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We know that entropy of a system only depends on heat energy (supplied to or extracted from system) not on any other work (expansion work + non expansion work) by mathmatical expression ∆S = dQrev / T.

Now suppose we a closed container (whose walls are adiabatically insulated), with a movable piston, is filled with an ideal gas. According to defination to entropy we can't increase or decrease entropy of this system as it's walls are thermally insulated so we can't supply heat. But if we compress piston , molecules gets closer and when we pull piston , there is more space available for molecules. Do these changes not affect the disorder or randomness of molecules ?

1.If yes then why entropy doesn't change ?

2.If no then why randomness of molecules is independent of work ?

And also why entropy depend on temperature, means if I supply same amount of heat to system at two different temperatures, why entropy change is different in both cases ?

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  • $\begingroup$ Note that S=f(T,p) // See also how-does-entropy-change-with-pressure // You should include in questions a short summary of what you already know or found (with eventual quotes/links), so others will not write what you are already aware of, or what you should know if you searched about it. If nothing was found, write at least what you have searched for. $\endgroup$
    – Poutnik
    Commented Aug 1, 2023 at 11:26
  • $\begingroup$ Note that more proper is $$\text{d}S_\mathrm{sys} = \text{d}S_ \mathrm{rev} + \text{d}S_ \mathrm{irrrev} = \frac{\delta Q_\mathrm{rev}}{T} + \text{d}S_ \mathrm{irrev}$$ $\endgroup$
    – Poutnik
    Commented Aug 1, 2023 at 11:33
  • $\begingroup$ Do you really think that the only way for the entropy of a system to change is by exchanging heat? $\endgroup$
    – Chet Miller
    Commented Aug 1, 2023 at 13:03
  • $\begingroup$ @ChetMiller no I also think we can change entropy of a reactant by converting into product. $\endgroup$
    – Govind Prajapat
    Commented Aug 2, 2023 at 0:59
  • $\begingroup$ What about changing entropy by doing irreversible work? $\endgroup$
    – Chet Miller
    Commented Aug 2, 2023 at 10:46

2 Answers 2

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Try not to think of entropy as 'randomness' but as the number of ways that energy levels in a molecule can be occupied or of the number of positions a molecule can have in space. (The 'spread' if you wish).

If the ideal gas expands adiabatically and reversibly, (infinitesimally slowly so equilibrium is always maintained) as you move the piston out then the entropy increases due to increasing the volume, i.e. more space available so more possible positions. However, the temperature falls because no heat can enter and work is done so entropy falls. These two effects exactly cancel so the entropy change is zero.

If the change is adiabatic and irreversible, i.e sudden change of piston, this balance no longer applies, the gas does less work and loses less internal energy and so has a higher final temperature than the reversible case, but the volume increase is the same. The entropy change is thus greater than zero.

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  • $\begingroup$ Could you mathmatically prove that in adiabatic irreversible expansion , net entropy change is positive ? $\endgroup$
    – Govind Prajapat
    Commented Aug 3, 2023 at 2:12
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    $\begingroup$ We know that it is always the case that the heat $q_{rev} > q_{irrev}$ and because the change is adiabatic then $q_{irrev}=0$ thus $q_{rev} > 0$ and as we use $\Delta S =q_{rev}/T$ to obtain the entropy change, it follows that $\Delta S >0$. $\endgroup$
    – porphyrin
    Commented Aug 3, 2023 at 12:46
  • $\begingroup$ but in adiabatic reversible process also , q will be equal to zero. $\endgroup$
    – Govind Prajapat
    Commented Aug 3, 2023 at 17:02
  • $\begingroup$ Yes quite so, but it is a different experiment. In any adiabatic process $q=0$, suppose when reversible we go from $V_1,T_1 \to V_2,T_2$ and the entropy increase due to expansion and decrease due to gas cooling exactly balance and $\Delta S=0$. In the irreversible case we have $V_1,T_1\to V_2,T_3$ where $T_3>T_2$ because less work is done and the cooling is less and the balance is lost and so $\Delta S>0$ as explained. $\endgroup$
    – porphyrin
    Commented Aug 4, 2023 at 7:55
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A similar question it was omited the fact that the temperature increases in the compression phase because the gas molecules change direction more frequently after an elastic collisions $T~\propto \langle v^2\rangle$, the motion is disordered $\langle v^2\rangle$ and trivially the entropy increases. It's the reason why the entropy depends on the temperature in this case.

The amount of heat before and after compression has changed the pressure has increased. At the expansion phase this pressure relaxes the piston. The irreversible comes from the fact that after an inelastic collision between the particles and the walls, a portion of the heat is absorbed $Q_{irr}<Q_{rev}$ by the walls, creating new velocity states and an irreversible entropy is created, the work for the expansion is less efficient.

Note than an ideal gas is at high temperture $T\gg$ otherwise even the collision between the particles will be inelastic and an irreversible entropy will be created.

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