A heat engine using 150 m g of helium as the working substance follows the cycle

Question

A heat engine using 150mg of helium as the working substance follows the cycle shown in the figure.

I have the correct answers in green for the first ones. I just need the last 2.

Part A: Determine the pressure of the gas at point 1.

1 atm

Part B: Determine the temperature of the gas at point 1.

330K

Part C: Determine the volume of the gas at point 1.

1000 cm^3

Part D: Determine the pressure of the gas at point 2.

5 atm

Part E: Determine the temperature of the gas at point 2.

1600K

Part F: Determine the volume of the gas at point 2.

1000 cm^3

Part G: Determine the pressure of the gas at point 3.

1 atm

Part H: Determine the temperature of the gas at point 3.

1600K

Part I: Determine the volume of the gas at point 3.

5000 cm^3

Part J: What is the engine's thermal efficiency?

Part K: What is the maximum possible efficiency of a heat engine that operates between and Tmax and Tmin?

A heat engine using 150mg of helium as the working substance follows the cycle shown in the figure. Determine the pressure of the gas at point 1. 1 atm Determine the temperature of the gas at point 1. 330K Determine the volume of the gas at point 1. 1000 cm^3 Determine the pressure of the gas at point 2. 5 atm Determine the temperature of the gas at point 2. 1600K Determine the volume of the gas at point 2. 1000 cm^3 Determine the pressure of the gas at point 3. 1 atm Determine the temperature of the gas at point 3. 1600K : Determine the volume of the gas at point 3. 5000 cm^3 What is the engine's thermal efficiency? What is the maximum possible efficiency of a heat engine that operates between and Tmax and Tmin?

Explanation / Answer

Here T max = 1600 K

Tmin = 330 K

max possibel efficiency = ( 1 - Tmin /Tmax ) x 100

Max Efficiency = ( 1- 330/1600) x 100

Max efficiency = 79.375

So Maximum possibel efficiency is 79.375 %

1) In process 1 to 2

Volume is constant ,isochoric process

work done =0

Heat = nCv dT

moles of Helium = 150 x 10-3 /4 = 0.0375

Cv = 3R/2

Heat H1 = 0.0375 x 3R/2 x ( 1600-330)

H1 = 593.93 J

From process 2 to 3

temperature is constant , isothermal process

wrok done = Heat = nRT ln (V3/V2)

work done W2 =H2 = 0.0375 x 8.314 x 1600 x ln ( 5000/1000)

W2=H2 = 802.85 J

  

from process 3 to 1

pressure is constant ,isobaric

work done = P( V1-V3)

work done W3 = 101325 ( 1000-5000) x 10-6

W3 = -405 .3 J

Heat H3 = nCpdT

H3 =0.0375 x 5R/2 x ( 330-1600)

H3 = -989.88 J

net wrok done = W1 + W2 +W3 = 0 + 802.85 -405.3 = 397.55 J

heat given to the engine = H1 + H2 = 593.93 + 802.85 = 1396.78 J

efficiency = ( net work / heat given ) x 100

eficiency = 397.55 x 100/ 1396.78

Efficiency = 28.46 %

so the efficiency is 28.46 %

the max efficiency is 79.375 %

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