# How the temperature used in second law of thermodynamics and ideal gas law equivalent? [closed]

I was reading Theory and Problems of Thermodynamics by Schaum's Outline series and there the author starts from some axioms (which are the thermodynamic laws, excluding the zeroth and third). Then he derived the efficiency of Carnot engine working between two absolute temperatures (but he did not define what absolute temperature is or how one calculates it). Now my first question is:

How is this absolute temperature measured?

He then derived the same Carnot efficiency using ideal gas (for which temperature is accurately defined ie; average kinetic energy of gas multiplied by 2/3$$k$$ factor). And then he said that the absolute temperature is consistent with the temperature used in ideal gas law!

How are the two equivalent?

• There are a few different ways to answer this question. In order to pick the most helpful one, where do you think of the ideal gas law as coming from? Mar 13 at 11:25

In a reversible process, the fundamental thermal quantity is entropy not heat or energy. When a mass $$m$$ drops in a gravitational potential from $$\phi_1$$ to $$\phi_2$$ it can deliver $$w_{12}=(\phi_1-\phi_2)m$$ work by, say, raising another mass $$m'$$ from $$\phi'_1$$ to $$\phi'_2$$ so that $$w_{12}=w'_{12}=$$ or $$(\phi_1-\phi_2)m$$=$$(\phi'_1-\phi'_2)m'$$. An engine performing a reversible Carnot cycle so that an amount of entropy $$S$$ is moved between empirical temperatures as seen by thermometers $$\theta_1$$ and $$\theta_2$$ delivers an amount of work $$w=(\theta_1 -\theta_2)S$$. Using the same temperature scale the zero point of the thermometer does not matter when the delivered work is calculated.

The efficiency of the Carnot cycle is referenced relative to the incoming thermal energy. To find the relationship between the experimental and absolute temperature scales one can notice that during heat conduction between two similar blocks of matter of different temperatures the total energy is constant and the amount of thermal energy one loses it must equal the amount being gained by the other, hence the loss of "workable" energy from one is gained by the other as the entropy is moved through the temperature drop. And what is moving during "heat" conduction is really entropy from a higher to a lower temperature.

In an irreversible process, in which a small amount of entropy $$\delta S^*$$ is generated the amount of "workable" energy is reduced by an amount $$\delta \mathcal E$$ that is proportional to the irreversibly and locally generated entropy $$\delta S^*$$, more specifically $$\delta \mathcal E =T \delta S^*$$. Experimentally, this proportionality factor $$T$$ is a function of the experimental temperature and is called the absolute temperature and you might as well use it everywhere. Since entropy is just that irrespective of its origin, if the "workable" energy loss is proportional to $$T$$ then so is the energy loss or gain $$\delta \mathcal E =T \delta S$$ is irrespective of the provenance of $$S$$.

This allows us to measure the efficiency of the Carnot engine operating between temperatures $$T_1$$ and $$T_2$$ that delivers work $$w=(T_1-T_2)S$$ against the absorbed thermal energy $$T_1S$$ resulting in $$\eta = \frac{(T_1-T_2)S}{T_1S}= 1-\frac{T_2}{T_1}$$.

As you can see, the local caloric equation $$\delta \mathcal E =T \delta S$$ expressing the relationship between energy, temperature and entropy, is universally valid. When you break it down further by specifying the behavior of the matter, such as being a noninteracting monatomic gas statistical mechanics tells us that for such gas $$\mathcal E$$ is proportional to the average kinetic energy of the atoms.

The logic can be presented in several ways, but I would single out two for special mention. Argument A and argument B outlined below have some elements in common but I offer them both because they have different styles and each can teach something useful.

Argument A

We begin by proving Carnot's theorem, which is the theorem that for given thermal reservoirs no engine is more efficient than a reversible engine (where we have in mind a cycle with two adiabatic and two isothermal stages i.e. the Carnot engine) and all reversible engines are equally efficient. Note, the proof of this does not require any definition of temperature beyond an ability to state which of two reservoirs is the hotter one. Also it does not require any particular example of a Carnot engine (it need not involve a gas or even a fluid, for example). The proof can be done by showing that the existence of a more efficient engine would lead to a machine which could break either the Clausius or the Kelvin statement of the Second Law.

With Carnot's theorem in hand, one can now define absolute temperature $$T$$. We define $$T$$ such that if $$Q_1$$ and $$Q_2$$ are the heats exchanged at two reservoirs during a cycle of a Carnot engine, then the reservoir temperatures are in the ratio given by $$\frac{T_1}{T_2} = \frac{Q_1}{Q_2}. \tag{1}$$ This is sufficient to define temperature, and Carnot's theorem will then guarantee that all reservoirs having the same temperature thus defined will be found to be in equilibrium with one another if they are brought into thermal contact. Note there is no need to mention any gas, whether ideal or not, in any argument so far.

Finally we come to Boyle's law which says $$pV = f(T)$$ for an ideal gas, with the function $$f(T)$$ to be discovered. If we also bring in Joule's law (that $$U=U(T)$$ for ideal gas) then one can show that for the ideal gas one must have that $$f(T)$$ is proportional to $$T$$. Thus one derives the ideal gas equation of state from the combination {Boyle's law, Joule's law, defn of temperature}.

Once one has the ideal gas equation of state, one can show that the heats exchanged for a Carnot cycle of an ideal gas obey the equation (1) given above. But this has to happen because that equation was used to derive the ideal gas equation of state! If instead one claims to know the ideal gas equation of state at the outset, then the argument amounts to showing that the $$T$$ mentioned in the ideal gas equation of state must be proportional to the absolute temperature whose definition is given above.

(By the way, arguments involving the ideal gas cannot be used to define temperature in general because the ideal gas is itself an impossible and inconsistent system at low enough temperature at any given density.)

Argument B

For argument B one takes as starting-point the entropy statement of the Second Law, which asserts that every isolated system has a function of state called entropy which is extensive and whose value is maximised in thermal equilibrium. One then defines $$T \equiv \left. \frac{\partial U}{\partial S} \right|_{V,N}$$ and one goes on to consider heat transfer. One notes that during reversible heat transfer, the entropy functions of the two systems involved change by $$\Delta Q/T$$ (one going up, the other going down), and that reversible process must involve no change in entropy for an isolated system. From this it follows that the heats transferred in a Carnot cycle must be in proportion to $$T_1/T_2$$. One can then invoke the same reasoning as in argument A to derive the equation of state of an ideal gas, or to show that the ideal gas temperature given by $$pV$$ must be proportional to the absolute temperture (in the limit where the ideal gas makes sense as a physical system).

• Thanks for the explanation, i learned something new. But i have a doubt: for two given heat reservoirs we have a quantity Q1/Q2 constant(for reversible heat engines). And using that ratio we defined absolute temperature scale. But to quantitatively define the temperature we need to measure the ratio Q1/Q2. How does one measure it? Mar 14 at 8:12
• In order to use the temperature scale in practice one would not normally directly measure the heat transfer in a Carnot cycle. Rather, one develops standard thermometers such as the ones uses in standards laboratories. An important one is based on extrapolating the behaviour of a gas to the low pressure limit. Mar 14 at 12:20