Consider the adiabatic expansion of 0.500 mol of an ideal monatomic gas with C v

Question

Consider the adiabatic expansion of 0.500 mol of an ideal monatomic gas with C v, m = 3R/2. The initial slate is described by P= 6.25 bar and T = 300. Calculate the final temperature if the gas undergoes a reversible adiabatic expansion to a final pressure P= 1.25 bar. Calculate the final temperature if the same gas undergoes an adiabatic expansion against an external pressure of P = 1.25 bar to a final pressure P= 1.25 bar. Explain the difference in your results for parts (a) and (b). Ans. T2 = 158 K T2 = 204 K

Explanation / Answer

For an Adiabatic Expansion; T^gamma * P^(1-gamma) = constant gamma = Cp / Cv = (5R/2) / (3R/2) = 5/3 300^5/3 * 6.25^(1-5/3) = T^5/3 * 1.25(1-5/3) => 3962.3 = T^5/3 * 0.862 => T = 4587.84^3/5 = 157.6 K Cv * (T2-T1) = -Pext * (RT2/P2 - RT1/P1) => 1.5 * 8.314 * (T2-300) = -1.25 * (8.314 * T2 / 1.25 - 8.314*300/6.25) => T2-300 = -0.8333*(0.8*T2 - 48) => T2 - 300 = -0.6667*T2 + 40 => T2 = 204 K c) for the same initial and final conditions work done by a reversible process is more than a irreversible process. This accounts for the difference

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