# Why is the entropy of a Reservoir Q/T?

I am currently strugling as to why the entropy of a Reservoir, is defined as the heat transfered to said reservoir divided by the temperature of said reservoir. Such that:

$$\frac{Q}{T}=\Delta S$$

Since that the integration of the fundamental expression of Thermodinamics will give the change in entropy as:

$$\Delta S=1.5*n*\ln{\frac{T_f}{T_i}}+n*R*\ln{\frac{V_f}{V_i}}$$

Under the assumption that the gas in the reservoir is an ideal monoatomic gas.

In my understanding since the volume and the temperature of the reservoir remain constant, during the heat tranfer process, this would imply that the change in entropy would be zero. How does this correlate with

$$\frac{Q}{T}=\Delta S$$

I must be doing something wrong :(

I apolagize in advance, if there are any grammar or spelling errors. My english is getting rusty.

• A reservoir is often taken to have an infinite specific heat so its temperature remains constant as heat flows in or out of it. It may be that this is the case here. – John Rennie Feb 2 at 12:23
• See my answer in this thread: chemistry.stackexchange.com/questions/90852/… and this thread: chemistry.stackexchange.com/questions/42681/… – Chet Miller Feb 2 at 12:59
• Much obliged @ChetMiller, the second link was really helpfull and answered just what i needed, so it all comes to the fact that when $m_{reservoir}*c_{reservoir} \rightarrow \infty$ then $\Delta S \rightarrow \frac {Q}{T}$ – miguel shyshark Feb 2 at 13:41
• Yes, that’s exactly right. – Chet Miller Feb 2 at 14:31

The temperature and volume cannot remain constant.

When you add heat to the gas, it means you are adding energy. This energy has to go somewhere: either it stays in the gas as internal energy (for an ideal gas, this always means a change in $$T$$) or it leaves the gas as work done on another system (a change in $$V$$). This is the first law $$\delta Q = dU + \delta W = \frac{3}{2} n R dT + p dV;$$ if $$\delta Q$$ is nonzero then you can't have both $$dT$$ and $$dV$$ being zero.

Indeed, if you divide the above equation by $$T$$ and use the ideal gas law, you get exactly $$\frac{\delta Q}{T} = \frac{3}{2} n R \frac{dT}{T} + nR \frac{dV}{V}$$ which integrates to the equation you have.

Edit: a reservoir made out of a huge amount $$n$$ of ideal gas will have a huge heat capacity $$C_V = \frac{3}{2} nR$$. Thus, for any small amount of heat $$Q$$ transferred to/from a smaller system, the change in the reservoir's temperature $$T$$ will be minuscule: $$\Delta T = \frac{Q}{C_V}$$ (assuming constant volume). So the expression for entropy change $$\Delta S = C_V \ln\left(\frac{T + \Delta T}{T}\right) \approx C_V \frac{\Delta T}{T}$$ becomes something huge times something tiny. In the limit $$n \to \infty$$ it becomes $$\infty \cdot 0$$ and can thus remains finite and nonzero.

You are right that entropy normally is a state function $$S(T, V)$$. But when $$V$$ is infinite, that notion is perhaps not so well defined.

• I understand your answer perfectly, but there is just one little detail. Wouldn't the infinite specif heat of a reservoir, mentioned by @John Rennie, make a change in temperature of no consequence? If we take the atmosphere for example, it does not change volume or temperature, if someone on the surface decided to make a campfire. I mean significantly – miguel shyshark Feb 2 at 12:46
• @miguelshyshark You are right, this becomes a bit ill-defined in the infinite limit, see my edit. Just imagine a reservoir that is very large but not infinite and then take the limit, and you will see what to do. – Elias Riedel Gårding Feb 2 at 13:33
• Your information in the edited section, was very helpfull in understanding the struggle associated with the Logic I was using. I deeply apreciate it! – miguel shyshark Feb 2 at 15:16
• Whether or not a heat source/sink can be considered a thermal reservoir is a relative matter. The smaller the amount of heat transferred the lower the heat capacity of the source/sink has to be so that its temperature does not measurably change. – Bob D Feb 2 at 17:52