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Concept of entropy.

  1. Nov 11, 2008 #1
    concept of "entropy."

    Hey, I am new to this concept of "entropy." And my book talks about it in terms of "the ways you can arrange something, more arrangements = more entropy."

    So, why is the molar entropy of Helium gas higher than that of Argon gas? There are the same number of particles. Shouldn't there be the same number of arrangements possible?

    Ok, I don't think I actually understand entropy, some help please? :P
  2. jcsd
  3. Nov 11, 2008 #2


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    Re: Entropy

    Hi psychedelia, welcome to PF. Can you cite your reference? The reference I'm using has a slightly higher entropy for argon, 59.8 J/mol-K vs. 54.4 J/mol-K.
  4. Nov 11, 2008 #3

    Andrew Mason

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    Re: Entropy

    The entropy of a gas at Standard Temp/Pressure (STP) is a theoretical number that is supposed to represent the integral of dQ/T for one mole of the substance in going from absolute 0 K to 273 K. Since dQ is the molar heat capacity multiplied by the change in temperature, the entropy of a gas at STP is:

    [tex]S = \int_0^{273} \frac{C_pdt}{T}[/tex]

    The problem is that Cp will change as temperature changes. For example, when the gas changes state from a solid to a liquid and from a liquid to a gas, one has to take into account the latent heats of fusion and vaporisation. These are different for each substance. There is not necessarily any relationship between atomic number and entropy.

  5. Nov 12, 2008 #4
    Re: Entropy

    Thanks for the replies. =)

    Yeah, sorry about that. Helium's S is higher than Argon's. Similarly, Fluorine's S is lower than Chlorine's.
    And S of
    Methane < Ethane < Propane
    Why is that?

    This kind of suggests that size plays a role in entropy. Probably because there are more number of ways energy could be arranged in a bigger molecule, within the molecule.
  6. Nov 20, 2008 #5
    Re: Entropy

    The entropy of an ideal mono atomic gas is given by the Sackur-Tetrode equation:

    S=N k\left\{\log\left[\frac{V}{N}\left(\frac{E}{N}\right)^{\frac{3}{2}}\right]}+\frac{5}{2}+\frac{3}{2}\log\left(\frac{4\pi m}{3 h^{2}}\right)\right\}

    This equation can be derived by counting the number of quantum states available for the gas molecules, precisely as the original poster suggested.

    Although Andrew Mason is also correct about entropy change being related to all the changes the substance undergoes when you het it from absolute zero the gas phase, it is wrong to suggest that once in the (dilute) gas phase, the entropy can only be computed from the knowledge of all these changes. In fact all these entropy changes will necessarily have to conspire in such a way to yield the Sackur-Tetrode formula.

    So, why has Argon a higher absolute entropy than Helium? This is because the energy levels of an argon atom in some volume are lower and closer spaced. This means that if you know that the system has some energy then there are more quantum states the system can be in.
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