Thermal Physics question help

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Intramoon
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#1
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#1
hello, i would appreciate if anyone could help me with this question:
2 containers of volume V and 4V are connected by a capillary tube of negligible volume. both are initially at room temperature 293K and contain air of atmospheric pressure 100kPa. the larger container is placed in boiling water and is heated to 373K. calculate the final pressure of the gas in the two containers.

thank you!
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Callicious
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What have you tried?

The problem amounts to container A connected to container B, volumes VA and VB, with a temperature difference between them (and a pressure difference- see the heating), a tube connecting them, being left to come to an equilibrium.
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Elderfury843
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Use PV=Constant
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Lizbae
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pv1=pv2
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Intramoon
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i tried (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 and got an answer of 1.12x10^5 Pa but apparently the answer is 1.21x10^5 Pa

(Original post by Callicious)
What have you tried?

The problem amounts to container A connected to container B, volumes VA and VB, with a temperature difference between them (and a pressure difference- see the heating), a tube connecting them, being left to come to an equilibrium.
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Callicious
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(Original post by Intramoon)
i tried (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 and got an answer of 1.12x10^5 Pa but apparently the answer is 1.21x10^5 Pa
Apparently a fine way to do it- disregard the stuff in the spoiler (sorry- didn't make much sense to me conceptually since I attributed p_2,T_2 as the steady-state pressure/temperature in the distant future, rather than T_2 being the heated temperature)

Edit: See Eimmanuels answer. Two interpretations to the problem- this method here is for the alternative interpretation.

