Suppose an amount of heat ΔQ supplied to a substance changes its temperature from T to `T + ΔT`. We define heat capacity of a substance to be

`color{orange} {S = (Delta Q)/(Delta T)}` .................12.4

We expect ΔQ and, therefore, heat capacity S to be proportional to the mass of the substance. Further, it could also depend on the temperature, i.e., a different amount of heat may be needed for a unit rise in temperature at different temperatures.

To define a constant characteristic of the substance and independent of its amount, we divide S by the mass of the substance m in kg:

`color{orange} {s = s/m = (1/m) (Delta Q)/(Delta T)}` ......................12.5

s is known as the specific heat capacity of the substance. It depends on the nature of the substance and its temperature. The unit of specific heat capacity is `J kg^-1 K^-1`.

If the amount of substance is specified in terms of moles μ (instead of mass m in kg ), we can define heat capacity per mole of the substance by

`color{purple} {C = S/ mu = 1/ mu (DeltaQ)/(Delta T)}` .......................12.6

C is known as molar specific heat capacity of the substance. Like s, C is independent of the amount of substance. C depends on the nature of the substance, its temperature and the conditions under which heat is supplied.

The unit of C is `J mol^-1 K^-1`. As we shall see later (in connection with specific heat capacity of gases), additional conditions may be needed to define C or s. The idea in defining C is that simple predictions can be made in regard to molar specific heat capacities.

Table 12.1 lists measured specific and molar heat capacities of solids at atmospheric pressure and ordinary room temperature. We will see that predictions of specific heats of gases generally agree with experiment.

We can use the same law of equipartition of energy that we use there to predict molar specific heat capacities of solids. Consider a solid of N atoms, each vibrating about its mean position. An oscillator in one dimension has average energy of `2 × 1/2 k_BT = k_BT`. In three dimensions, the average energy is `3 k_BT`.

For a mole of a solid, the total energy is

`color{purple} {U = 3 k_BT × N_A = 3 RT}`

Now, at constant pressure, `color{green} {ΔQ = ΔU + P ΔV ≅ ΔU}`, since for a solid ΔV is negligible. Therefore,

`color{orange} { C = (DeltaQ)/(DeltaT) = (Delta U)/(Delta T) = 3 R}` ........................12.7

As Table 12.1 shows, the experimentally measured values which generally agrees with predicted value 3R at ordinary temperatures. (Carbon is an exception.) The agreement is known to break down at low temperatures.



`color{navy}bbul("Specific heat capacity of water")`

The old unit of heat was calorie. One calorie was earlier defined to be the amount of heat required to raise the temperature of 1g of water by `1°C`. With more precise measurements, it was found that the specific heat of water varies slightly with temperature. Figure 12.5 shows this variation in the temperature range 0 to `100 °C`.



For a precise definition of calorie, it was, therefore, necessary to specify the unit temperature interval. One calorie is defined to be the amount of heat required to raise the temperature of `1g` of water from 14.5 °C to 15.5 °C.

Since heat is just a form of energy, it is preferable to use the unit joule, J. In SI units, the specific heat capacity of water is `4186 J kg^1 K^1` i.e. `4.186 J g^-1 K^-1`. The so called mechanical equivalent of heat defined as the amount of work needed to produce `1 c a l` of heat is in fact just a conversion factor between two different units of energy : calorie to joule.

Since in SI units, we use the unit joule for heat, work or any other form of energy, the term mechanical equivalent is now superfluous and need not be used.

As already remarked, the specific heat capacity depends on the process or the conditions under which heat capacity transfer takes place. For gases, for example, we can define two specific heats : `color{lime} "specific heat capacity at constant volume"` and `color{lime} "specific heat capacity at constant pressure"`. For an ideal gas, we have a simple relation.

`color{orange} {C_p = C_v = R}` ...............12.8

where `C_p `and `C_v` are molar specific heat capacities of an ideal gas at constant pressure and volume respectively and R is the universalgas constant. To prove the relation, we begin with Eq. (12.3) for 1 mole of the gas :

`color{orange} {ΔQ = ΔU + P ΔV}`

If ΔQ is absorbed at constant volume, ΔV = 0

`color{purple} {C_v = ((DeltaQ )/(DeltaT) )_v = ((DeltaU)/(Delta T))_v = ((Delta U)/(Delta T ))}` ...................12.9

where the subscript v is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ΔQ is absorbed at constant pressure,

`color{orange} {C_p = ((Delta Q )/(Delta T))_P = ((DeltaU)/(Delta T))_P + P ( (Delta V)/(Delta T))_P }` ..................12.10

The subscript p can be dropped from the first term since U of an ideal gas depends only on T. Now, for a mole of an ideal gas

`color{purple} {PV = RT}`

which gives

`color{navy} { P ((DeltaV)/(Delta T) )_P = R}` ..........12.11

Equations (12.9) to (12.11) give the desired relation, Eq. (12.8).

Every `color{green} "equilibrium state"` of a thermodynamic system is completely described by specific values of some macroscopic variables, also called state variables.

For example, an equilibrium state of a gas is completely specified by the values of pressure, volume, temperature, and mass (and composition if there is a mixture of gases). A thermodynamic system is not always in equilibrium. For example, a gas allowed to expand freely against vacuum is not an equilibrium state [Fig. 12.6(a)].

During the rapid expansion, pressure of the gas may not be uniform throughout. Similarly, a mixture of gases undergoing an explosive chemical reaction (e.g. a mixture of petrol vapour and air when ignited by a spark) is not an equilibrium state; again its temperature and pressure are not uniform [Fig. 12.6(b)].

