In: Chemistry
Argon at 400 K and 50 bar is adiabatically and reversibly expanded to 1 bar through a turbine in a steady process. Compute the outlet temperature and work derived per mole.
Argon is monoatomic gas, so gamma, r = 1.66
Consider 1 mol
Pi = 50 bar = 49.35 atm
Vi = n*R*Ti / Pi = 1*0.0821*400/49.35 = 0.67 L
Pf = 1 bar = 0.99 atm
find Vf using:
Pi*Vi^r= Pf*Vf^r
49.35 * (0.67)^1.66 = 0.99 * (Vf)^1.66
(Vf)^1.66 = 25.64
Vf = 7.06 L
K = Pi*Vi^r = 49.35 * (0.67)^1.66=25.38
work done = K {Vf^(1-r) - Vi^(1-r)} / (1-r)
= 25.38* {7.06^(1-1.66) - 0.67^(1-1.66)} / (1-1.66)
= 25.38* {7.06^(-0.66) - 0.67^(-0.66)} / (-0.66)
=39.5 atmL
= 39.5 * 101.3 J
=4001.6 J
Outlet Temperature, Tf= Pf*Vf/(n*R) = 0.99*25.64 / (1*0.0821) = 309.2 K
Answers are:
Work done = 4001.6 J
Temperature= 309.2 K