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Is entropy change of the system for reversible process same as entropy change of the system for irreversible change? Is Q for reversible change equal to Q for irreversible change? Or the change in entropy of the surrounding is same in any case because surrounding is bigger than system, so the change in surrounding is small? And finally, does internal energy of the system depend on entropy of the system and how?

I registered just so I could ask this question. I would be very grateful for answer.

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  • $\begingroup$ There are a lot of questions here. Can you narrow it down a bit? $\endgroup$ – Zhe Oct 16 '16 at 18:33
  • $\begingroup$ Supposing the heat is transferred from surroundings to the system irreversibly, if we want to calculate entropy change of the system we have to find reversible path for the change, so that means that change in entropy for the reversible path found and previous irreversible is the same. But shouldn't for spontaneous irreversible process entropy change be greater than for reversible path? $\endgroup$ – NeneS Oct 16 '16 at 18:46
  • $\begingroup$ For the same initial and final states of the system, the entropy change doesn't depend on whether it is a reversible path or an irreverisible path. However, for an irreversible path, the integral of dq/T (where T is the temperature of the surface through which the heat transfer is occurring) will be less than for the reversible path (because entropy is being generated within the system to make up the difference). $\endgroup$ – Chet Miller Oct 16 '16 at 21:29
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Being a state function, entropy is independent of path, so the entropy change of the system is indeed the same for both a reversible and an irreversible process. The transferred heat, however, is not, and does depend on process.

You've probably heard mentioned that the entropy change through a reversible process is less than that through an irreversible process. At first glance this proposition seems suspicious, because we've just said that as a state function entropy is independent of path$-$the resolution is that the entropy change of the system remains the same regardless of process, but the entropy change of the environment is indeed different, such that the total entropy change is greater for an irreversible process than for a reversible process. [Interpreted differently: the environment is in a different state after the system undergoes a reversible process than after it undergoes an irreversible process.]

The internal energy does depend on the system's entropy$-$by the first law we may write $$\mathrm{d}U = T\,\mathrm{d}S + \mathbf{f\cdot\mathrm{d}X}$$ (where $\mathbf{f\cdot\mathrm{d}X}$ represents a generalized work term; commonly this is $-p\,\mathrm{d}V$), so we see that $U$ is a natural function of $S$ and $\mathbf{X}$; $U = U(S,\mathbf{X})$. To give a functional form for $U$ as a function of $S$ and $\mathbf{X}$, we'll need in general an equation of state, an equation for energy, and a type of process$-$let's restrict ourselves to the case of an ideal gas, for which the two equations are $$pV = nRT \quad \text{and} \quad U = \frac{3}{2}nRT$$ (assuming also a monatomic gas), and work out how $U$ varies over an isobaric process. We want to evaluate $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$; let's start by working out $-p\,\mathrm{d}V$: $$-p\,\mathrm{d}V = -p\,\mathrm{d}\left(\frac{nRT}{p}\right) = -nR\,\mathrm{d}T = -\frac{2}{3}\,\mathrm{d}U.$$ Now plug that into our starting equation, substitute, and rearrange... $$\frac{5}{3}\,\mathrm{d}U = T\,\mathrm{d}S = \frac{2U}{3nR}\,\mathrm{d}S \quad \Longleftrightarrow \quad \frac{5}{2}nR\,\frac{\mathrm{d}U}{U} = \mathrm{d}S,$$ which integrates to $$\Delta S = \frac{5}{2}nR\ln\left(\frac{U_2}{U_1}\right) = \frac{5}{2}nR\ln\left(\frac{T_2}{T_1}\right).$$ This is no different from working out $q_\text{rev}$ for such a process and then using $\Delta S = q_\text{rev}/T$; the role of energy is simply more explicit in this approach.

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