Clausius inequality

 

Let system I be an arbitrary thermodynamic system, which is able to exchange heat with thermal reservoirs R1 and R2, having the temperatures T1 and T2 respectively. In this derivation we will not specify which of these two reservoirs is a heater and which is a cooler. We will consider the heat lost by a thermal reservoir to be positive and the heat gained by a reservoir to be negative. This will allow formulating the final result symmetrically with respect to both reservoirs.

            Let the system I perform a cyclic process – reversible or irreversible – during which it received heat Q1 from reservoir R1 and Q2 from reservoir R2. Since the system has returned to its initial state the total amount of heat gained by the system Q1+Q2 is equal to the work it has performed.

Lets now take a Carnot engine and make it work using the same heat reservoirs R1 and R2. In order to the presence of the Carnot engine not to influence the amounts of heat Q1 and Q2 the system I has received from the heat reservoirs during its cycle, we can run the Carnot engine after the system I has completed its cycle. If at this point we thermally isolate the system I, the reservoirs R1 and R2 will exchange heat with the Carnot engine only. The presence of the Carnot engine does not affect the cyclic process in the system I since it has already finished by the time the Carnot engine was introduced.

Let the Carnot engine perform a cyclic process, during which it receives heat Q1’ from reservoir R1 and Q2’ from R2 respectively. It is essential for the further derivation to note that the Carnot engine is reversible. It can work either as an engine or a refrigerator. Apart from that the isotherm 12 in Carnot cycle can be taken as short as desired (see Fig. 1). Therefore, the work done by the Carnot engine can be as small as desired. One can also make the Carnot engine perform as much work as desired via performing many cycles. Thus the Carnot engine can perform any specified amount of positive or negative work. This allows choosing the value of either Q1’ or Q2’. It is always possible to make one of these heats have a chosen positive or negative value.

 

 

Figure 1. Carnot cycle

 

According to Carnot theorem

 

(Q1’/T1) + (Q2’/T2) = 0

 

Let us combine the system I and the Carnot engine into one complex system. The cycles performed consecutively by the system I and Carnot engine can be combined into one cyclic process. During this process the combined system

 

has received the heat Q1 + Q1’ from reservoir R1,

has received the heat Q2 + Q2’ from reservoir R2,

has performed the work W = Q1 + Q1’ + Q2 + Q2’.

 

Let us choose Q1’ so that Q1’ = -Q1 then

 

Q2’ = T2 (Q1/T1).

 

As a result of the cycle the state of the reservoir R1 remains unchanged. The reservoir R2 looses the amount of heat

 

 Q2 + Q2’ = Q2 + t2 (Q1/T1) = T2 [(Q1/T1) + (Q2/T2)]

 

An equivalent work W = Q2 + Q2’ has been performed using this amount of heat. If this work were positive, we would have the Thomson-Plank process, which is forbidden by the II law of thermodynamics. Therefore, the work should be negative or equal to zero (W ≤ 0). Since the absolute temperature T2 is always positive this leads to

 

(Q1/T1) + (Q2/T2) ≤ 0.

 

This inequality is called Clausius inequality.