Why does heat added to a system cause an increase in entropy that is independent of the amount of particles in the system?

Say we have two gas containers of $N_{2}$ at the same temperature of $300 ~\text{K}$, one containing $10^{23}$ particles and the other containing $10^{13}$ particles. If we add a quantity of heat to both containers, and the volume of these containers remains constant, then the rise in entropy is given by $d\mathtt{S} = \dfrac{dQ}{T}$. Why will both containers, with a different particle count, experience the same increase in entropy?

I tried to prove that the entropy increase is equal by working out some equations but couldn't follow through and also fell short of a conceptual explanation. My attempt at an equation:

Where volume is constant: $dQ = C_{v}\cdot dT$ and $C_{v} = \alpha N_k$, where $N$ is the number of particles in some container. Thus, $d\mathtt{S} = \frac{\alpha N_k\cdot dT}{T}$ and $\Delta \mathtt{S} = \alpha N_k \ln\left(\frac{T_{f}}{T_{i}}\right).$ My attempt was to show that two different $\Delta S$'s are really the same, assuming that these different processes have the same $T_{i}$ and $\alpha$. My attempt didn't produce anything meaningful.

Both perspectives, one from an equation and one conceptual, appreciated!

Why does heat added to a system cause an increase in entropy that is independent of the amount of particles in the system?

The systems won't end up with the same entropy. Your intuition is correct that the change in entropy depends on the number of particles. The reason why you can't just reason directly from $dS = \delta Q/T$ is that, first of all, the temperature is changing, and secondly, the temperature is changing at a different rate relative to the amount of energy added via heat, because the systems have different heat capacities $C_V$. That is, the same amount of heat added causes a different change in temperature, which makes $\Delta S$ different.

Calculation

We will assume an ideal gas for these calculations (or, at least, that the specific heat at constant volume is independent of temperature).

During an isochoric process, the change in entropy is given exactly by $$\Delta S = \int_i^f \frac{\delta Q}{T} = \int_{T_i}^{T_f} \frac{n\bar{c}_V}{T}dT = n\bar{c}_V\ln\left(\frac{T_f}{T_i}\right),$$ where $\bar{c}_V$ is the molar specific heat of the system. The final temperature is related to the initial temperature via $$Q = n\bar{c}_V(T_f - T_i),$$ in which case we can write $$\Delta S = n\bar{c}_V\ln\left(1 + \frac{Q}{n\bar{c}_VT_i}\right).$$ It should be pretty clear that this expression definitely depends on the value of $n$, and therefore the change in entropy will be different for the two cases.

As march pointed out, your reasoning is incorrect for finite $Q$ and $\Delta S$. However, the equation is true for differentials, so we have to address that.

• Stat mech: the increase in $S$ is related to how much extra disorder is produced. When there are less particles, they each get a bigger share of the $dQ$, so they each speed up more. This balances out the fact that there are less of them.
• Thermodynamics: the change in entropy should be independent of anything besides $T$. Heat transfer between two objects of the same temperature can go either way, so the resulting change of entropy has to be zero. If it weren't, we could violate the Second Law.
Technical answer: there is none, really. If you want to think microscopically (i.e. with stat mech), temperature is defined as $\partial S/\partial E = 1/T$ shortly after defining $S = k \log \Omega$. After some rearranging, this becomes $dS = dQ/T$.
If you want to think macroscopically (i.e. thermo), well, thermodynamics just starts with defining $dS = dQ/T$, basically for the reason given in the second conceptual answer.
• Thanks for the explanation. One aside: you wrote Heat transfer between two objects of the same temperature can go either way, so the resulting change of entropy has to be zero. Why is heat transfer a reversible process? If I place a hot cup in my fridge, there is no way for the heat dissipated to one day return? – Muno Nov 8 '15 at 15:06