Eventually, the gas attains a uniform temperature and pressure and comes to thermal and mechanical equilibrium with its surroundings.



In short, thermodynamic state variables describe equilibrium states of systems.

The various state variables are not necessarily independent. The connection between the state variables is called the equation of state. For example, for an ideal gas, the equation of state is the ideal gas relation

`color{purple} {P V = μ R T}`

For a fixed amount of the gas i.e. given μ, there are thus, only two independent variables, say P and V or T and V. The pressure-volume curve for a fixed temperature is called an isotherm. Real gases may have more complicated equations of state.

The thermodynamic state variables are of two kinds: extensive and intensive. Extensive variables indicate the ‘size’ of the system. Intensive variables such as pressure and temperature do not.

To decide which variable is extensive and which intensive, think of a relevant system in equilibrium, and imagine that it is divided into two equal parts. The variables that remain unchanged for each part are intensive. The variables whose values get halved in each part are extensive.

It is easily seen, for example, that internal energy U, volume V, total mass M are extensive variables. Pressure P, temperature T, and density ρ are intensive variables. It is a good practice to check the consistency of thermodynamic equations using this classification of variables. For example, in the equation

`color{orange} {ΔQ = ΔU + P ΔV}`

quantities on both sides are extensive*. (The product of an intensive variable like P and an extensive quantity ΔV is extensive.)

`color{blue}• color{lime} " Quasi-static process"`
`color{blue}• color{lime} " Isothermal process"`
`color{blue}• color{lime} " Adiabatic process"`
`color{blue}• color{lime} " Isochoric process"`
`color{blue}• color{lime} " Isobaric process"`
`color{blue}• color{lime} " Cyclic process"`

Consider a gas in thermal and mechanical equilibrium with its surroundings. The pressure of the gas in that case equals the external pressure and its temperature is the same as that of its surroundings.

Suppose that the external pressure is suddenly reduced (say by lifting the weight on the movable piston in the container). The piston will accelerate outward. During the process, the gas passes through states that are not equilibrium states. The nonequilibrium states do not have well-defined pressure and temperature.

In the same way, if a finite temperature difference exists between the gas and its surroundings, there will be a rapid exchange of heat during which the gas will pass through non-equilibrium states.

In due course, the gas will settle to an equilibrium state with well-defined temperature and pressure equal to those of the surroundings. The free expansion of a gas in vacuum and a mixture of gases undergoing an explosive chemical reaction, mentioned in section 12.7 are also examples where the system goes through nonequilibrium states.

Non-equilibrium states of a system are difficult to deal with. It is, therefore, convenient to imagine an idealised process in which at every stage the system is an equilibrium state. Such a process is, in principle, infinitely slow-hence the name quasi-static (meaning nearly static).

The system changes its variables `(P, T, V )` so slowly that it remains in thermal and mechanical equilibrium with its surroundings throughout. In a quasi-static process, at every stage, the difference in the pressure of the system and the external pressure is infinitesimally small.

The same is true of the temperature difference between the system and its surroundings.

To take a gas from the state `(P, T )` to another state `(P ′, T ′ )` via a quasi-static process, we change the external pressure by a very small amount, allow the system to equalise its pressure with that of the surroundings and continue the process infinitely slowly until the system achieves the pressure P ′.

Similarly, to change the temperature, we introduce an infinitesimal temperature difference between the system and the surrounding reservoirs and by choosing reservoirs of progressively different temperatures T to T ′, the system achieves the temperature `T ′`.



A quasi-static process is obviously a hypothetical construct. In practice, processes that are sufficiently slow and do not involve accelerated motion of the piston, large temperature gradient, etc. are reasonably approximation to an ideal quasi-static process.

We shall from now on deal with quasi-static processes only, except when stated otherwise.A process in which the temperature of the system is kept fixed throughout is called an `color{green} "isothermal process"`.

The expansion of a gas in a metallic cylinder placed in a large reservoir of fixed temperature is an example of an isothermal process. (Heat transferred from the reservoir to the system does not materially affect the temperature of the reservoir, because of its very large heat capacity.)

In `color{orange} "isobaric processes"` the pressure is constant while in `color{purple} "isochoric processes"` the volume is constant. Finally, if the system is insulated from the surroundings and no heat flows between the system and the surroundings, the process is `color{lime} "adiabatic"`.

The definitions of these special processes are summarised in Table. 12.2

For an isothermal process (T fixed), the ideal gas equation gives

`PV = "constant"`

i.e., pressure of a given mass of gas varies inversely as its volume. This is nothing but Boyle’s Law.

Suppose an ideal gas goes isothermally (at temperature T ) from its initial state `(P_1, V_1)` to the final state `(P_2, V_2)`. At any intermediate stage with pressure P and volume change from V to `V + ΔV` (ΔV small)

`color{purple} "ΔW = P Δ V"`

Taking (ΔV → 0) and summing the quantity ΔW over the entire process,

`color{green} { W = int_(V_1)^(V_2) P dV}`

`color{green} { = mu RT int_(V_1)^(V_2) (dV)/V = mu RT In\ \ V_2/V_1}` .............12.12

where in the second step we have made use of the ideal gas equation `color{orange} {PV = μ RT}` and taken the constants out of the integral.

For an ideal gas, internal energy depends only on temperature. Thus, there is no change in the internal energy of an ideal gas in an isothermal process. The First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas `: Q = W.`

`"Note"` from Eq. (12.12) that for `V_2 > V_1, W > 0`; and for `V_2 < V_1, W < 0`. That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal compression, work is done on the gas by the environment and heat is released.