A heat engine carries 3.20 mol of an ideal monatomic gas around the cycle shown

Question

A heat engine carries 3.20 mol of an ideal monatomic gas around the cycle shown in the figure. Process 1-2 takes place at constant volume and takes the gas from 300 K to 600 K, process 2-3 is adiabatic and takes the gas from 600 K to 455 K, and process 3-1 takes place at a constant pressure and returns the gas to 300 K.

Compute the heat Q for the whole cycle.

Calculate the work done by the gas for the cycle. Compute the change in internal energy DeltaE for the cycle as a whole.

If the initial pressure at point 1 is 3.4 atm, find the pressure at point 2 in atm.

If the initial pressure at point 1 is 3.4 atm, find the volume at point 3 (in m^3

Explanation / Answer

for the path 1 to 2

Q = (3/2) n R delta T

= (3/2) * 3.2 * 8.314 * (600-300)

= 11972.16 J

for path 2 to 3

Q= 0

for path 3 to 1

Q = (5/2) n R delta T

= (5/2) * 3.2 * 8.314 * (455-300)

= -10309.36 J

total energy in cyclic process

Q = 11972.16 J-10309.36 J =1662.8 J

for cyclic process dU = 0

from teh first law of thermodynamics

delQ = d U + W

1662.8 J = W

from the ideal gas equation

P1/T1 = P2/T2

P2 = P1T2/T1 = 3.4((600)/300 = 6.8 atm

volume at point 3 is

V = nRT/ P = 0.08207 ( 3.2) ( 455)/3.4 = 35.14 L = 0.03514 m^3

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