Carnot Cycle

Carnot Cycle

Carnot Cycle

Carnot Cycle

Carnot used a reversible cycle to show the maximum convertibility of heat into work. It is a cyclic process carried out in four reversible steps alternatively isothermal and adiabatic. The first two steps being of expansion and the last two of compression. The engine that works by the above process is known as Carnot engine or ideal gas engine.
The working substance of the engine is one mole of an ideal gas.
The engine works between two temperature(T2) and (T1). It takes heat from a source at higher temperature (T2) and does work and gives out the unused heat into a sink at lower temperature (T1).
Stroke-1
Isothermal Expansion
The gas is allowed to expand reversibly and isothermally at the temperature T2 so that the volume increases from V1 to V2.
We know that in isothermal expansion of an ideal gas ΔE = 0. So, from first law of thermodynamics-
ΔE = q - w
or, q = w
Let q2 be the heat absorbed by the system at temperature T2 and work w be the work done by the system on the surroundings.
then- q2 = w1
= RT2 ln V2/V1 -----(equation-1)
Stroke-2
Adiabatic Expansion
The gas is now allowed to expand reversibly and adiabatically from volume V2 to V3.
Since work has been done by the system adiabatically, where heat is not absorbed. So, the temperature of the system falls from T2 to T1. Therefore, q = 0. Thus, the 1st law of thermodynamics becomes-
ΔE = - w
or, -ΔE = w
we know that-
Cv = (δE/δT)v
or, Cv . dT = dE
or, Cv . (T1 - T2) = -w
or, Cv . (T2 - T1) = w
Let work done by the system in this step is denoted by w2 then-
w2 = Cv . (T2 - T1) -----(equation-2)
Stroke-3
Isothermal Compression
Here gas is subjected to a reversible and isothermal compression at the lower temperature T1 so the volume decreases from V3 to V4.
In this case work is done on the system and hence heat will be produced and given up to the surroundings. Since compression takes place isothermally and reversibly,
so, ΔE = 0
and if q1 be the heat given out to the surrounding at temperature T1 and work w3 be the work done on the system.
so in this process-
q1 = -w3
or, -w3 = RT1 ln V4/V3 -----(equation-3)
Stroke-4
Adiabatic Compression
Finally by adiabatic and reversible compression, the gas is restored to the original volume V1 and temperature T1. In this case, work is done on the system. Hence, w is negative. Then, first law of thermodynamics becomes-
ΔE = q - (-w) = q + w
In adiabatic process, q = 0
Hence- ΔE = w = Cv (T2 - T1)
Let w4 be the work done in this stage-
then- w4 = Cv (T2 - T1)
or, -w4 = -Cv (T2 - T1) -----(equation-4)
where, T2 - T1 is increase in temperature produced by the adiabatic compression.
The net heat absorbed q by an ideal gas in the whole cyclic process is-
q = q2 + (-q1)
q = RT2 ln V2/V1 + RT1 ln V4/V3
or, q = RT2 ln V2/V1 - RT1 ln V3/V4 -----(equation-5)
According to the expression governing adiabatic changes-
(T2/T1) = (V3/V2)γ-1---for adiabatic expansion
(T1/T2) = (V1/V4)γ-1---for adiabatic compression
or, V3/V2 = V4/V1 Therefore, substituting the value of V3/V4 in (equation-5), the Net Heat may be give as-
q = RT2 ln V2/V1 - RT1 ln V2/V1
or, q = R(T2 - T1)ln V2/V1 -----(equation-6)
Net work done in one cycle-
W = w1 + w2 + (-w3) + (-w4)
W = RT2 ln V2/V1 + Cv(T2 - T1) + (-RT1 ln V4/V3) + (-Cv(T2 - T1)

or, W = RT2 ln V2/V1 + RT1 ln V3/V4 -----(equation-7)

Equation-7 is work done in one cycle.
or, q = RT2 ln V2/V1 - RT1 ln V2/V1

or, q = R(T2 - T1)ln V2/V1 (as V2/V1 = V3/V4) -----(equation-8)

Equation-8 is Net Heat absorbed in one cycle.

 Share