The figure below shows a cyclic thermodynamic process followed by n=1mol of a mo
ID: 2245785 • Letter: T
Question
The figure below shows a cyclic thermodynamic process followed by n=1mol of a monatomic ideal gas. The initial pressure is p1=2.2E5 and the initial temperature is T1=560 K. The process 1-2 is isothermal, the process 2-3 is constant pressure with final temperature T3=360 K, and the process 3-1 is adiabatic.
a) Determine pressure p2 and volumes V1, V2, and V3
b) Find work W12, W23, W31
c) Find change in internal energy E12, E23, E31
Figure: http://imgur.com/Ax92bVd
Thank you!
The figure below shows a cyclic thermodynamic process followed by n=1mol of a monatomic ideal gas. The initial pressure is p1=2.2E5 and the initial temperature is T1=560 K. The process 1-2 is isothermal, the process 2-3 is constant pressure with final temperature T3=360 K, and the process 3-1 is adiabatic. Determine pressure p2 and volumes V1, V2, and V3 Find work W12, W23, W31 Find change in internal energy E12, E23, E31Explanation / Answer
V1 = nRT1/P1 ; V1 = 2.116*10^-2 m3 = 21.163 L
For adiabatic process, (y means gamma, the ratio of Cp and Cv)
P^(y-1)/T^y = constant
(P3/P1)^(y-1) = (T3/T1)^y
P1, T1 and T3 are known, y = 1.67 for mono atomic gas
P3 = 0.729*10^5 Pa = P2
V2 = nRT2/P2 {(T1=T2)} ; V2= 6.386*10^-2 m3 or 63.86 L
From 2-3, Pressure same.
V2/T2 = V3/T3 ; V3 = 4.1*10^-2 m3 or 41.05 L.
W12 = -nRT1 ln(P2/P1) = 5.143 kJ
W23 = P2*(V3-V2) = -1.67 kJ
W31 = -(P1V1 - P3V3)/(y-1) = -2.487 kJ
E12 = 0. Temp same.
E23 = nCp(T3-T2) = -4.157 kJ
E12 + E23 + E13 = 0 (state function)
E13 = 4.157 kJ
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