Consider 1 mole of an ideal gas with constant heat capacity C degree = 5/2R that

Question

Consider 1 mole of an ideal gas with constant heat capacity C degree = 5/2R that changes state from T_1 = 500K and P_1 = 0.6 MPa to T_2 = 200K and P_2 = 0.1 MPa. A possible way of achieving this change is to first at constant pressure P_1 the temperature is changed to T_2 and the pressure is changed to P_2 at constant temperature T_2. Calculate delta U, Q, and W for each step and the total change. Assume that each step is carried out reversibly. Another possibility to make these changes is to first at constant temperature T_1 the pressure is changed to P_2 and then the temperature is changed to T_2 at a constant pressure P_2. Calculate delta U, Q, and W for each step and the total change. Assume that each step is carried out reversibly. Which way is more efficient and why

Explanation / Answer

(a) at constant pressure P1,

w = -nRdT = -1 x 8.314 (200-500) = 2.49 kJ

q = nCpdT = 1 x 2.5 x 8.314 (200-500) = -6.23 kJ

dU = q - w = -6.23 - 2.49 = -8.72 kJ

at constant Temperature T2, dU = 0

w = -nRT2lnP1/P2 = -1 x 8.314 x 200 ln(0.6/0.1) = -2.98 kJ

q = dU - w = -w = 7.45 kJ

Overall change in,

w = 2.49 - 2.98 = -0.49 kJ

q = 7.45 - 6.23 = 1.22 kJ

dU = -8.72 kJ

(b) at constant Temperature, dU = 0

w = -nRT1lnP1/P2 = -1 x 8.314 x 500 ln(0.6/0.1) = -7.45 kJ

q = dU - w = -w = 7.45 kJ

at constant pressure,

w = -nRdT = -1 x 8.314 (200-500) = 2.49 kJ

q = nCpdT = 1 x 2.5 x 8.314 (200-500) = -6.23 kJ

dU = q - w = -6.23 - 2.49 = -8.72 kJ

Overall change in,

w = -7.45 + 2.49 = -4.96 kJ

q = 7.45 - 6.23 = 1.22 kJ

dU = -8.72 kJ

(c) So path b is more efficient as more heat is generated as work is done, that is process is more spontaneous.

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