The figure below shows the Stirling cycle followed by n=1 mol of a monatomic ide

Question

The figure below shows the Stirling cycle followed by n=1 mol of a monatomic ideal gas. The initial volume is V1=1 L and the initial temperature is T1=480 K. The process 1-2 is isothermal with final volume V2=2.2 L. The process 2-3 is constant volume with final temperature T3=320 K. The process 3-4 is isothermal. Finally the process 4-1 is constant volume.

a) Determine p1, p2, p3, p4

b) Find Q12, Q23, Q34, Q41

c) Find change in entropy S12, S23, S34, S41

d) Find efficiency of Stirling engine, where |Q+| is the magnitude of heat which enters system during those segments of the cycle for which Q > 0 and |Q-| is the magnitude of heat which exits the system during those segments of the cycle for which Q < 0. How does this compare with a Carnot engine operating between the same temperatures?

Figure: http://imgur.com/XzelSQm

Thank you!

The figure below shows the Stirling cycle followed by n=1 mol of a monatomic ideal gas. The initial volume is V1=1 L and the initial temperature is T1=480 K. The process 1-2 is isothermal with final volume V2=2.2 L. The process 2-3 is constant volume with final temperature T3=320 K. The process 3-4 is isothermal. Finally the process 4-1 is constant volume. Determine p1, p2, p3, p4 Find Q12, Q23, Q34, Q41 Find change in entropy S12, S23, S34, S41 Find efficiency of Stirling engine, where |Q+| is the magnitude of heat which enters system during those segments of the cycle for which Q > 0 and |Q-| is the magnitude of heat which exits the system during those segments of the cycle for which Q

Explanation / Answer

Given V1= 1L , T1 =480 ,V2 = 2.2 L, T2= 320 K

1-2 isothermal process

2-3 isochoric

3-4 isothermal

4-1 isochoric

R = 0.082 L.atm/K . mol

a) p1 = n RT/V

= 1*0.082*480/1= 39.36 atm

p2 = 1*(0.082)*480/2.2=17.89atm

p3= 1*(0.082)*320/2.2=11.92 atm

P4 = 11.92 atm

b) Q12 - (isothermal )= 0

Q23 - (isochoric)= n Cv dT

Cv for monoatomic gas = 3/2 R = 1.5 * 0.082 = 0.123

Q23 = 1*0.123*(480-320)=19.68 J

Q34 = 0 (isothermal)

Q41 = 19.68 J

C) S12 = 0

S23 = 19.62/160 =0.122 J/K

S34= 0

S41 = 0.122 J/K

d) efficiency = 1- Tc/Th = 1- 320/480= 0 .33

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