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29.- A quantity of neon, confined a container at 2.50 atm and 298 K, is allowed to expand reversibly and adiabatically satisfying the relation R/CV 7 - C) to a final pressure of 1.0 atm. Find the final temperature of the gas. (A) 107 K Given, P1 = 2.50 atm and P2 = 1.0 atm For Ne, Cp/Cv = 1.667 Also it is given that T1 = 298 K (B) 207 K So, Cp = 1.667Cv We also know for adiabatic process (C) 407 K Also we know Cp-Cv = R PVY = constant (where, y=Cp/C) Putting the value of Cp, we get, So, P V Y = PzV2Y (D) 103 K (1.667 – 1) Cv=R (V4N2) = (P2/P1)1/V = (1/2.5)1/1.667 = 0.578 So, 0.667 Cv=R (E) 241 K Thus, R/Cv = 0.667 From the given equation we can write, (A (B) O O O O C T2 = T1 (V2/V)R/CV = 298 * (0.578)0.667 = 206.7 K Thus answer is 207 K Option B is correct. ( (2

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