Argon at 400 K and 50 bar is adiabatically and reversibly expanded to 1 bar through...

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.

Solutions

Expert Solution

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

queen_honey_blossom answered 8 months ago

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