Ideal gas law: Will the piston move at all? We have the following experimental setup:

Before the experiment starts: $$p_1=p_2; \space V_1=V_2; T_1=T_2+\Delta T$$
The experiment starts and both the containers are heated so that the temperature difference $\Delta T$ remains constant (Edit: I wanted this to mean that there is no physical deformation, expansion or contraction of the containers. The piston is still allowed to move). The Volume of the containers also stays constant. Does the piston move to the left or to the right?
We have been discussing this question for an hour in our study group and we haven't really come to a conclusion. There are basically two hypothesis:


*

*Since the piston didn't move before the containers were heated, the piston won't move after the containers are heated because the tempereature will be increased by the same amount in both containers.

*Applying the ideal gas law $$pV=Nk_bT$$
and taking into consideration the fact that before the experiment started, both the containers had the same volume ($V_1=V_2$)and pressure ($p_1=p_2$) but different temperatures, it follows that there must be more particles/molecules in container 2 to "compensate" the higher temperature in the first one. Heating both containers by the same amount (equal temperature increase) implies more energy supplied to container 2 which will cause the piston to move left.


Can you give us a hint if we are going in the right direction with any of these hypothesis?
 A: Your second thesis is right. At least the result. The explanation not, as the example in the answer of Diracology might illustrate. 
Assume, that the piston does not move (we fix it). So $V$ didn't change on either side. Obviuosly, $n$ didn't change on either side. Therefore, $p$ is proportional to $T$ (in each side separately!).
So if $T$ rises by the same factor, the pressures will also remain equal. If $T$ rises by the same amount, the pressure will change. That's wrong in the fist argument.
If the pressure does not remain equal, the piston will move after release. $^1$  

The amount of temperature does not matter here, the ratio does.
To put it more general, and to explain why you felt, that the first argument might hold:
There is a linear dependence $p(T)$ on either sides. The slope of this linear function is different, but the offset is zero. So the same ratio in $T$ will give the same ratio in $p$. ... Well if on the contrary the slope would be equal and the offset were different, than the same difference would give the same difference.
I hope that explains the "psychology" :)

