Let's say we have two vessels connected by a valve. Both the vessels have the same volume. Initially, there is vacuum in one vessel and the other is maintained at temperature $T_0= 300K$ and pressure $p_0=1 atm$. The condition on the valve is such that it opens only when the pressure difference $\triangle p \ge 1.10atm$ The finally both the vessels were heated to a temperature $T'=380K$ and I want to find the increment in the pressure in the vessel having volume.
Here it's clear that the number of moles initially in the first vessel at temperature $T_0$ and pressure $p_0$ are
$$n = {p_0V}/{RT_0}$$
The gas distributes in the two vessels. Lets say finally there are $n_1$ moles and pressure $p_1'$ in first vessel and $n_2$ moles and pressure $p_2'$ in the second vessel(that had vacuum initially), then we can write
$$n_1=p_1'V/n_1RT'$$
$$n_2=p_2'V/n_2RT'$$
Here one this that we can also use is $$n_1+n_2=n$$
$$ {p_1'V \over n_1RT'}+ {p_2'V \over n_2RT'} = {p_0V \over RT_0}$$
Now here comes my question.
IS THERE ANY RELATION BETWEEN THE FINAL AND INITIAL PRESSURE OF THE TWO VESSELS?
My solution manual says that $p_2'=p_1'- \triangle p$. Now I don't get the fact as to how can this be true. After the $\triangle p$ factor has been reached by the first vessel, the two vessels exchange the number of moles freely then how can this be possible that they show pressure increment with the same difference?
Best Answer
I will consider the process to be quasi-static: this assumption doesn't reflect what's really going on, but it is the only way to solve this problem; indeed, if it wasn't quasi-static, then there could be pressure inhomogeneity which would make the problem unsolvable without knowing the geometry of the vessels.
So, let's consider that at any given time, pressure is homogeneous in any of the two vessels. You know that there will be a gas exchange only if $\Delta p \geq 1.1$ atm. In other words, the process stops when $\Delta p < 1.1$ atm, or shortly after $\Delta p$ reaches $1.1$ atm. Then, you know that, by definition of $\Delta p$, $p_2' = p_1' - \Delta p$.
On the other hand, as you noticed, you know that $n_1 + n_2 = n$, so these two equations give you the value of pressures depending on the temperature (which appears because the gas is perfect).
Now, you can find a link between initial and final pressures.