Cycles and heat pumps

Second law: It is impossible to construct a device that operates in cycles that converts heat into work without producing some other change in the surroundings.

Consider an idealized cycle. In this cycle there are the following four steps

I. Isothermal expansion

II. Adiabatic expansion

III. Isothermal compression

IV. Adiabatic compression

This is known as a Carnot cycle.

Figure 1. A thermodynamic cycle

Clearly, the volume ratios at points 1, 2, 3, and 4 are established by

so that

The heat q for the adiabatic steps is zero.

Therefore,

For the isothermal steps. The red curve represents the isothermal expansion (hot gas) and the blue curve represents the isothermal compression (cold gas). Since these processes are isothermal we know that

DU = 0 and q_total = -w_total.

The total work of isothermal expansion and compression is given by

Note that V₂ > V₁ and so w_hot < 0 represents work done by the system.

Note that V₄ < V₃ and so w_cold > 0 represents work done on the system.

Note also that q_total = -w_total ¹ 0 and neither heat nor work is a state function.

Since

And furthermore V₄/V₃ = V₁/V₂ we have

which means that

The latter equation is very important since it defines a thermodynamic temperature scale. The ratio of the temperatures is equal to the ratio of the heats transferred

Note further that we define the thermodynamic efficiency as

Using this definition of efficiency we can prove that the cycle shown in Figure 1 applies not only to an ideal gas but any working substance. The logic is as follows. We show two engines operating such that one is a heat pump and the other is a cooler (i.e. it is a refrigerator which is the reverse of a heat pump). We can show that no matter what the working fluid of engines 1 and 2 (ideal gas, real gas etc.) the thermodynamic efficiency of two ideal engines must be the same. Our negative proof is as follows:

Suppose the efficiencies are not the same, e.g. h₁ > h₂.

This implies w₁ > w₂ and since

This implies that we are able to convert a net work |w₁| - |w₂| into heat |q_cold1| - |q_cold2| without changing the surroundings in any way. This violates the second law and therefore all engines that operate in a reversible Carnot cycle have the same efficiency.

We can use the combined first and second law expression to derive a large number of important thermodynamic expressions.

dU = TdS - PdV

For example,

This means that

We can integrate the above expression to determine the entropy change at constant volume. The entropy change at constant pressure can be derived by an exactly analogous procedure starting with

dH = TdS + VdP

Finally, we have

Thus, the entropy change for the system can be calculated by

DS = nRln(V₂/V₁) constant T

DS = C_vln(T₂/T₁) constant V

DS = C_pln(T₂/T₁) constant P