1. The problem statement, all variables and given/known data Here's a link to the pV-diagram I am using - http://session.masteringphysics.com/problemAsset/1396784/2/p16.7.jpg A heat engine takes 2.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram shown in the figure. The path bc is an isothermal process. The temperature at c is 820 K, and the volumes at a and c are 0.010 m3 and 0.16 m3, respectively. The molar heat capacity at constant volume, of the gas, is 37 J/mol·K, and the ideal gas constant is R = 8.314 J/(mol·K). The thermal efficiency of the engine is closest to Answers) A - 0.026 B - 0.53 C - 0.40 D - 0.26<--- This is the right answer, but I'm not sure how to get to it E - 0.33 2. Relevant equations [tex] ΔU = Q - W [/tex] [tex] ΔU = nC_vΔT [/tex] [tex] Work = ∫PdV [/tex] [tex] Isothermal Work = nRT*ln(V_f/V_i) [/tex] 3. The attempt at a solution I'm getting really lost on differentiating the positive/negative signs and thus finding Qin and Qout. I understand that the efficiency of a thermal engine is [tex] e = W_out/Q_H = 1 - Q_c/Q_h [/tex] I started finding some of the areas, but I am not sure where to subtract and add them to find Q_c and Q_h. Isothermal Work = nRTln(.010-.16) = 37804. J ΔU = 0, therefore W=Q --> So since W = 37804 J, there is 37804 J of heat added to the gas here? Isobaric Work = p(ΔV) = p(.16-.01) To solve for the pressure I re-arranged PV=nRT. First I solved for the temperature at A. [tex] V_A/T_A = V_C/T_C [/tex] This gave me the temperature at A as 51.25 Kelvin. Using PV=nRT.. [tex] P_A = (2.0mol)(8.314J/mol-K)(51.25 K)/(.010m^3) [/tex] This gave me the pressure P_A = 85218.5 Pa. Then the work under the Isobaric Line is p(ΔV) so I plugged in the pressure of A.. [tex] (85218.5 Pa)(.010 m^3-.16 m^3) = -12782.775 J [/tex] For the isobaric line I also said that, [tex] ΔU = nC_vΔT = (2.0mol)(37J/mol-K)(820K - 51.25 K) = 56887.5 J [/tex] I think I'm close to the answer, but I'm not sure which numbers to use for the efficiency.