Question

Question 18 1 pts Problems 18-20 are based on the following description An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 10 kg of an ideal gas with molar mass M-16 kg/kmol, a pressure of 330 kPa, and a temperature of 50 oC. The other part is evacuated. The partition is now removed and the gas fills the entire tank. (R. - 8.31447 kJ/kmol.K) The expansion work in kJ is closer to: 1678 725 0 -356 267 820 3215 512

Answer 1

Partition 5.089 ssr Vi=5.089 M = 10kg M-16 kg 1 kmol P, = 330kPa Ti = 50°C =323K (Evacauted) li partion i removed so it ei freee expansion Vacuum state I Ru=81314 ks/kmol-k Free Expansion Preocess As the tank ei rigid and the partion is removed State a V = Vitva = 10.178 m3 T2=323k m, =lokg M = 16 kg 1 kmol since the tank i insulated and the Insulated Rigid tank process

( since in freee Expansion the expansion Worele ii zerw Wica =ol And 20 option ared į oreret las dince in Freel Expansion process the intial and find temperature is equal Ans Ta = Final temperature = 323k So option a i corrent PV = nRT [ Ideal Geas Equation] R M Piviam, RTI =8,314 = 0.52 Kg-k =) Viam, RTL P = 10x 0.52X323 330 = 5.08am

As am[coln (II) + R b (t] Sin Ta= Ti AS = m R ln (V) 5.089 = 10x0o52 ln ( 10 178 = 3.6 ks lk Ang AS 6 ů correeet So the option