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Very hard ideal gas law problem!

  • Thread starter NasuSama
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1. Homework Statement

A cylinder of height H = 1.2 m high rests on a table. The bottom half of the cylinder is filled with an ideal gas; on top of that is a frictionless, movable disk of negligible mass and thickness. Above the disk, the top half of the cylinder is completely filled with liquid mercury, and the top is open to the air. Initially, the gas is at temperature [itex]T_{1}[/itex] = 319 K; air pressure is [itex]P_{0}[/itex] = 1.013 x 10^5 Pa, and the density of mercury is [itex]\rho_{Hg}[/itex] = 13600 kg/m3. Now the temperature of the gas is raised until one-half of the mercury spills out of the cylinder. Assume you can ignore the expansion of the mercury and the cylinder due to change in temperature, and evaporation of mercury is negligible. Find [itex]T_{2}[/itex], the new temperature of the system.

2. Homework Equations

→ Ideal Gas Law is the start.
→ [itex]P_{1} = P_{0} + \rho hg[/itex]

3. The Attempt at a Solution

I answered the question incorrectly by following this method..

[itex](P_{0} + \rho_{Hg}gh/2) * V_{1}/T_{1} = (P_{0} + \rho_{Hg}gh * 3/4) * V_{2} /T_{2}[/itex]
[itex](P_{0} + \rho_{Hg}gh/2) * A * h/2 /T_{1} = (P_{0} + \rho_{Hg}gh * 3/4) * A * 3h/4 /T_{2}[/itex]

I only considered the volume of the mercury on the top of the disk. I thought that the cylinder is filled with half mercury and half gas, so I included h/2. That is the height of the mercury in the cylinder. Then, since the temperature is raised until more than a half of the mercury is spilled out of the cylinder, I believe I include 3h/4. I solved for [itex]T_{2}[/itex], but I have the wrong answer.

Don't know what goes wrong here.
 

TSny

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What is the depth of the mercury after half spills out? Is it 3/4 of h?
 
326
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What is the depth of the mercury after half spills out? Is it 3/4 of h?
Then, it's h/4, but wouldn't this mean that the temperature decreases? I checked that value already, and it seems invalid to me. The problem says that the temperature is supposed to increase.

Maybe, I should use the new formula to work out the calculation.
 
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Did you get an answer using TSny's advice? I just worked it out using your formula and got a positive answer.

What was the negative answer you got and maybe we can work out where you're going wrong?

If you got a T2 of 219K, then you need to remember that 0.25 + 0.75 = 1 - i.e. the height of the mercury column isn't the height of the gas column - the [itex]\rho_{Hg} g h[/itex], the h is the height of the mercury column exerting a pressure. Whereas the Volume is made up of h*A, where h is the height of the gas column.
 
Last edited:

rude man

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1. I answered the question incorrectly by following this method..

[itex](P_{0} + \rho_{Hg}gh/2) * V_{1}/T_{1} = (P_{0} + \rho_{Hg}gh * 3/4) * V_{2} /T_{2}[/itex]

Change "3/4' TO "1/4".
 
123
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Change "3/4' TO "1/4".
Since you want to spill half of the mercury, in the final state, Hg's height will be reduced by half, so it is h/2. Also, the initial height is h...
 
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... and V1 = A h , V2 = 3 A h / 2
 

rude man

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Since you want to spill half of the mercury, in the final state, Hg's height will be reduced by half, so it is h/2. Also, the initial height is h...
My correction assumes h = height of the entire vessel. That is how the problem is stated.
So the initial Hg height is h/2 and it ends up h/4.
 
326
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Then, the expression becomes:

[itex](P_{0} + \rho_{Hg}gh/2) * Ah/T_{1} = (P_{0} + \rho_{Hg}gh/4) * 3Ah/2 / T_{2}[/itex]
 

rude man

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Then, the expression becomes:

[itex](P_{0} + \rho_{Hg}gh/2) * Ah/T_{1} = (P_{0} + \rho_{Hg}gh/4) * 3Ah/2 / T_{2}[/itex]
That is correct.

EDIT: sorry, I was wrong here. V1 = Ah/2 and V2 = 3Ah/4 if h is the height of the whole cylinder.
 
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rude man

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Now, if you want an even harder problem - what is the heat Q added? :eek:
 

TSny

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Then, the expression becomes:

[itex](P_{0} + \rho_{Hg}gh/2) * Ah/T_{1} = (P_{0} + \rho_{Hg}gh/4) * 3Ah/2 / T_{2}[/itex]
It appears that you are using h for the entire height of the cylinder. But then, I don't think the final volume is expressed correctly in terms of h.
 
326
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It appears that you are using h for the entire height of the cylinder. But then, I don't think the final volume is expressed correctly in terms of h.
Why is that? Is it because there will be half the volume of the mercury? But someone just said that it's right. I don't understand why.. :\
 

TSny

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Then, the expression becomes:

[itex](P_{0} + \rho_{Hg}gh/2) * Ah/T_{1} = (P_{0} + \rho_{Hg}gh/4) * 3Ah/2 / T_{2}[/itex]
There's confusion because some here are using h for the initial height of the mercury while others are using h for the entire height of the cylinder. Either way is fine, but once you decide on the definition of h, then you have to stick with it.

On the left side of the equation above, the initial height of the mercury is written as h/2. So, that means h is the height of the entire cylinder. But also on the left you have the initial volume of the gas as Ah. But that means you are now taking h to be half the height of the entire cylinder, so there's an inconsistency in notation. Same for the right hand side.
 

rude man

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It appears that you are using h for the entire height of the cylinder. But then, I don't think the final volume is expressed correctly in terms of h.
Right. I made a mistake earlier - see my edited post above.

V1 = Ah/2 and V2 = 3Ah/4 if h is height of entire cylinder.
 
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rude man

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There's confusion because some here are using h for the initial height of the mercury while others are using h for the entire height of the cylinder. Either way is fine, but once you decide on the definition of h, then you have to stick with it.

On the left side of the equation above, the initial height of the mercury is written as h/2. So, that means h is the height of the entire cylinder. But also on the left you have the initial volume of the gas as Ah. But that means you are now taking h to be half the height of the entire cylinder, so there's an inconsistency in notation. Same for the right hand side.
This is exactly right. The OP's two opening equations are actually mutually contradictory in this sense.

I picked h to be per the problem definition (OK, they used H insead of h) and corrected his second equation accordingly.
 
326
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This is exactly right. The OP's two opening equations are actually mutually contradictory in this sense.

I picked h to be per the problem definition (OK, they used H insead of h) and corrected his second equation accordingly.
Actually, he is right....
 

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