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First law of thermodynamics at constant volume (ideal gas)

  1. Jul 28, 2015 #1
    If [itex]c_v[/itex] is the specific heat at constant volume, authors substitute this into the first law as follows:

    [itex]c_v d\theta + pdv = dq[/itex]

    How can one deduce that equation for any case? Since the specific heat at constant volume is used, the equation would be valid only where there is no expansion i.e. when dv=0?
     
    Last edited: Jul 28, 2015
  2. jcsd
  3. Jul 28, 2015 #2
    The specific heat at constant volume is defined as the amount of thermal energy that needs to be supplied (or removed) from unit mass of a substance whose volume is held constant during the process to raise (or bring down) its temperature by unit degree.In such case the thermal energy (or heat as commonly used) supplied or removed results in an increase or decrease in the internal energy of the substance alone.Thus, the amount of thermal energy supplied or removed from a substance at constant volume is a measure of the change in internal energy (a property) of the substance between any two states (in this case between any two temperatures).

    One need not to ensure that the volume of the system be kept constant to measure the change in internal energy during a process as internal energy is a property of the substance and the change in internal energy would be the same regardless of kind of process used to take the substance from an initial state to a final one (say from temperature T1 to T2).For example, lets consider unit mass of an ideal gas which is initially at a temperature 25°C.We might heat the gas in a rigid tank (at constant volume) so that its temperature rises to 26°C and measure the amount of heat or thermal energy supplied to do so.This would be the specific heat of the ideal gas at constant volume at that temperature which is 25°C.If we wish to heat the gas again to 27°C (a unit rise in temperature) the amount of heat that would be needed to do so would differ slightly than the previous attempt and this would happen as we continue to increase the temperature of the gas by unit degree.Thus, Cv of an ideal gas is a function of temperature as well.Now say we heat the same gas which was initially at 25°C this time at constant pressure(of 1atm) until its temperature rises to 26°C the amount of heat needed to do so would be the enthalpy of the gas at the initial temperature.The enthalpy values would again differ as we subsequently heat the gas in increments of unit degree rise in temperature because enthalpy for an ideal gas is again a function of temperature.But say if one requires to calculate the rise in internal energy of the gas as the gas was heated from 25°C to 26°C at constant pressure he doesn't need to calculate the change in internal energy by heating the gas again at constant volume rather the rise in internal energy is equal to Cv(T1-T2) or CvdT in differential form.Why?

    Because the change in internal energy for a substance between any two end states would be the same regardless of the nature of process which could be constant volume or constant pressure or poly-tropic.The only difference is during a constant volume process thermal energy supplied exactly equals the change in internal energy and for other processes it is not.In fact heating(or cooling) a substance at constant volume through a unit degree rise(or fall) in temperature and measuring the amount of thermal energy supplied(or removed) to do so is the way used to obtain the value for the specific heat at constant volume for a substance.A similar process is used to obtain the values for specific heat at constant pressure where heating is done under constant pressure of 1 atm and the amount of thermal energy supplied exactly equals a change in the enthalpy of the substance.
     
  4. Jul 28, 2015 #3
    Thanks, that makes sense. Fair enough, whether or not the volume is constant does not affect the change in internal energy.

    Given this, for any process (volume not constant), we can write the first law as above:

    [itex]c_v dT+pdV=dq[/itex]

    Given that the internal energy is a function of temperature alone, implying that the specific heat is also a function of temperature, which is true for an ideal gas. Otherwise, hypothetically if we did not have the internal energy as a function of temperature alone, e.g. u=u(T,v), then we cannot use that specific heat constant in the first law.

    We can still assume that c_v is independent of temperature to get the first law in the form of Boltzmann's law for entropy in statistical mechanics.
     
  5. Jul 28, 2015 #4
    This is a very restricted version of the first law, based on the following two assumptions:
    • The process is reversible
    • The system is an ideal gas
    We know that the equation is assuming that the process is reversible because it is written in terms of differentials and, more significantly, the differential work dW is expressed as pdv, where, presumably, p is the thermodynamic pressure of the system. For an irreversible process, the internal energy can be established confidently only in the initial and final equilibrium states of the system, which is described by ΔU and not dU.

    The system is assumed to be an ideal gas because, for an ideal gas, the internal energy U is a function only of temperature. For real gases (i.e., at pressures beyond the ideal gas limit), liquids, and solids, the internal energy is a function also of volume (or pressure).

    For more discussion of these points, please see my Physics Forums Insights article at: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/. This article includes a discussion of the first and second laws.

    Chet
     
  6. Jul 28, 2015 #5
    Thanks Chet, that makes sense about the restricted first law and reversible processes.

    I will post my understanding, please let me know of any flaws:

    For a reversible process, we can use two variables to define the state of the system (choosing temperature T and specific volume v in this case), then the specific energy:

    [itex]du = (\frac{\partial u}{\partial T})_v dT + (\frac{\partial u}{\partial v})_T dv[/itex]

    The restricted first law is (pressure is constant throughout the system as the process is reversible):

    du = dq - pdv

    Setting dv=0 for a constant volume process:

    du = dq

    Comparing the two equations:

    [itex]dq = du = (\frac{\partial u}{\partial T})_v dT[/itex]

    Then:

    [itex](dq/dT)_v = (du/dT)_v = (\frac{\partial u}{\partial T})_v[/itex]

    Since we know the specific energy depends on temperature alone for a reversible process:

    [itex]c_v = du/dT = dq/dT [/itex]

    Now, for any reversible process:

    du = dq - pdv

    Since the specific heat does not depend on volume, we can write for any reversible process:

    [itex]du = c_v dT - pdv[/itex]

    Recalling dq=du for a constant volume process, but the internal energy depends only on temperature, we can sub du=dq:

    [itex]dq = c_v dT - pdv[/itex]

    For an ideal gas: Pv=RT (R is the specific gas constant), we have:

    [itex]dq = c_v dT - \frac{RT}{v} dv[/itex]

    We can divide this by T, so that the entropy is an exact differential for an ideal gas:

    [itex](dq/T)_{rev} = \frac{c_v}{T} dT - \frac{R}{v} dv[/itex]
     
  7. Jul 28, 2015 #6
    Don't you mean "for an ideal gas" here? For a reversible process in general, specific internal energy depends on temperature and specific volume or pressure.

    Only for an ideal gas. The specific heat depends on volume for materials other than ideal gases.

    I'm going to cut off the discussion here to give you a chance to respond. Have you looked at the link I referred you to? Almost certainly not, or else you would not have posted what you did post.

    Chet
     
  8. Jul 28, 2015 #7
    I read the part in your link on the first law. Ok I see, I have confused the temperature dependence, which I missed in your link. I think I will just leave it at that to avoid more confusion, thank you for your time.
     
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