QH = TH In | The heat on segment cd is QC = - TC In | By inspection, we can say that THVE - Vr-1 and Y-1_ Y-1 TH Thus, Vb/ then |QC/QH| = Which solves the efficiency of the Carnot engine.

As shown in the figure, a chamber with a moveable piston and containing a monatomic ideal gas in an initial state A undergoes an isovolumetric, then an isothermal, and finally an isobaric process to complete the cycle. P (atm) LV (L) When the gas is in the initial state, the volume is 3.00 L, the pressure is 5.00 atm, and the temperature is 200 K. The gas is first warmed at constant volume to a pressure of 4 times the initial value (state B). The gas is then allowed to expand isothermally to some new volume (state C). Finally it is compressed isobarically to its initial state. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.)

Calculate the net work output of a heat engine following path ABCDA in the figure below, where  V1 = 2.6 ✕ 10−3 m3  and  V2 = 10.4 ✕ 10−3 m3.

Calculate the efficierncy of a Carnot reversible heat engine working between the temperatures 127 degree C- (-73 degree C).