1.80 mol sample of an ideal monatomic gas, originally at a pressure of 1.40 atm

Question

1.80 mol sample of an ideal monatomic gas, originally at a pressure of 1.40 atm , undergoes a three-step process: (1) it is expanded adiabatically from T1 = 593 K to T2 = 398 K ; (2) it is compressed at constant pressure until its temperature reaches T3; (3) it then returns to its original pressure and temperature by a constant-volume process.

Part A

Determine T3.

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Part B

Calculate the change in internal energy for each process.

Enter your answers numerically separated by commas.

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Part C

Calculate the work done by the gas for each process.

Enter your answers numerically separated by commas.

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Part D

Calculate the heat added to the gas for each process.

Enter your answers numerically separated by commas.

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Part E

Calculate the change in internal energy, the work done by the gas, and the heat added to the gas for the complete cycle.

Enter your answers numerically separated by commas.

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Explanation / Answer

a.P=101300*1.40 Pa

8.31451 J/K·mol

V=nRt1/P=0.0625 m^3

T3=PV/nR=593 K

b.delta E=q+w

w=-Px*deltaV

since v is constant w=0

Cv=Cp-nR=49.22

deltaU=CvdeltaT=-9598.44

q=deltaU-w=-9.59*10^3

deltaE=q+W=-9.59*10^3

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