$^1$ - You could also do the argument with equal pressure on both sides, as the other answers do. It's mathematically the same, since $p$ and $V$ are symmetric in the ideal gas law. But I found it more intuitive like this, probably because in your picture the piston would leave the tube in the middle if it moves too much :)
A: Let $T_{20}$ be the initial temperature of tank 2 and $T_{10}=T_{20}+\Delta T$ be the initial temperature in tank 1.  Let $\delta T$ be the equal rise in the temperature of both thanks.  Assuming that the piston does not move, we would have $$p_{2f}=p_2\frac{T_{20}+\delta T}{T_{20}}$$and$$p_{1f}=p_1\frac{T_{20}+\Delta T+\delta T}{T_{20}+\Delta T}$$
Since $p_1=p_2$, if we divide one equation by the other, we obtain:
$$\frac{p_{2f}}{p_{1f}}=\frac{(T_{20}+\delta T)(T_{20}+\Delta T)}{T_{20}(T_{20}+\Delta T+\delta T)}=1+\frac{(\Delta T)( \delta T)}{T_{20}^2+T_{20}(\Delta T+\delta T)}$$
So, if the piston doesn't move, the final pressure in chamber 2 will be higher than in chamber 1.  The piston must move in the direction from chamber 2 to chamber 1.
A: $$P_1V_1=m_1RT_1$$
$$P_2V_2=m_2RT_2$$
$P_1=P_2$,$V_1=V_2$,$T_1=T_2+\Delta T$
Then, we have:$$\frac{m_2}{m_1}=\frac{T_1}{T_1-\Delta T}$$
After heating:
$$P'_1V'_1=m_1RT'_1$$
$$P'_2V'_2=m_2RT'_2$$
$P'_1=P'_2$,$T'_1=T'_2+\Delta T$
Then, we have:$$V'_2=V'_1\left(\frac{m_2(T'_1-\Delta T)}{m_1T'_1}\right)=V'_1\left(\frac{T_1}{T_1-\Delta T}\right)\left(\frac{T'_1-\Delta T}{T'_1}\right)$$
$$T'_1>T_1$$
Thus:$$V'_2>V'_1$$
And the piston moves to the left.
A: Rigorous calculation
The first hypothesis is hand-wavy Aristotelian logic.  Whatever does happen has to be explainable by the Ideal Gas Law.
At all times, $p_1 = p_2$ must hold, otherwise, the piston would move to equalize the pressures.  So,
$$\begin{eqnarray}
p_1 =& p_2 \\
\frac{N_1 k_b T_1}{V_1} =& \frac{N_2 k_b T_2}{V_2} \\
\frac{V_2}{V_1} =& \frac{N_2}{N_1} \frac{T_2}{T_1}
\end{eqnarray}$$
Since no gas is being added or removed, $\dfrac{N_2}{N_1}$ remains constant, so
$$\frac{V_2}{V_1} \propto \frac{T_2}{T_2 + \Delta T}$$
If both sides were initially at the same temperature ($\Delta T = 0$), then of course the piston won't move as you apply the same temperature increase to both sides.
If the left side was initially hotter ($\Delta T > 0$), then $V_1$ decreases as the temperatures increase, and the piston moves left.
If the right side was initially hotter ($\Delta T < 0$), then $V_1$ increases as the temperatures increase, and the piston moves right.
Intuitive interpretation of the calculation
How hard each side pushes is based on the number of particles multiplied by the momentum of each particle strike (temperature).
Assuming that $\Delta T > 0$, that means that the piston is initially in  equilibrium mostly because the fewer particles on the left are each striking harder.  As both sides are subjected to the same temperature rise, the constant $\Delta T$ becomes less and less significant, so the left side loses out to the right side.
A: The piston moves to left if $P_2\gt P_1$ and to right if $P_2\lt P_1$.
In above answers, Lucas's beginning part is fine. 
$$P_1V_1=m_1RT_1$$ $$P_2V_2=m_2RT_2$$ with $P_1=P_2$,$V_1=V_2$,$T_2=T_1-ΔT$
We then have,
$$\frac{m_2}{m_1}=\frac{T_1}{T1-ΔT}$$
When temperature on both side is raised by $\delta T$
$$P′_1V′_1=m_1RT′_1$$
$$P′_2V′_2=m_2RT′_2$$
with $V'_1=V'_2$ and $T_2+\delta T=T_1+\delta T-ΔT$
The pressure ratio $$\frac{P'_1}{P'_2}=\frac{m_1}{m_2}\times \frac{T'_1}{T'_2}=\frac{T1-ΔT}{T_1}\times \frac{T'_1}{T'_2}$$  $$\frac{P'_1}{P'_2}=\frac{T1-ΔT}{T_1}\times \frac{T_1+\delta T}{T_1+\delta T-ΔT}$$
$$\frac{P'_1}{P'_2}=(1-\frac{ΔT}{T_1})\times \frac{1}{1-\frac{ΔT}{T_1+\delta T}}$$
Therefore, if one increase $\delta T$, the ratio $\frac{P'_1}{P'_2}$ will decrease and the piston will tend to move towards left. 
A: I think the idea is basic. In the experiment you say that "The Volume of the containers also stays constant." Which the tubes in the diagram are part of the containers, the piston can not move in order to keep the volume constant.
Now if the piston could move, then the volume would change, considering the Ideal Gas Law, the piston would move to the left and allow section on the right to expand. Pressure will increase too in booth containers to account for the increase in temperature.
A: The piston cannot move if the pressures are maintained equal (assuming same cross section are at both extremes). Otherwise you would violate the second Newton's law. The fact that the second container has more energy does not imply the piston should move. Imagine two containers filled with the same gas linked by a piston. Assume same pressure and same temperature but consider the second container larger than the first one. It would have again more particles hence more energy. However the piston would remain static.
A: I actually did some of the algebra work and using the equations given, I managed to show that the pressure from 1 is less than 2. The volume being held constant is important here. Just to note I'm purely using algebra here, so please check through my working and see if it makes sense to you. 
For notation sake, I've set anything with b as before the tube is heated, and a with anything after the tube is heated
$$\frac{P_{b1}V_{b1}}{T_{b1}}=N_1k,  \frac{P_{b2}V_{b2}}{T_{b2}}=N_2k $$
Rearranging for both $$P_{b1},P_{b2}$$
We find that
$${P_{b1}=\frac{N_1kT_{b1}}{V_{b1}}}, {P_{b2}=\frac{N_2kT_{b2}}{V_{b2}}} $$
Since $$T_{b1}=T_{b2}+ΔT$$
$${P_{b1}=\frac{N_1kT_{b1}}{V_{b1}}}, {P_{b2}=\frac{N_2k(T_{b1}-ΔT)}{V_{b2}}} $$
And we know both pressures are the same at the moment, we are equate the pressures, giving us the result:
$$ ΔT = \frac{(N_2-N_1)T_{b1}}{N_2} $$
Now looking at the eqns after heating has taken place. And going through the same process again, we find that
$$\frac{P_{a1}V_{a1}}{T_{a1}}=N_1k,  \frac{P_{a2}V_{a2}}{T_{a2}}=N_2k $$
Let us assume that the rise in temperature is the same $$ΔT$$ (the difference between Tb1 and Tb2).
Rearranging the equations again for pressure,
$${P_{a1}=\frac{N_1kT_{a1}}{V_{a1}}}, {P_{a2}=\frac{N_2kT_{a2}}{V_{a2}}} $$
We can rewrite them as: 
$${P_{a1}=\frac{N_1k(T_{b1}+ΔT)}{V_{a1}}}, {P_{a2}=\frac{N_2k(T_{b2}+ΔT)}{V_{a2}}} $$
Since we are assuming the volume is constant as mentioned in the question,
$${P_{a1}=\frac{N_1k(T_{b1}+ΔT)}{V}}, {P_{a2}=\frac{N_2k(T_{b1})}{V}} $$
Substituting ΔT:
$${P_{a1}=\frac{(2-\frac{N_1}{N_2})(N_1)kT_{b1}}{V}}, {P_{a2}=\frac{(N_2)kT_{b1}}{V}} $$
$${\frac{P_{a1}}{P_{a2}}=(2-\frac{N_1}{N_2})(\frac{N_1}{N_2})}$$
Corrected the algebra, Pa1 should be less than Pa2, and it should move to the left. You will notice that the expression will always tend to a value that is less than 1 as long as N1 is less than N2, which should be the case as mentioned in the original post.