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<div><br></div><div class="spoiler">&lt;div&gt;Both volumes have the same temperature in the steady state at the end in the second scenario, too, if you let the clock flow to infinity and assume no losses from the combined system. This has to be considered.&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;If we look at your method of solving, it relies on conservation of number of the particles-&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;$pV = NkT$&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;$\frac{pV}{T} \propto N$&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;Your equation is&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;$(4Vp_1)/T_1 + (Vp_1)/T_1 \stackrel{?}{=} (4Vp_2)/T_2 + (Vp_2)/T_1$&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;You're trying to do something along the lines of&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;$N_{4V,initially} + N_{1V,initially} = N_{4V,final} + N_{1V,final}$&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;Now, you can see that we went from something fairly symmetric on the left &lt;em&gt;in your equation&lt;/em&gt;, to something asymmetric in the right- this implies that some particles have flowed (otherwise the RHS would look a bit more symmetric and wouldn't be mixing p2/T1.) In this case, does the form of the RHS make sense? The answer is no. &lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;Think it through logically. If particles have indeed flowed (as the asymmetry here implies) then the other volume would not be at the initial temperature- these particles will have transported heat to this volume (being at the higher temperature from the heated volume), and heated it up. &lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;As a result, this equation can't make any sense- the temperature denominator the second term on the RHS won't be the initial temperature in any scenario. This points to a problem in the method (not withstanding the fact that the denominator on the left term on the RHS won't be the final heated temperature, either.) &lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;Now, while the above equation might not work, you might be tempted to also try something along the lines of&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;$(4Vp_1)/T_1 + (Vp_1)/T_1 = (4Vp_2)/T_2 + (Vp_2)/T_2$&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;br&gt;&lt;/div&gt;&lt;div&gt;for some steady-state solution far in the future after temperature and pressure equilibrium has been reached. How do you do this, though? You see, you have two unknowns here: temperature and pressure. This method isn't going to work out for you, at least not just slapping two equations together and hoping for the best.&lt;br&gt;&lt;/div&gt;&lt;div&gt;&lt;/div&gt;</div><div><br></div><div>Personally, I would try to attack the problem by considering energy and how pressure relates to an energy density/equation of state.<br></div><div><br></div><div>What pV effectively calculates is the energy of the system, or something proportional to it. p is an energy density (convince yourself of this if you have to) that satisfies<br></div><div><br></div><div>$p = \frac{2}{3}U(n,T) <br></div><div><br></div><div>for number density and temperature. You can see where I'm going with this- a solution leveraging energetic equilibrium.<br></div><div><br></div><div>In the final steady-state, the energy density is uniform between the two volumes, alongside number density and temperature, at least for this equation of state/case. We're only heating one of the volumes in this case. We're using a capillary tube- this implies that over the period of the heating, virtually no flow occurs from the heated volume to the unheated one. Consequently, we can state that the increase in the energy density on the heated volume corresponds to, since <br></div><div><br></div><div>$U = \frac{3}{2}nk_b{T}$<br></div><div><br></div><div>is the energy density for the ideal gas, <br></div><div><br></div><div>$U_{4V, heated} = \frac{T_{heat}}{T_0}U_0$<br></div><div><br></div><div>where <br></div><div><br></div><div>$U_0$<br></div><div><br></div><div>is the initial energy density. <br></div><div><br></div><div>Now, multiply this by the volume of the 4V to get total energy of the 4V, and also note the total energy of the 1V as initially,<br></div><div><br></div><div>$K_{4V, heated} = 4V\frac{T_{heat}}{T_0}U_0$<br></div><div><br></div><div>$K_{1V, initially} = VU_0$<br></div><div><br></div><div>Their sum will be spread equally across both in the final steady state. If you solve through this, and note that <br></div><div><br></div><div>$K = UV$<br></div><div><br></div><div>...<br></div><div><br></div><div>$5VU_{final} = 4V\frac{T_{heat}}{T_0}U_0 + VU_0$<br></div><div>$U_{final} = \frac{4}{5}\frac{T_{heat}}{T_0}U_0 + \frac{1}{5}U_0$<br></div><div><br></div><div>We expect both sides to have the same number density in the end (no addition in particle count) and can thus find the equilibrium temperature<br></div><div><br></div><div>$T_{final} = \frac{4}{5}\frac{T_{heat}}{T_0}T_0+ \frac{1}{5}T_0$<br></div><div><br></div><div>Since pressure is proportional to temperature (see our equation of state above) we can expect<br></div><div><br></div><div>$p_{final} = \frac{T_{final}}{T_0}p_{initial}<br></div><div><br></div><div>which does give the correct answer (close enough, at least.)<br></div><div><br></div><div>That's just an alternative way of looking at the problem and the way I would have solved it- it's quite quick to do mentally once you get the concept. All of this is A2-doable method mind you- and fairly fast once you understand the concepts mentally. <br></div><div><br></div><div><strong><em>TLDR:</em></strong><br></div><div><em>Energy of the 4V increases by T_final/T_initial </em><br></div><div><em>A fifth of this "new energy" must go to the other side for equilibrium (so that both sides have same energy density hence pressure)</em><br></div><div><em>Number density unchanged through process by steady state </em><br></div><div><em>The "new energy" is proportional to the temperature difference of the 4V, i.e. T_heated - T_unheated, multiplied by 4V</em><br></div><div><em>A fifth of this energy goes to the other side, i.e. 4V/5*(dT)</em><br></div><div><em>Other side is of volume V, hence the temperature change of the other side is (4/5)dT = 0.8*(T_heated-T_unheated)</em><br></div><div><em>P propto T hence get new P via ratios</em><br></div><div>Very fast method if you understand all the science involved, although it only applies to this kind of situation (though can be adapted for other forms where number densities may be different, etc.</div>
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Intramoon
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#7
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#7
(Original post by Callicious)
Both volumes have the same temperature in the steady state at the end in the second scenario, too, if you let the clock flow to infinity and assume no losses from the combined system. This has to be considered.

If we look at your method of solving, it relies on conservation of number of the particles-

$pV = NkT$

$\frac{pV}{T} \propto N$

Your equation is

$(4Vp_1)/T_1 + (Vp_1)/T_1 \stackrel{?}{=} (4Vp_2)/T_2 + (Vp_2)/T_1$

You're trying to do something along the lines of

$N_{4V,initially} + N_{1V,initially} = N_{4V,final} + N_{1V,final}$

Now, you can see that we went from something fairly symmetric on the left in your equation, to something asymmetric in the right- this implies that some particles have flowed (otherwise the RHS would look a bit more symmetric and wouldn't be mixing p2/T1.) In this case, does the form of the RHS make sense? The answer is no.

Think it through logically. If particles have indeed flowed (as the asymmetry here implies) then the other volume would not be at the initial temperature- these particles will have transported heat to this volume (being at the higher temperature from the heated volume), and heated it up.

As a result, this equation can't make any sense- the temperature denominator the second term on the RHS won't be the initial temperature in any scenario. This points to a problem in the method (not withstanding the fact that the denominator on the left term on the RHS won't be the final heated temperature, either.)

Now, while the above equation might not work, you might be tempted to also try something along the lines of

$(4Vp_1)/T_1 + (Vp_1)/T_1 = (4Vp_2)/T_2 + (Vp_2)/T_2$

for some steady-state solution far in the future after temperature and pressure equilibrium has been reached. How do you do this, though? You see, you have two unknowns here: temperature and pressure. This method isn't going to work out for you, at least not just slapping two equations together and hoping for the best.

