Connecting two vessels of equal volume, one with an ideal gas and other having vacuum

Question

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?

Answer 1

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.

Answer 2

I fully didnt understand what did you mean by your last line "After the △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 temperature increment with the same difference?"

But let me try.The free exchange of gas only happens as long as the pressure difference is greater than the valve's limit. As the temperature increases the pressure in the occupied chamber increases , while the other being empty has 0 pressure. When the difference in pressure becomes more than the valve's limit, gas start to leak from the first chamber to second, reducing the pressure difference. When the difference in pressure falls below the limit, the valve prevents any gas to move.So the low pressure zone stops receiving flow of gas, when its pressure becomes p2 = p1 -limit.

Answer 3

Initially suppose that chamber 1 has $n$ moles of gas and suppose that the volumes of both chambers is $V$. Then from ideal gas law we have,

$$P_1 V=nR T_1$$

Now both chambers are heated to a temperature of $T_2$ and then the valve let open. Now, gas from chamber 1 starts to flow into chamber 2, until the pressure difference between the chambers is $\Delta p$. Suppose that finally, chamber 1 has $n_1$ moles of the gas and chamber 2 has $n_2$ moles of the gas. Let the final pressures in chambers 1 and 2 be, $P'_1$ and $P'_2$. Then we have, $$p'_1V=n_1RT_2 \tag{2a}$$ $$p'_2V=n_2RT_2 \tag{2b}$$

Since the total mass of the gas is conserved we have, $$n_1 + n_2 = n \tag{3}$$

Also $$V=\frac{nRT_1}{P_1} \tag{3a}$$

From (2a),(2b),(3a) and (3) we obtain, $$(p'_1+p'_2)= \frac{T_2 P_1}{T_1} \tag{4}$$

Since the valve opens only when the pressure difference is $\Delta p$, we have, $$p'_1-p'_2=\Delta p \tag{5}$$ From (4) and (5) we obtain, $$p'_2=\frac{1}{2}\left(\frac{T_2P_1}{T_1} - \Delta p\right). \tag{Ans}$$ here it is assumed that valve remains open initially and finally