# Pressure of two flasks connected to each other at different temperatures [closed]

Two flasks of equal volume connected by a narrow tube (of negligible volume) are at $$\pu{27 °C}$$ and contain $$\pu{0.7 mol}$$ of $$\ce{H2}$$ at $$\pu{0.5 atm}$$. If one of the flasks is then immersed into a bath kept at $$\pu{127 °C}$$ while the other remains at $$\pu{27 °C}$$. What is the final pressure?

Will the pressure of the two flasks which are now kept at different temperatures be the same or will they have different pressures?

In the solution I found of this question, the pressure of the two flasks at different temperatures are taken to be equal. Can someone please explain why this is so?

• Because they are connected. You don't have the two gases equally divided per amount of particles. May 13 at 14:51
• The gist here is that in order for each flask to be at equilibrium then there must not be any net flow of gas through the tube which connects the two flasks. Thus the whole system must be at constant pressure. That means that there must be a 100 degree C temperature gradient along the tube. Thus the gas is denser at the 27 C end than at the 127 C end. But since the tube was deemed to be "of negligible volume" you can ignore the gas in the tube when calculating the solution.
– MaxW
May 13 at 18:40

## 1 Answer

The easiest way to prove the equality in pressure is by taking a cross-section of the narrow tube and finding the forces acting on it. By extending this, we can prove that the pressures are equal.

Take a cross-section area of the narrow tube. We assume this is at steady state and there is no average motion of particles in the entire system.

Now, since there is no net average motion, this means that there is no pressure gradient across any cross-section. Expanding this cross-section into the tube, we see that the ends of the tube have to be at the same pressure since there is no net average motion of the gas within the two flasks at steady state (which is where the parameters are asked to be measured).

Therefore $$p_\mathrm{left} = p_\mathrm{right}$$