Personally, I would try to attack the problem by considering energy and how pressure relates to an energy density/equation of state.

What pV effectively calculates is the energy of the system, or something proportional to it. p is an energy density (convince yourself of this if you have to) that satisfies

$p = \frac{2}{3}U(n,T)

for number density and temperature. You can see where I'm going with this- a solution leveraging energetic equilibrium.

In the final steady-state, the energy density is uniform between the two volumes, alongside number density and temperature, at least for this equation of state/case. We're only heating one of the volumes in this case. We're using a capillary tube- this implies that over the period of the heating, virtually no flow occurs from the heated volume to the unheated one. Consequently, we can state that the increase in the energy density on the heated volume corresponds to, since

$U = \frac{3}{2}nk_b{T}$

is the energy density for the ideal gas,

$U_{4V, heated} = \frac{T_{heat}}{T_0}U_0$

where

$U_0$

is the initial energy density.

Now, multiply this by the volume of the 4V to get total energy of the 4V, and also note the total energy of the 1V as initially,

$K_{4V, heated} = 4V\frac{T_{heat}}{T_0}U_0$

$K_{1V, initially} = VU_0$

Their sum will be spread equally across both in the final steady state. If you solve through this, and note that

$K = UV$

...

$5VU_{final} = 4V\frac{T_{heat}}{T_0}U_0 + VU_0$
$U_{final} = \frac{4}{5}\frac{T_{heat}}{T_0}U_0 + \frac{1}{5}U_0$

We expect both sides to have the same number density in the end (no addition in particle count) and can thus find the equilibrium temperature

$T_{final} = \frac{4}{5}\frac{T_{heat}}{T_0}T_0+ \frac{1}{5}T_0$

Since pressure is proportional to temperature (see our equation of state above) we can expect

$p_{final} = \frac{T_{final}}{T_0}p_{initial}

which does give the correct answer (close enough, at least.)

That's just an alternative way of looking at the problem and the way I would have solved it- it's quite quick to do mentally once you get the concept. All of this is A2-doable method mind you- and fairly fast once you understand the concepts mentally.

Hate to sound like a rep-*** but for the love of christ give me some rep for this- this took way too long to write! I want "Clever Cogs" 5 gosh darn it! T_T
Thank you for taking the time to write such a detailed explanation. Using your Pfinal equation I got 1.27x10^5, which I also got with the (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T + (V*p2)/T2 equation you suggested, substituting T2 with 373K, is that the answer you calculated?
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Callicious
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(Original post by Intramoon)
Thank you for taking the time to write such a detailed explanation. Using your Pfinal equation I got 1.27x10^5, which I also got with the (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T + (V*p2)/T2 equation you suggested, substituting T2 with 373K, is that the answer you calculated?
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<div>Alright I'm on my PC now. I did not get that answer- I got closer to 1.21E5 Pa (which you have already stated is the answer, for anyone claiming I'm a hypocrite for posting answers :lol: )<br></div><div><br></div><div>I've demonstrated that using that equation (pv1/T1=pv2/T2) in a summation like you've done isn't going to work, at least not in the original equation you gave, and at least not in the way of just doing the above, since you don't have the final pressure or temperature until you work the problem (since this is the steady state solution you're after, at least in my derivations and what the numerical answer implies.) (edit- disregard this, see Eimmanuels) <br></div><div><br></div><div>I didn't suggest using that equation- if anything I advised against using it explicitly as it's provided.</div>
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Eimmanuel
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#9
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#9
(Original post by Intramoon)
i tried (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 and got an answer of 1.12x10^5 Pa but apparently the answer is 1.21x10^5 Pa
How do you actually compute using (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 to get 1.12x10^5 Pa?



(4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 is a correct way of doing this problem.

Check your calculation.
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Callicious
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(Original post by Eimmanuel)
How do you actually compute using (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 to get 1.12x10^5 Pa?



(4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 is a correct way of doing this problem.

