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Enthalpy of an ideal gas

  1. Apr 25, 2014 #1
    I have read that enthalpy of an ideal gas is a function of temperature only, I am having a little trouble getting it.
    Consider a cylinder filled with an ideal gas so if we increase the pressure by pushing the piston down on it , a/c to ideal gas equation the temperature also rises, which in return must increase the enthalpy , so haven't we changed enthalpy by changing pressure?
  2. jcsd
  3. Apr 25, 2014 #2
    Enthalpy is a state function where only PV is allowed to do work. The only way to measure enthalpy is at constant pressure. H<0 exothermic, H>0 endothermic. Exothermic and endothermic is the heat exchange though the container to the outside environment. exothermic is heat absorbed from the outside environment, vise versa for endothermic.
    The only way to measure a change in enthalpy is at a specific thermodynamic state.
    a state function refers to a property of the system that depends only on its present state. This article will help

    http://memo.cgu.edu.tw/hsiu-po/chemistry/lecture 6.pdf
    Last edited: Apr 26, 2014
  4. Apr 25, 2014 #3


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    Enthalpy is H=E+PV. For an ideal gas, E is dependent on the number of particles and the temperature only. By application of the equipartition theorem and the kinetic theory of particles, one finds:


    Additionally, we know the ideal gas law:


    Giving us immediately:


    Which is dependent, if N is constant, on only the temperature.

    Your confusion arises from the fact that there are often too many thermodynamic variables, more than required to define the state of the system. This is why we have equations of state relating some of the variables to others.

    For an ideal gas, for example, we cannot independently change T, N, V, and P at the same time, and so T, N, V, and P are not independent variables.

    When we make statements such as "X is dependent only on the temperature", we usually mean we can determine X, if you only give us the temperature. Changes in other thermodynamic variables can of course change X, but the change will most readily manifest as a change in T.
  5. Apr 26, 2014 #4
    So , we can change any other variable as long as Temp. is constant the enthalpy wont change , like if we change pressure isothermally don't effect 'H' of an ideal gas. Am I right?
  6. Apr 26, 2014 #5
    not quite, lets try this definition

    Enthalpy is the amount of heat content used or released in a system at constant pressure. Enthalpy is usually expressed as the change in enthalpy. The change in enthalpy is related to a change in internal energy (U or E) and a change in the volume (V), which is multiplied by the constant pressure of the system.

    From the above definition you can see that pressure is constant. Isothermal heat exchange is a change in entropy, however a change in temperature will change the internal energy (E or U) and will result in a change in entropy. Now some examples of various types of entropy change

    When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal to the change in enthalpy. Enthalpy is a state function which depends entirely on the state functions T, P and U. Enthalpy is usually expressed as the change in enthalpy, energy is a state function, but work and heat are not state functions

    [tex] \Delta H=\Delta E+\Delta PV [/tex]

    If temperature and pressure remain constant through the process and the work is limited to pressure-volume work, then the enthalpy change is given by the equation:
    [tex] \Delta H=\Delta E+ P \Delta V [/tex]

    Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined by the equation:

    [tex]H=q [/tex]

    When the temperature increases, the amount of molecular interactions also increases. When the number of interactions increase, then the internal energy of the system rises. According to the first equation given, if the internal energy (U) increases then the change of enthalpy increases as temperature rises.

    Also you can have enthalpy of phase transitions, such as,

    1)enthalpy of vaporization/condensation
    2)enthalpy of fusion/freezing
    3)enthalpy of sublimation

    all three are at constant pressure, The enthalpy change of a reverse phase transition is the negative of the enthalpy change of the forward phase transition.

    edit:forgot to mention the heat exchange is either exothermic or endothermic depending on the direction of the heat exchange, see post 2
    Last edited: Apr 26, 2014
  7. Apr 26, 2014 #6
    I just thought of a curious question myself on enthalpy change. Is quantum tunneling a form of enthalpy change?
    Last edited: Apr 26, 2014
  8. Apr 26, 2014 #7
    The ideal gas equation does not predict that the temperature must rise. If we compress the gas, then the volume will decrease and the pressure will correspondingly rise. The temperature can be held constant by removing heat. Under the these conditions, the enthalpy is constant.

    So, if you increase the pressure holding the temperature constant, the enthalpy will not rise. So, enthalpy of an ideal gas is not a function of pressure.

  9. Sep 16, 2016 #8
    I am assuming what you are referring to, Chet is the case in the compression-condensation stage of a heat pump. I'm curious to know how the ideal gas laws remain valid in this situation? How can the process be isochoric and isobaric while still undergoing a temperature change?
  10. Sep 16, 2016 #9
    No. I was referring to a plain old ideal gas. Nothing more.
  11. Sep 16, 2016 #10
    How can you increase the temperature of the gas while the pressure and volume remain constant? Wouldn't that be violating the ideal gas laws?
  12. Sep 16, 2016 #11
  13. Oct 7, 2017 #12
    (∂H/∂v) at constant T for an ideal gas is zero correct? Because there is no temperature change, there will be no change in enthalpy?
  14. Oct 8, 2017 #13
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