Question

15.0 L of an ideal monatomic gas at 3.00 atm and 450 K are contained in a cylinder with a piston. The gas first cools isochorically to 270 K (step 1). It then expands isobarically back to its original temperature (step 2), and then contracts isothermally back to its original volume (step 3).

a) Show the series of processes on a pV diagram.

b) Calculate the temperature, pressure, and volume of the system at the end of each step in the process. Indicate the p and V values on the pV diagram.

c) Compute the total work done by the gas on the piston during each step of the cycle in L-atm, and the total work done by the gas for one complete cycle.

d) Compute the heat added during each step of the cycle in L-atm, and the net heat added for one cycle. Compare the total work done with the net heat added.

e) Is this an engine or a refrigerator? If it is an engine, what is its efficiency; if it is a refrigerator, what is its coefficient of performance?

Answer 1

initial pressure=P1=3 atm=3*101325 N/m^2

initial volume=V1=15 L=15*0.001 m^3

initial temperature =T1=450 K

number of moles=n=P1*V1/(R*T1)

=1.2187

step 1:

cooling isochorically:

volume remains constant.

hence volume =V2=V1=0.015 m^3

temperature=T2=270 K

pressure=P2=n*R*T2/V2=1.8238*10^5 Pa

work done=0

heat added=n*Cv*change in temperature

=n*(3/2)*R*(T2-T1)

=-2735.8 J

step 2:

expanding isobarically to 450 K:

temperature=T3=450 K

isobaric expansion ==>pressure is constant

P3=P2

then volume V3=n*R*T3/P3

=0.025 m^3

work done=P3*(V3-V2)

=1823.8 J

heat added=n*Cp*(T3-T2)

=n*(5/2)*R*(T3-T2)

=4559.6 J

step 3:

isothermal contraction:

temperature=T1

volume=V1

then pressure=n*R*T1/V1

=P1

work done=heat added=n*R*T*ln(final volume/initial volume)

=n*R*T1*ln(V1/V3)

=-2329.2 J

hence P-V diagram:

part b:

temperature , pressure and volume are as calculated above.

part c:

total work done during the gas on the piston:

step 1: work done=0

step 2:work done=1823.8 J=18 L.atm

step 3: work done=-2329.2 J=-22.987 L.atm

total work done by the gas for one complete cylce

=1823.8-2329.2

=-505.32 J

=-4.9872 L.atm

part d:

heat added :

step 1:-2735.8 J=-27 L.atm

step 2:4559.6 J=45 L.atm

step 3:-2329.2 J=-22.987 L.atm

total heat added=-505.32 J=-4.9872 L.atm

total work done=total heat added as for a closed process, change in internal energy =0