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The Carnot Cycle (for ideal gases)

Carnot Cycle (specific for Ideal Gases)

If we repeat the Carnot cycle between the temperatures T2 and T1  with  n moles fan ideal gas, then we have in line with the following Figure (1)

 

Step 1: Isothermal reversible expansion at T2



Step 2: Adiabatic reversible expansion from T2 to T1



Step 3: Isothermal reversible compression at T1

 ∆E3 = 0    W3= -q1 = nRT1 In (V4/V3)

Step 4: Adiabatic reversible compression from T1 to T2


 


For the complete cycle ∆E =  0.  So  the total maximum work done, W, must be given by 

 W = q2 – q1 Further,

 W = W1 + W2 + W3 + W4




   W  = nRT2 ln (V1/V2) + nRT1 1n (V4/V3)

because the two integrals cancel each other. so

W = q2-q1  =   nRT2 ln (V1/V2) + nRT1 1n (V4/V3)                             ………..(1)

 Again, since   q2 = nRT2 ln (V1/V2), division of Eq. (1) by q2 yields

 W/q2 = (q2- q1)/q2 =  RT2 In (V1/V2) + RT1 In (V4/V3) /RT2 In (V1/V2 …(2)

simplifying eq 2 . Since the points (P1 , V1) and (P3, V3) lie on the same adiabatic, then a relation can be used  

V1/V2=V3/V4 ....................(3)

And consequently,

ln(V4/V3) = -ln (V1/V2) ...........(4)

substituting this value for ln (V4/V3)  in eq (2) we get finally

W/q2 = (q2- q1)/q2 =  RT2 In (V1/V2) + RT1 In (V1/V2) /RT2 In (V1/V2      .......(5) 

           = T2-T1/T2  .........(6)

 If we compare any substance and an ideal gas operating in a Carnot cycle between 0 and 100°C, then (T2 – T1) = 100°C. From this argument the thermodynamic and ideal gas temperature scales are identical, and we may write thus for all substances

W/q2=q2-q1/q2

        =T2-T1/T2

Equation (6) we can assume the maximum work that may be recovered from the process is equal to the heat absorbed at T2 multiplied by the ratio (T2 -T1)/T2. It is important to observe that the absorption takes place at a higher temperature and that the heat passes from the higher temperature to the lower. The magnitude of the work involved is given by the enclosed area in Figure (1). Furthermore, the thermodynamic efficiency must be the same for all processes operating under given temperature conditions.

From this equation, it may be seen that 100% thermodynamic efficiency could be obtained only by making T1 = 0. To reach this temperature, we would require surroundings at least dT below absolute zero in order to absorb the heat rejected by the system at T = 0. However, since the latter temperature is the lowest possible, we are forced to the conclusion that absolute zero is approachable but not attainable. Actually, temperatures several thousandths of a degree above absolute zero have been obtained experimentally, but not 0°K. Consequently, 100 % conversion of heat into work is impossible in any cyclic process.

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