# Resultant temperature of mixture

**Question: ** There are 2 flasks of equal volume, connected by a narrow tube of negligible volume, filled with nitrogen gas. When both are immersed in boiling water, the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water mixture, keeping the other in boiling water.

Problem: There are two ways that I am going to describe that I can think of to solve the question, but they give different answers.

Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry. As both the flasks are identical and at 100 and 0 so it should be 50. Calculating the answer with this assumption and the formula P1/V1=P2/V2 The answer comes out to be about 0.432 atm

Way 2: The sum of the moles of gas in both the flasks would remain same. And using the ideal gas equation to get n = number of moles = PV/RT After equating the values. The pressure come out to be about 0.422 atm

I was bothered with the 0.1 change. So I took the liberty to calculate the temperature from way 2, and it came out to be about 41.82 degree Celsius.

Note : calculations were done in Kelvin itself.

So we assumed that temperature would be 50 but way 2 contradicts that. I could not find a fault in either of the ways so could not understand what temperature would it really be.

Attempt to solve: is it something about Gibbs Paradox? I was Googleing and that came, but I couldn't understand...

• I see no reason to believe that the gas temperature in both flasks will be the same. – Ivan Neretin Jan 27 '16 at 12:30
• @IvanNeretin why not? Wouldn't it exchange the gas until it reaches equilibrium in temperature – Daksh Shah Jan 27 '16 at 12:32
• It would, if the flasks were not heated and cooled. The way we have it, though, there would be a constant stream of heat from boiling water into one flask, and from another flask to icy water. Would the effects of those be totally negated by the heat exchange between the flasks though the tube (narrow tube, mind you)? – Ivan Neretin Jan 27 '16 at 12:38
• @IvanNeretin ahh I am afraid I do not completely understand you. This is a high school question, and the reason that they say that the tube is narrow is I think needed to neglect the volume. But you might be right, so could you write more simple explanation? – Daksh Shah Jan 27 '16 at 12:41
• I think the tube is there just to equate the pressure; as for the temperature, it will be $0^\circ\rm C$ and $100^\circ\rm C$, respectively. Oh, and this has nothing to do with Gibbs paradox, BTW. – Ivan Neretin Jan 27 '16 at 12:45

In the first condition when both flasks are boiling, the temperature of the gases in both flasks are the same (100 °C). Using the ideal gas law on the system that comprises both flasks, where $n$ is the number of moles initially in each flask and $V$ is the volume of one flask.

$$P_{init} = \frac{(2 n) R\;T_{boil}}{2 V} = \mathrm{0.5~atm}$$

$$2n = \frac{P_{init}(2 V)}{R\;T_{boil}}$$

Now if Flask #2 is immersed in ice water, we should assume that the pressure in the system will be constant but that the flasks will have different temperature. For the pressure to be constant at different temperatures, material will have to flow from the hot flask to the cold flask; now, $n_1$ and $n_2$, the moles of gas in flask #1 and flask #2, will not be the same, but since no gas leaves the system comprising both flasks, then we know $n_1 + n_2 = 2n$

$$n_1 = \frac{P_{final}V}{R\;T_{boil}}$$ $$n_2 = \frac{P_{final}V}{R\;T_{ice}}$$ $$n_1 + n_2 = 2n$$

Substituting:

$$\frac{P_{final}V}{R\;T_{boil}} + \frac{P_{final}V}{R\;T_{ice}} = \frac{P_{init}(2 V)}{R\;T_{boil}}$$

$$P_{final}\frac{V}{R}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}V}{R\;T_{boil}}$$

$$P_{final}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}}{T_{boil}}$$

$$P_{final} = 2 P_{init}\frac{\frac{1}{T_{boil}}}{\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)} = 2 P_{init} \frac{\frac{1}{T_{boil}}}{\frac{T_{ice} + T_{boil}}{T_{ice}{T_{boil}}}}$$

$$P_{final} = 2 P_{init} \frac{T_{ice}}{T_{ice} + T_{boil}}$$

This answer makes sense in that (i) if the temperature $T_{ice}$ is actually the same as $T_{boil}$, then the final pressure would be the same as the initial pressure, and (ii) the answer doesn't depend on the volume or moles of gas in the flasks, just the temperatures and the initial pressure.

Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry.

This way is incorrect because a flask immersed in boiling water ($T=$100°C) is unlikely to have a temperature of 50 °C. Similarly for a flask immersed in ice.

Way 2: The sum of the moles of gas in both the flasks would remain same.

This is the correct method. It does require assuming that the rate of heat transfer from the boiling / ice baths to the flasks is much much greater than the rate of heat transfer between flasks via diffusion through the tube. However since the tube is small, this assumption is probably OK.

The gas law is applicable when both elements are at same temperature but it isnt the case here so your way 2 is right

• No, I am doing that for the system. And the temperature in the system. Is the same :/ – Daksh Shah Jan 27 '16 at 11:14