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: 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.
