Question

Problem 5: A rigid tank contains 3.52 kg of an ideal gas. The volume of the tank is V 1.20 m3 and initially the gas is at a pressure of Pi 225 kPa. The tank is charged by adding 0.74 kg of the same gas at which point the new pressure is 450 kPa and the new temperature is 153.94 °C. Determine the gas constant, R, a possible identity of the gas, the original temperature, T, the original density, pi, and the original specific volume, V. Lastly, express the final temperature in each of the other common temperature units: °F, K, and R (Hint: while the identity of the gas cannot be conclusively proven based on the given data, give one option) possible

Answer 1

Initial binal Po V B, m. I D = 225kpa mg = 3.52 +0.74 = 4.26164 me = 3.52kg = 4,26 kg V = 1.2 m3 % - 153 (c 12 = 450kna. V2 = 4, = 1.2m3 brom a deal gach, equation P Q V2 = M RG2 >> R= P2 Ve Me Je R = 450 x103) (1.2) 14,26) (153.94 +273) (R = 0.2969 Kiligh o Diviam,RO, (225 x103) (1.2) 0,- Min - 72.52) (0,2969610) P. Vi MIR T= 258.35k - 258.35-272.15 Ri= 14.7000

= e, m1 = 3.52 = 2.933 x 13/1631 6= Ye = 1/2.933 = 0.341 m3lrg = - 14.79°C » 92 = 153.900 = 5.378°F - 768.762 R = 258.30K = 309.092°F - 465.048R, - 427.09k.