the P-V work of a specific system is calculated, it is calculated according to the following equation. What are the assumptions used in the calculation? To calculate the P-V work, a path linking state 1 and state 2 needs to be provided. Explain which step in the calculation indicates the path. 1. 1 mole of gas. 2. Ideal gas (or any answer explain ideal gas), Reversible. 3. Isothermal. Step 2 tells us it is a reversible process, while step 3 tells us the path is isothermal. So the path is an isothermal reversible process. 2. Two moles of monatomic ideal gas at 25 C and 10 atm pressure (state A) expand under a reversible adiabatic process to reach pressure 2 atm (B) and expand further to double the volume of V B under a constant pressure to reach state C. Calculate the overall work done and the amount of heat adsorbed by the system, and the variation of enthalpy during this two steps process. c v =3/2R and c p =5/2R. V A = nRT A / P A =2×0.082057×(25+273.15)/10=4.89 L P A V A 5/3 = P B V B 5/3 V B =(P A /P B ) 3/5 V A =(10/2) 3/5 4.89=12.84 L T B = P B V B /nR=2×12.84/2/0.082057=156.54 K T c = P c V c /nR= 2×2×12.84/2/0.082057=312.95K From A to B ∆U AB =− w AB = A B nc v dT = nc v ( T B T A ) = 2 × 1.5 × 8.314 × ( 156.54 298.15 ) =− 3528.3 J w BC = B C PdV = P B ( V C V B ) = 2 × 101325 × 12.84 × 10 3 = 2602.0 J
w total =3528.3+2602= 6130.3J q total = q BC = B C nc p dT = nc p ( T C T B ) = 2 × 2.5 × 8.314 × ( 312.95 156.54 ) = 6501.96 J ∆ H AC = ∆U AC + ( P C V C P A V A ) = nc v ( T C T A ) + ( P C V C P A V A ) = 2 × 1.5 × 8.314 × ( 312.95 298.13 ) + ( 2 × 12.84 × in which 1 liter atm is 101.3 Joules. 3. Two moles of an ideal gas, in an initial state P = 10 atm, V = 5 liters are taken reversibly in a clockwise direction around a circular path given by ( V -10) 2 +( P -10) 2 =25. Calculate the amount of work done by the gas as a result of the process and calculate the maximum and minimum temperatures attained by the gas during the cycle. The amount of work done by the system is just the area cover by the circle. w = π R 2 = 3.14 × 5 2 × 101.3 = 7956 J Tmax is at the point where PV has the maximum value. It is the point P=V due to symmetry. P=13.5355 atm and V= 13.5355 liters T=PV/nR=13.535*13.535/2/0.082507=1110 K Tmin is at the point where PV has the minimum value. It is the point P=V due to symmetry. P=6.4645 atm and V= 6.4645 liters T=PV/nR=6.4645 *6.4645 /2/0.082507=253 K 4. Monatomic ideal gas expands from A to B to C through reversible isothermal and adiabatic processes. Calculate overall work and internal energy change. V 0 P A B C A P B P C P A V B V C V (2 mole ideal gas, P A = 80 atm, T A = 800 C) Isothermal adiabatic ( P B = 40 atm) ( P C = 20 atm) w AB = A B PdV = nRTln P A P B = 2 × 8.314 × ( 800 + 273.15 ) × ln 80 40 = 12368.75 J V B = nRT / P B =2×0.082057×1073.15/40=4.403 L P C V C 5/3 = P B V B 5/3 V C =( P B /P C ) 3/5 V B =(40/20) 3/5 4.403=6.67 L
T c = P c V c /nR = 20×6.67/2/0.082057=812.85K=539.7 C w BC =-∆ U =- nc V T = 2×1.5×8.314×(1073.15-812.85)=6492.4J for this step, students may also use w BC = B C PdV = B C P B V B 5 / 3 V 5 / 3 dV to calculate, but the final answer should be the same. w AC = w AB+ w BC =12368.75+6492.4= 18861.15 J ∆U AC = ∆U AB + ∆U BC = 0 w Bc =− 6492.4 J 5. Demonstrate PV γ = constant for ideal gas under a reversible adiabatic process. = c p /c v (the ratio of molar heat capacities under constant pressure and constant volume). Show at least 8 steps in your derivation, starting from dU =- w . For adiabatic process
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