Question

A heat engine with 0.252 moles of a monatomic as undergoes the cyclic procedure shown in the pV diagram on the right. Between stages 3 and 1 the gas is at a constant temperature, and between 2 and 3 no heat is transferred in or out. The temperature of the gas at stage 2 is 345 K. 535 What is the type of each process in the cycle? Between 1 and 2 is isachoric Between 2 and 3 is adiabatic 395 Between 3 and 1 is sothermal em atebetiween slages 3 and 17 1350 1620

1) What is the work per cycle of this heat engine?

2) Find the efficiency of this heat engine.

Answer 1

Between I and 2 is rightarrow isochoric rightarrow (volume + constant). Between 3 and 3 is rightarrow Adiabotic rightarrow Between 3 and 1 is rightarrow isothemal rightarrow (Temperature constant) rightdoublearrow number of moles rightdoublearrow 0.252 mole. For monoatomic gas rightdoublearrow y = 5/3 rightdoublearrow T(v)^y - I = const rightdoublearrow T_1 = 345k rightdoublearrow T_1v_1^(y - I) = T_2.V_2(y - I) rightdoublearrow 345 (1350/1620)^2/3 = T_2 T_3 = 305 k. T_1 = 305 K + (Temp is const. between 3 + I.) rightarrow Work per cycle of this heat engine? n = 1 - T_2/T_1 rightarrow efficiency of heat engine. \r\n W_I - 2 = 0 rightarrow (isochoric process) so W = integral p.dv dv = 0 W_2 + 3 = (Adlabatic process) delta theta = d_0 + delta W rightdoublearrow (delta theta = 0) rightarrow In Adlabatic process). 0 = du + dw increment u = -increment W so increment W = nc_vdT = 0.253 times 3/2 (RT_2 - RT_1)