Problem set 3

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SUNY Empire State College *

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1201

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Chemistry

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Feb 20, 2024

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docx

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3

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Problem set #3 9.   Predict/Calculate   Energy from Gasoline   Burning a gallon of gasoline releases   1.19×108   J   of internal energy. If a certain car requires   5.20×105   J   of work to drive one mile,   (a)   how much heat is given off to the atmosphere each mile, assuming the car gets 25.0 miles to the gallon?   (b)   If the miles per gallon of the car is increased, does the amount of heat released to the atmosphere increase, decrease, or stay the same? Explain. Given: Energy released while burning 1 gallon gasoline: 1.19 x 10^8J/gal Energy required to drive one mile: 5.20 x 10^5 J/mile The mileage of the car: 25.0 mile/gal Δ U = Q – W Q = Δ U + W = (-1.19x10^8 J/gal/25.0 mile/gal) + 5.20 x 10^5 J/mile = -4.24MJ/mile b) If the miles per gallon were to be increased, the heat release to atmosphere would decrease due to the efficiency with miles per gallon being increased. The efficient engine converts a higher percentage of fuel into work which produces less heat. Higher MPG = less heat released. 13. The volume of a monatomic ideal gas triples in an isothermal expansion. By what factor does its pressure change? PV = constant P is the pressure V is the volume So, if the volume is tripled: it would follow as = Pinitial x Vinitial = Pfinal x Vfinal(3Vinitial) Pi/Pf = Vi/3Vi = 1/3 So the pressure of the monatomic ideal gas changes by a factor of 1/3
35. A system consists of 3.5 mol of an ideal monatomic gas at 315 K. How much heat must be added to the system to double its internal energy at   (a)   constant pressure or   (b)   constant volume? Monatomic gas: 3.5 mol Monatomic gas temp: 315 K The constant pressure would follow as: Q=nR(Tf−Ti )=5nRΔT Q=5nR(Tf −Ti )=5nR(2Ti −Ti )=5nRTi a) Q=5×3.5×8.31×(2×315−315)=23kJ b) Q = nCvΔT = 3nRΔT Q = 3 x 3.5 x 8.31 x (2x 315 – 315) = 14 kJ So, the heat added at constant pressure is 23 kJ, and at constant volume, it's 14 kJ. 46.   A nuclear power plant has a reactor that produces heat at the rate of 838 MW. This heat is used to produce 253 MW of mechanical power to drive an electrical generator.   (a)   At what rate is heat discarded to the environment by this power plant?   (b)   What is the thermal efficiency of the plant? Heat produced (Qh ) = 838 MW Heat used (Qc ) = 253 MW work by engine (W) = Qh −Qc 838MW−253MW= 585MW is the rate heat is discarded to environment. b) To find the thermal efficiency: e = Qh – Qc/ Qh 838 MW – 585 MW / 838 MW = 0.302MW is the thermal efficiency.
58. To keep a room at a comfortable 20.5 °C, a Carnot heat pump does 315 J of work and supplies the room with 4180 J of heat.   (a)   How much heat is removed from the outside air by the heat pump?   (b)   What is the temperature of the outside air? Heat supplied: 4180 J Work: 315 J Room temp: 20.5 C a) Qc = Qh – W = 4180 J – 315 J = 3865 J = 3.9kJ is the amount of heat removed from outside air by the pump. b) To find the outside air temperature: Tc = Th (Qc/Qh) = [(20.5 C + 273.15 K)] (3865J/4180J) = 271.5 K 271.5 K – 273.15 K = -1.6C is the outside air temperature. 69. An 88-kg parachutist descends through a vertical height of 380 m with constant speed. Find the increase in entropy produced by the parachutist, assuming the air temperature is 21 °C. Parachutist weight: 88kg Vertical height: 380m Air temp: 21 C Δ S = Q/T = U / T = mgh/T Δ S = (88kg)(9.81 m/s^2)(380m) / 21 C +273.15K = 1100 J/K = 0.11 kJ/K is the increase in entropy produced from the parachutist at air temperature of 21 C.
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