Question

A rigid, insulated tank contains a certain amount of argon at room temperature 25^o C and 500 kPa. Now the valve is opened and the argon is allowed to vent into atmosphere until 80% argon remains in the tank. Determine the final temperature of the argon in the tank. It is assumed the argon leaving the tank is at room temperature.

Answer 1

1 given data initial condition T= 25°C P. = 500 kpa. final condition m = 0.8 m kg T2 = ? S m. = m kg hot valve. levalve. Control Volume two A 0.8m T m arion 2590 0 500 kpa TE? དངདེགོངོ། ང I given gas is cp ED Augon. R E M3 molecular weight Ra 8.314 Man = 40. An 40. IR = 0.2078 kJ gasest so adiabatic inder ,8=1.67 [mono atomic = 1.67 x 0.2018 = 0.518 kJ - kgk. C = - 0,6% = 0.310 kender Co specific heat at constant pressure. a specific heat at comtant volume.

I it is unsteady state Com). = m; - me Mi = mass entering me = mass leaving porno = -0.2m Energy balance ama -0.2m thler (4). --me Christine o gut (neglect kinchic de potential energy. enviny) given rigid & insulated Q =0, x=0. change of energy in . a-meho control volume us internal U-U = - me hi mu; -mu, = a me hi us specific internal 0.8 m uz mu, = -0.2 no hi energy. 6: 8 u, - u = -0-2 h, [u = GT] The GTI considering argon behavious to be ideal. ACAU lu=mul

oscuta - cut, 2012 2011 Ist, because, it is given mass leaning at same temperature. cu [ 0.8 Tz - 1] = -0.2 (p. 7 0.875 – 298 = -0.2 (1248 0.8.72 -298 = - C (i <=Y) 0.872-298 = -99.532 0.89, = 198.96€ © = 198.965 og IT, = 248.0856] * final temperature of argon = 248.08 K m=0.8m T = 248.085 ki

N07E: if presswie was alno osked. 6 Pi = 500 kpa... T= 298k. Mam - ideal gan equation 1. PIE MRT 500 v= m (0.201) 298.40 m₂=0.8m T, ? PV = m₂ RTL pr=0.8m (0.207) T2 - Ľ Rigid container; V= constant] . 0.8 xmx0.201 x 12 mx 0207 x298 (T₂= 248.08k). Pz 5oo P = 0.8*248.08 x 500 - 298 [P₂ = 333 kpal