A 1.00-mol sample of a monatomic ideal gas is taken through the cycle shown in t

Question

A 1.00-mol sample of a monatomic ideal gas is taken through the cycle shown in the figure below. At point A, the pressure, volume, and temperature are P_i, V_i, and T_i, respectively. (Answer the following questions in terms of R and T_i.) Find the total energy entering the system by heat per cycle. times nRt_i Find the total energy leaving the system by heat per cycle. times nRt_i Find the efficiency of an engine operating in this cycle. How does the efficiency compare with that of an engine operating in a Carnot cycle between the same temperature extremes? The Carnot efficiency is much higher. The Carnot efficiency is much lower. The Carnot efficiency is about the same.

Explanation / Answer

For one mole of ideal gas we have P0 V0 = R T0

(a) total heat entering equals the work done ON the system along the parts where heat enters:

Qin(A->C) = -(2n-2) P0 * V0 - (2n-1)P0 * (m-1)V0 = (m+1) P0 V0 = 4 P0 V0 = 4 R T0

(b) total heat leaving equals the work done BY the system along the parts where heat leaves:

Qout(C->A) = (2n-2) m P0 V0 + m P0 V0 = (2n + m - 2) P0 V0 = 7 P0 V0 = 7 R T0

So it is a refrigerating machine
(c) the work done (area inside the rectangle) by the engine is

W = (2n - 2) (m -1) P0 V0 = 8 P0 V0 = 8 R T0

Therefore the efficiency is:
efficiency = Qout/W = 7/8 = 87.5%

d) The extreme temperatures are (at A and C)
Tlow = P0V0/R
Thigh = (2n-1) m P0 V0/R = 15 P0 V0/R
So the Carnot efficiency is

1 - 1/15 = 14/15 = 93.3%

so option a) the carnot efficiency is much higher.

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