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

Question

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

Explanation / Answer

The heat transfferred to the gas can be calculated via the internal energy change:
U = Q + W
<=>
Q = U - W

The internal energy of an ideal gas is given by:
U = nCvT
Molar hat capacity at constant volume of an ideal monatomic gas is:
Cv = (3/2)R

The temperatures can be found from ideal gas law:
pV = nRT
=>
T = pV/(nR)

You can calculate two temperatures explicitly, but you don't need it.
Simply substutite the expression for T into the internal energy definition:
U = (3/2)nR pV/(nR) = (3/2)pV
Hence:
U = (3/2) (pV)

The the internal energy change of the the processes is
U_AB = 0 because it is an isothermal proceass
U_BC = (3/2) (p_CV_C - p_BV_B)
(because p_C = p_B)
= (3/2) p_B (V_C - V_B)

U_CA = (3/2) (p_AV_A - p_CV_C)
(because V_C = V_A)
= (3/2) V_C (p_A - p_C)

So the transferred to the gas in each step is:
Q_AB = U_AB - W_AB

Q_BC = U_BC - W_BC

Q_CA = U_CA - W_CA

The net energy added to the gas by heat is
Q_in = Q_AB + Q_CA

The net energy exhausted by the gas by heat is
Q_out = -Q_BC


(c)
efficeny is th ratio of useful work done by the gas t heat added to the gas, i.e.
= -W/Q_

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