# Can the change in the entropy of the surroundings always be obtained by dividing heat transferred by the temperature at which the transfer occurs?

Consider $\pu{1 mol}$ of an ideal monoatomic gas going through reversible isochoric heating from $\pu{100 K}$ to $\pu{1000 K}$.

Calculate $\Delta S_\pu{sys}, \Delta S_\pu{surr}.$

$$\Delta S_\pu{sys} = nC_v \int_{\pu{100 K}}^{\pu{1000 K}} \frac{dT}{T} = \frac{3}{2}R\ln10$$ Now, as I have $q_\pu{sys} = 1350R$. I can put $\displaystyle \Delta S_\pu{surr} = \frac{-q_\pu{sys}}{T_\pu{surr}}$ to get $$\Delta S_\pu{surr} = -\frac{27}{20}R$$ but in a reversible process $\Delta S_\pu{total} = 0$ but that gives me $$\Delta S_\pu{surr} = -\frac{3}{2}R\ln10$$

Where am I wrong? Here, $\displaystyle q_\pu{sur} =$ heat supplied to surroundings.

The surroundings that are assumed in the question are not the same surroundings as in the image. The surroundings in the image are essentially at a constant temperature while it is assumed in the question that the surroundings are heating the gas reversibly. The gas is heated by some heat source which is in thermal equilibrium with the gas.

Let us assume that the surroundings which are not at a constant temperature are heating the gas. We assume that our universe consists of only gas plus the surroundings. The surroundings have some internal source of energy distributed uniformly so that no local hot spots occur. Now,

$$\Delta S_\pu{gas} = n C_V \int_{\pu{100 K}}^{\pu{1000 K}} \frac{\operatorname{d}T}{T} = \frac{3}{2} R\ln10 \tag1\label{(1)}$$

And for a reversible change, $\Delta S_\pu{total} = 0$ and thus $\Delta S_\pu{surr} = - \Delta S_\pu{gas}$.

We can also verify this in the following way:

$\pu{d}q_\pu{gas} = n C_{V} \cdot \operatorname{d}T$ and from the First law of thermodynamics, $$\pu{d}q_\pu{surr} = - n C_{V} \cdot \operatorname{d}T = - \pu{d}q_\pu{gas} \tag{2}\label{(2)}$$ Thus, $$\Delta S_\pu{surr} = \int^{\pu{1000 K}}_{\pu{100 K}} \frac{\pu{d} q_\pu{surr}}{T} = - n C_V \int_{\pu{100 K}}^{\pu{1000 K}} \frac{\operatorname{d}T}{T} = - \frac{3}{2} R\ln10 \tag{3}\label{(3)}$$

I disagree with the explanation given in the text. The entropy change of the surroundings is given by the equation presented in your title only for an ideal constant temperature reservoir. Such an ideal reservoir is characterized by (a) an infinite capacity to absorb heat without its temperature changing and (b) infinite thermal conductivity, so that temperature is uniform throughout the reservoir, including at the interface with the system (where all the heat transfer occurs). So all the heat transfer between the system and the reservoir occurs at the reservoir temperature $T_{Res}$. In this way, all the irreversibility of the process occurs within the system, irrespective of whether the process is reversible, and the entropy change of the system always satisfies $ΔS≥Q/T_{Res}$.

None of the entropy generation for the process occurs within the ideal reservoir, and the entropy change of the ideal reservoir is thus always $-Q_{sys}/T_{Res}$, irrespective of whether the process is reversible. I should also mention that, although often not stated, the Clausius inequality is supposed to be expressed in terms of the temperature at the interface with the surroundings $T_{Res}$ where all the heat transfer is occurring.

Typically, all the surroundings reservoirs you will encounter in most thermodynamics texts will be ideal constant temperature reservoirs.

The description in your post is inappropriate because, if the reservoir has a constant volume (i.e., constant mass), its temperature can change between the beginning and end of the process. Its entropy change will not be given then by the equation presented. In addition, the description omits the requirement that the reservoir most have infinite thermal conductivity (so that the temperature in the reservoir is spatially uniform).

• If I assume it to be constant temperature reservoir and use that equation then also there is a discrepancy in the two values of $\Delta S_{surr}$. The equation seems to hold only for irreversible processes. – Sachin Feb 18 '18 at 6:17
• Oh. I forgot to mention the other key characteristic of an ideal reservoir: the fluid is assumed to have infinite thermal conductivity, so that the temperature is uniform throughout the reservoir, including at the interface with the system (where all the heat transfer occurs). So all the heat transfer between the system and the reservoir occurs at the reservoir temperature $T_R$. In this way, all the irreversibility of the process occurs within the system, irrespective of whether the process is reversible or not, and the entropy change of the system always satisfies $\Delta S\geq Q/T_{R}$. – Chet Miller Feb 18 '18 at 11:48
• None of the entropy generation for the process occurs within the ideal reservoir, and the entropy change of the ideal reservoir is always $-Q_{sys}/T_R$, irrespective of whether the process is reversible or not. I should also mention that, although often not stated, the Clausius inequality is always supposed to be expressed in terms of the temperature at the interface with the surroundings $T_R$ where all the heat transfer is occurring. – Chet Miller Feb 18 '18 at 11:57
• @ChesterMiller I think you should add those comments to the main answer. They're important. – Apoorv Potnis Feb 18 '18 at 12:01
• @ApoorvPotnis OK. I've done that. I like the answer much more now. – Chet Miller Feb 18 '18 at 12:22