Question

3. A Sterling engine using 1.00 mol of an ideal gas initially at a volume of 24.6L and a temperature of 400K performs a cycle consisting of four steps: (1) An isothermal expansion at a temperature of 400K to twice its initial volume. (2) Cooling at constant volume to a temperature of 300K (3) An isothermal compression to its original volume (4) Heating at constant volume to its original temperature of 400. Assume that C, = 21.01/K. (a) Sketch a PV diagram for the Stirling cycle. (b) Find the net work done for an entire cycle し4.6 1 4.6L -1723 (c) Find the heat input during one cycle (Qim 2100 L4u0 vf (d) Find the efficiency of the engine. 400-300 리 1(.34) ne t) (e) How much energy was lost due to irreversible processes? n0 0 574

part C and E. Please label each part with its answer.

Answer 1

Heat is input is 1st and 4th process Q_1 = nRT ln v_2/v_1 = 1 times 8.31 times 400 ln2 = 2304 J Q_4 = nc_v delta T = 1 times 21 times (400 - 300) = 2100 J therefore Q_in = 2304 + 2100 = 4404 J for reversible process = 0.75 times qin = 3303 J now in irreversible process heat lost = win 0- w = 4404 - 576 = 3828 J therefore heat lost due to irreversible process = 3828 - 3303 = 525 J