A heat engine consists of a heat source that causes a monatomic gas to expand, p

Question

A heat engine consists of a heat source that causes a monatomic gas to expand, pushing against a piston, thereby doing work. The gas begins at a pressure of 285 kPa, a volume of 136 cm3, and room temperature, 20 °C. On reaching a volume of 421 cm3, the piston is locked in place, and the heat source is removed. At this point, the gas cools back to room temperature. Finally, the piston is unlocked and used to isothermally compress the gas back to its initial state.

(a) Sketch the cycle on a pV-diagram.

(b) Determine the work done on the gas and the heat flow out of the gas in each part of the cycle.

(c) Using the results of part (b), determine the efficiency of the engine.

Explanation / Answer

Well I have been trying for a while and cant seem to upload a picture. Your P-v diagram will be a triange with the following verticies: P1 = 285 KPa V1 = 136 cm^3 P2 = 285 KPa V2 = 421 cm^3 P3 = 131.7 KPa V3 =421 cm^3 Here is how I got those numbers: P1 and V1 are given. P2, since you are not told the piston has a spring behind it you can assume it will move lineary with any change in pressure. i.e. your pressure will be constant and your volume will change (this assumption is confirmed since the table you have asks about the isochoric process which is constatn volume) V2 Given P3, since you are told its a monoatomic gass you can assume ideal gass law. P2*V2/T2 = P3*V3/T3 V2 is equal to V3 so... P3 = P2*T3/T2 T3 is give, 20C, and T2 is found by the same process.... P1*V1/T1 = P2*V2/T2 P1 = P2 so... T2 = V2*T1/V1 T2 = .000421(m^3)*(293 K) / .000136 m^3 T2 = 907 K = 634 C P3 = 285000 Pa * 293 K / 907 P3 = 92067 Pa = 92.067 KPa Now part b. Isobaric process (const pressure) 1-->2 Note, positive work meas work was done to the gas, energy was added to it. W = integral (pdV) since p is constant you can take it out of the integral, and itegral of dV is just the change in V so W = 285000 Pa * (.000421 m^3 - .000136m^3) W = 81.225 J Isochoric proces (const volume) 2-->3 Since there is no change in volume work is 0 W = 0 J Isothermic process (const temp) 3-->1 You could find the equation for the pressure as a function of volume and it would follow P = mV + b just find the slope and y-int then integrate, however since the work is just the area under the p-v diagram I prefer just to find the area by using geometry (area of a trapizoid) W =(V3-V2) * (P3 + P1)/2 W = -53.73 I just realized I'm not srue how to find the heat since Cp or n is given... Hopefully this helped you a little bit. Post back if you can find how to get the heat input from here, I'd like to know mysefl.

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