A heat engine is composed of of a monatomic ideal gas which runs in the followin

Question

A heat engine is composed of of a monatomic ideal gas which runs in the following cycle: At point A the pressure is 2.6 atm, the volume is 2.7 liters and the temperature is 300 K. During the transition A to B the pressure is constant and the volume doubles to 5.4 liters. From B to C the pressure decreases at constant volume to 1.3 atm. From C to A there is a compression at constant temperature until the pressure is again 2.6 atm, where again the volume is 2.7 liters and the temperature is 300 K. What is the temperature at point B? What is the work done BY the gas in the transition A to B? What is the heat input into the gas for the transition A to B? What is the work done in the transition B to C? What is the heat input into the gas for the transition B to C? For the constant temperature transition C to A, 486.58932 J of heat are taken out of the gas. What is the work done ON the gas in the transition? What is the efficiency of this heat engine? What is the efficiency of a Carnot cycle running between the lowest and highest temperature of the cycle above?

Explanation / Answer

aNSWER:-

GIVEN
P_A = 2.6 atm, V_A = 2.7 lt, T_A = 300 K
P_B = P_A = 2.6 atm, V_B = 5.4 lt , T_B =?
P_C = 1.3 atm, V_C = V_B = 5.4 lt, T_C=?
P_A' = 2.6 atm, V_C =2.7 lt, T_A'=T_C=300 K?

Using the ideal gas equation,
P*V/T = const
which gives,
P_A*V_A/T_A = P_B*V_B/T_B
T_B = 600 K

Work done at constant pressure is W = p*dv
Work done from A to B transition = 2.6*(5.4-2.7) = 7.02 L atm = 0.7113 kJ

Change in internal energy for ideal gas (dU) = c_v*dT = 3/2 * 300 = 450
and heat supplied (dQ) = dU + dW = 450 + 711.3 = 1161.3 J

Work done at constant volume is zero.
so work done from B to C transition = 0

To calculate heat from B to C, First calculate T_C
P*V/T = const gives,
P_B*V_B/T_B = P_C*V_C/T_C
T_C = 300 K
So heat change from B to C, (dU') = 3/2* 300 = 450
which implies, (dQ') = 450 + 0 = 450 J

from C to A, dQ'' = 486.58932 J
Work done (dW'') = dQ'' - dU'' = 486.58932 J
Because for ideal gas, change in internal energy at constant temperature is zero.
So dU'' = 0

efficiency (eta_1) = 1 - Q/Q'' = 1-486.58932 /1161.3 = 1 - 0.419003978 = 0.580996022

Efficiency of carnot engine will be (eta) = 1- T_min/T_max = 1 - 300/600 = 0.5

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