A New Thermodynamics

Blog: Isobaric vs Isometric Specific Heats and ideal gas constant

By Kent W. Mayhew


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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. This is also discussed in my blog Isobaric vs Isometric heating.  Similar conceptualization is discussed in my blog Latent heat of vaporization vs condenstaion.


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)   Also see my peer reviewed paper titled "Lost work and second law": My papers 


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. 

Ideal Gas Constant (R) Reconsidered 

Note: see my peer reviewed paper titled "New perspective for  kinetic theory and heat capacity": My papers

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)

Ideal Gas Constant, Boltzmann’s Constant, and Work

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 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.

In order to understand the 2/3 see (Work vs Energy)


     Copyright Kent W. Mayhew

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