Check your calculation.
I'll be damned it does work- I thought the OP was trying to put in p_2 as the pressure of the volume after it was heated, not the pressure at infinity (which is what they were trying to find) since they were mixing indices of post-heating as T_2 too :cry:

PRSOM

(Could I ask if you could explain it physically how that equation works, though- why does it work to just take the ideal gas = constant and sum them up, then mix and match p2/T1 in that way?)
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Eimmanuel
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(Original post by Callicious)
I'll be damned it does work- I thought the OP was trying to put in p_2 as the pressure of the volume after it was heated, not the pressure at infinity (which is what they were trying to find) since they were mixing indices of post-heating as T_2 too :cry:

PRSOM

(Could I ask if you could explain it physically how that equation works, though- why does it work to just take the ideal gas = constant and sum them up, then mix and match p2/T1 in that way?)
I am sorry for the late reply.
I believe you know that we can use the ideal gas law to determine the amount of gas in both containers. This is what we are doing here.

Initially, both containers are at the same pressure and temperature and we use the ideal gas law to find out the total amount of gas (number of moles) in the 2 containers.

 \text{total amt (moles)} = \dfrac{p_{1i}V_{1i}}{R T_{1i}} + \dfrac{p_{2i}V_{2i}}{R T_{2i}}

The subscript of 1 and 2 is to denote the different containers and the subscript of i is for the initial condition.

Then we are told the larger container is placed in the boiling water and is heated to 373K. As the gas is not leaking, the total amount of gas in both containers must be the same as before. However, the gas will redistribute themselves for both containers to be at the same pressure but at different temperatures.

 \text{total amt (moles)} = \dfrac{p_{1f}V_{1f}}{R T_{1f}} + \dfrac{p_{2f}V_{2f}}{R T_{2f}}
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Callicious
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(Original post by Eimmanuel)
I am sorry for the late reply.
I believe you know that we can use the ideal gas law to determine the amount of gas in both containers. This is what we are doing here.

Initially, both containers are at the same pressure and temperature and we use the ideal gas law to find out the total amount of gas (number of moles) in the 2 containers.

 \text{total amt (moles)} = \dfrac{p_{1i}V_{1i}}{R T_{1i}} + \dfrac{p_{2i}V_{2i}}{R T_{2i}}

The subscript of 1 and 2 is to denote the different containers and the subscript of i is for the initial condition.

Then we are told the larger container is placed in the boiling water and is heated to 373K. As the gas is not leaking, the total amount of gas in both containers must be the same as before. However, the gas will redistribute themselves for both containers to be at the same pressure but at different temperatures.

 \text{total amt (moles)} = \dfrac{p_{1f}V_{1f}}{R T_{1f}} + \dfrac{p_{2f}V_{2f}}{R T_{2f}}
Wouldn't the gas redistributing itself transfer heat between the containers and thus alter the temperature profile of the containers, though? (and since we're assuming equilibrium as t->inf, the number densities should level out, too, so for pressure equilibrium there should be a thermal equilibrium too?)
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Eimmanuel
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(Original post by Callicious)
Wouldn't the gas redistributing itself transfer heat between the containers and thus alter the temperature profile of the containers, though? (and since we're assuming equilibrium as t->inf, the number densities should level out, too, so for pressure equilibrium there should be a thermal equilibrium too?)
I agree with your deductions.
OP’s question does not seem to want us to look into such details. That is why I did not speculate beyond what it is asking.
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Callicious
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(Original post by Eimmanuel)
I agree with your deductions.
OP’s question does not seem to want us to look into such details. That is why I did not speculate beyond what it is asking.
Why did the equation work out in the end though? I mean... if we didn't speculate into that then sure but if someone speculated and guessed that, does the equation actually make sense? ;-;
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Eimmanuel
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(Original post by Callicious)
Why did the equation work out in the end though?
The equation works as what it supposes to work. Not really sure what you are expecting.

(Original post by Callicious)
… if someone speculated and guessed that, does the equation actually make sense? ;-;
Yes and no.
Yes, is that we are still using the ideal gas law equation while no, is that we need to modify to account for the same final temperature and same pressure.

IMO, the phrasing of the question can lead to different answers. The more common interpretation is based on having the 2 different containers maintained at 2 different temperatures.
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Intramoon
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#16
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#16
(Original post by Eimmanuel)
How do you actually compute using (4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 to get 1.12x10^5 Pa?



(4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1 is a correct way of doing this problem.

Check your calculation.
I re-did it and got to the correct answer:
(4V*p1)/T1 + (V*p1/)T1 = (4V*p2)/T2 + (V*p2)/T1
5V*P1/293 = (4V*p2)/373 + (V*p2)/293
1*10^5*5V/293 = (1/293 + 4/373)V*P2
The V’s cancel
1706 = 0.01414P2
P2 = 1706/0.01414 = 1.21*10^5 Pa
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