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Molar Specific Heat (gas) at varying pressure and volume?

  1. Dec 17, 2015 #1
    I've read in my texts that the there are two kinds of Molar Specific Heat Capacities for gases:
    1. Molar Specific Heat Capacity at constant Volume ----- ##C_v##
    2. Molar Specific Heat Capacity at constant Pressure ---- ##C_p##

    And in case of Constant temperature there is no point in introducing Specific Heat Capacities.

    ##Doubt 1: ## So, we have Molar Specific Heat Capacities for cases when a gas is at a constant pressure or volume. But what if both the 'Volume' and 'Pressure' are varying?

    Moreover my text says that, if a certain amount of energy is transferred to a gas the internal energy of a gas is same for any sort of process (isochoric, isobaric, isothermal, adiabatic). And it often uses the term ##nC_v\Delta T## for internal energy.

    According to the first law of thermodynamics ##\Delta U = Q-W## (##W## - Work done 'by' the system).
    So for an 'isochoric' process, as ##\Delta V = 0##, Work done is also zero (##W=0##).
    Therefore: ##\Delta U=Q-0=nC_v\Delta T##
    And in case of an 'isobaric' process it can be proved that ##\Delta U=Q-W=nC_p\Delta T - nR\Delta T=n\Delta T (C_p - R)=nC_v\Delta T##

    But let's consider an adiabatic process. In the derivation of ##PV^\gamma=constant## the text gives the following relation:

    ##nC_v\Delta T = \Delta U = Q-W=0-W=-pdV## and proceeds.
    ##Doubt 2: ## But I don't understand the relation ##\Delta U = nC_v\Delta T##. In an adiabatic process the only condition is that no heat is transferred ##from## or ##to## the system. Both the 'volume' and 'pressure' and even the 'temperature' can be varying.

    If I consider a gas in an adiabatic container, and if the gas expands, it does a positive work at the cost of the internal energy. This results in the reduction of the temperature of the gas. But how can the reduction of the temperature be related as ##\Delta U = nC_v\Delta T##?
     
  2. jcsd
  3. Dec 17, 2015 #2
    The subscripts v and p on the heat capacities refer to how the heat capacity is measured experimentally, not how it can thereafter be used in practice. For an ideal gas, the internal energy and enthalpy are observed to be functions only of temperature, irrespective of how the pressure or volume is varying. So, for an ideal gas ΔU is always equal to nCvΔT and ΔH is always equal to nCpΔT, irresepctive of how the gas got from its initial thermodynamic equilibrium state to its final thermodynamic equilibrium state.
    Yes. This is always the case for an ideal gas.

    In freshman physics, you were taught that ##Q = nC_v\Delta T##. But this is not quite the correct definition of heat capacity. In the cases you studied in freshman physics, there was no work W being done. So Q was equal to ΔU. But now that you are taking thermodynamics, you are learning that there is a more appropriate and more general definition of heat capacity that does not even involve Q. This definition is related directly to the internal energy and the enthalpy, rather than to the heat flow. For heat capacity to be a property of a material, it must be a state function, and, unfortunately, Q is not a state function (it depends on process path). The new, more general and appropriate definitions of the heat capacities that you are now learning in Thermodynamics are:
    $$nC_v=\left(\frac{\partial U}{\partial T}\right)_V$$
    $$nC_p=\left(\frac{\partial H}{\partial T}\right)_P$$
    In conjunction with the first law, these equations reduce to the freshman physics definition when W = 0.
    You need to get accustomed to these new more precise definitions of heat capacities because they are used throughout Thermodynamics.

    Chet
     
  4. Dec 20, 2015 #3
    Thanks for your reply sir!

    So far I understand that:
    (1) in an 'isochoric' process the heat supplied to the system (containing gas) is stored as the Internal Energy without any amount of work being done. So we define in such process that ##C_v=\frac{Q}{n\Delta T}##
    (2) in an 'isobaric' process the heat supplied is partly stored as the Internal Energy and partly used up for doing work to obey the ideal gas law ##pV=nRT##

    (3) in an 'isothermal' process the heat supplied is completely used up to do work so as to obey the ideal gas law ##pV=nRT=constant##

    But, what exactly is going on in case of an adiabatic process? Can you give me a physical picture by taking an example of a 'gas' in a closed container?

    And what about a thermodynamic closed system where the pressure, volume and temperature can vary and the system is not adiabatic (i.e heat can flow from or to the system)?
     
  5. Dec 20, 2015 #4
    For an isothermal process in a closed system, the head supplied is used to do work, and there is no change in the internal energy of the gas.
    The closed container is a cylinder with a piston. You compress the gas using the piston, but, in an adiabatic process, no heat enters or leaves, so the change in internal energy of the gas is equal in magnitude to the work.
    Are you familiar with the first law of thermodynamics? If so, please write the equation for the first law, and define the terms in the equation.

    Chet
     
  6. Dec 21, 2015 #5
    Alright! So according to the first law of thermodynamics (for an adiabatic process) : [tex]\Delta E=Q-W=0-W=-p(dV) [/tex]
    From Ideal Gas Law: [tex]pV=nRT[/tex] Differentiating both sides we have: [tex]p(dV)+V(dp)=nR(dT)[/tex] Therefore [tex]p(dV)=nR(dT)-V(dp)[/tex]
    Substituting the above result in the former equation we have: [tex]\Delta E=V(dp)-nR(dT)=V(dp)-nR\Delta T[/tex]
    Now for an adiabatic process the Internal Energy to be equal to ##nC_v\Delta T##, we should have [tex]V(dp)-nR\Delta T=nC_v\Delta T[/tex] or [tex]V(dp)=nR\Delta T+nC_v\Delta T[/tex] (i.e) [tex]V(dp)=n\Delta T(R+C_v)=n\Delta T(C_p)[/tex]

    So I reach at a result [tex]Vdp=nC_p \Delta T[/tex] To be frank, what I intent to ask is that; is there any alternate way to prove the above formula so that we can prove that for an adiabatic process ##\Delta E=nC_v \Delta T## (so that I can be 'mathematically satisfied' with the concept?). I know that it's quite mad to ask this. But is there 'again' any 'physical meaning' for ##Vdp=nC_p \Delta T## ?

    Sure.
    Equation of first law of thermodynamics : [tex]\Delta E=Q-W[/tex] (I am little unsure about writing ##\Delta H## (enthalpy) or ##Q##(heat transferred)).
    Where
    1. ##\Delta E## is the Internal Energy of the system
    2. ##Q## is the Heat transferred 'to'(positive) or 'from'(negative) the system.
    3. ##W## is the Work done 'to'(positive) or 'from'(negative) the system.
     
  7. Dec 21, 2015 #6
    In post #2, I gave you the precise definition of Cv that is used in thermodynamics, in terms of the partial derivative of the internal energy with respect to temperature at constant volume. For an ideal gas, where the internal energy is a function only of temperature, independent of pressure and volume, this partial derivative becomes an exact differential. For a material where the internal energy is a function only of T, the equation you wrote for dE is always true, as long as the heat capacity is not a function of temperature. If it is, then the equation becomes an integral over temperature.

    Yes. This equation can be interpreted physically. It is the equation for the change in enthalpy for an adiabatic reversible process. For an ideal gas in an adiabatic reversible process, ##dH=VdP##. But also, for an ideal gas in any process (since H is a function only of temperature), ##dH=nC_pdT##. So, combining these equations, we have ##dH=VdP=nC_pdT## for the change in enthalpy in an adiabatic reversible process.
     
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