By Kent W. Mayhew

*Isobaric vs Isometric Specific Heat*

In order to appreciate the theoretical aspects
of the accepted differences between isobaric and isometric specific heats, consider Figs 1.10.3 and 1.10.4. Both show System 2 giving
the same amount of heat (dQ2) into System 1, that being a piston-cylinder apparatus.

For the isobaric case (Fig. 1.10.3)(AKA isochloric case), the heating of the gas causes a kinetic energy increase of the gas, which then drives the piston outwards by a distance, dx. Considering that the piston is both frictionless, and massless, then all the work that is actually done by the heating is done onto the atmosphere’s displacement [W=(Patm)dV]. (see LOST WORK)

Conversely, for the isometric case (Fig. 1.10.4), the piston
is locked in place. Thus the heating of the gas results in a kinetic energy increase of the gas, which increases the pressure within
the apparatus. Since no work is actually done, then the amount of heat required to raise the temperature of the gas inside the piston-cylinder
must be less than it would be for the isobaric case.

If the input of energy is the same for both the isometric and isobaric
cases, then the temperature increase in the isometric case will be greater than in the isobaric case.

Of course, if the
piston in the isometric case is unlocked then work will be done onto the atmosphere, and either: 1) the gas will cool if the heat
(dQ2) is turned off. Or: 2) Extra heat will be required in order for the gas to remain isothermal. We now begin to understand the
differences between isometric and isobaric heat capacities for gases.

For isobaric
specific heat (Cp) applied to n moles of gas we have:

Cp= (1/n)(dQ/dT)P
P means at constant pressure 1)

While for isometric specific heat (Cv) applied to n moles of gas we
have:

Cv= (1/n)(dQ/dT)V V means at constant volume 2)

Now the
ideal gas constant (R) is the difference between isobaric and isometric specific heats

R=Cp-Cv 3)

Substituting eqn 1) and eqn 2) into eqn 3, we obtain:

R=(1/n)[(dQ/dT)P-(dQ/dT)V] 4)

If the temperature change in
the denominator is the same for both the isobaric and isometric case, then eqn 4 becomes:

R=[(dQ)P-(dQ)V]/ndT 5)

Since the difference between the energy required for isobaric specific heat, and isometric specific heat, is the work done onto the atmosphere [W(atm)], therefore:

Watm = [(dQ)P-(dQ)V] 6)

Substituting eqn 6) into eqn 5), gives:

R= Watm/(ndT) 7)

Based upon our new perspective of work we wrote for work W=d(PV). Substituting back in we

R=d(PV)/(ndT) 8)

Obviously, eqn 8 could have been calculated by simply differentiating the ideal gas law, and then dividing both sides by: ndT.
Eqn 7) and 8) are a reiteration that the ideal gas constant is nothing more than a relation for a gaseous system’s ability to perform
work per degree temperature change, or, if you prefer, *PV* is simply the work that was required in forming the volume of gas at its
current pressure.

Again we must emphasize that work, and the total thermal energy of a gas, are two different things. Expressing
the total energy contained within a volume of gas

dEtotal= Einternal+3PV/2 9)

The total thermal energy change, then becomes:

dEtotal= dEinternal+3d(PV)/2 10)

For an ideal gas:dE=0 in which case becomes:

dEtotal= 3d(PV)/2 11)

Eqn 8) implies: The “__ideal gas constant ( R)”, is the ability of a mole
of ideal gas molecules, to perform work, per degree Kelvin__. And this relates to the total energy contained in a volume of ideal gas
within a closed system, because the ability of an ideal gas to perform work is 2/3 of that gas’s total kinetic energy.

The ideal gas
constant “*R*”, is equivalent to Boltzmann’s constant “*k*”, for a mole of gas molecules, i.e. *N*=1 mole. Therefore: “__Boltzmann’s
constant ( k)” is the mean ability of a solitary ideal gas molecule, to perform work, per degree Kelvin__. Again this relates to an ideal
molecule’s mean kinetic energy in a closed system by the same factor 2/3.