InterviewSolution
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InterviewSolution
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A weightless piston divides a thermally insulated cylinder into two equal parts. One part contains one mole of an ideal gas with adiabatic exponent `gamma`, the other is evacuated. The initial gas temperature is `T_0`. The piston is released and the gas fills the whole volume of the cylinder. Then the piston is slowly displaced back to the initial position. Find the increment of the internal energy and entropy of the gas was resulting from these two processes. |
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Answer» The process consists of two parts. The first part is free expansion in which `U_f = U_i`. The second part is adiabatic compression in which work done results in change of internal energy. Obviously, `0 = U_F - U_f + int_(V_f)^(V_0) pdV, V_f = 2 V_0` Now in the first part `p_f = (1)/(2) p_0, V_f = 2 V_0`, because there is no change of temperature. In the second part, `p V^gamma = (1)/(2) p_0 (2 V_0)^gamma = 2^(gamma -1) p_0 V_0^gamma` `int_(2 V_0)^(V_0) pdV = int_(2 V_0)^(V_0) (2^(gamma - 1)p_0 V_0^gamma)/(V^gamma) dV = [(2^(gamma -1) p_0 V_0^gamma)/(- gamma + 1) V^(1 - gamma)]_(2 V_0)^(V_0)` =`2^(gamma -1) p_0 V_0^gamma V_0^(-gamma + 1) (2^(- gamma +1) -1)/(gamma -1) = -((2^(gamma -1) -1))/(gamma -1) RT` Thus `Delta U = U_F - U_i = (R T_0)/(gamma - 1) (2^(gamma -1) - 1)` The entropy change `Delta S = Delta S_1 + Delta S_(II)` `Delta S_I = R 1n 2` and `Delta S_(II) = 0` as the process is reversible adiabatic. Thus `Delta S = R 1n 2`. |
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