# Simple Gas Pressure Question

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1. Sep 14, 2017

### john13

Before anything, my apologies for offering such an elementary question about gas pressure. Any help would be appreciated.

1. The problem statement, all variables and given/known data

There are two chambers. One is 100 L, containing air at a pressure of 200 kPa. The other, joined to it, is 10 L and contains air at a pressure of 20 kPa. The airtight seal between the chambers is broken. What is the final, stable pressure of the combined chambers?

2. Relevant equations

Basic arithmetic.

3. The attempt at a solution

I've assumed that it's the greater pressure (200 kPa) that matters, not the lesser. (That is, the pressures aren't additive.) So I get 100 L / 110 L = 0.909. Then: 0.909 * 200 kPa = 181.818 kPa for the whole, combined system.

I suspect that I'm wrong, but I don't know how or why. Anyone care to put me on the right path?

2. Sep 14, 2017

### jbriggs444

The result you obtained does not account for the gas that was already present in the smaller, lower pressure chamber. It cannot be completely accurate. It should be close.

Let's try to do better. Suppose that you count up how much air there is to start with. You have 100 liters times 200 kPa worth of air in the large chamber. You have 10 liters times 20 kPA worth of air in the small chamber.

3. Sep 14, 2017

### Staff: Mentor

Are you familiar with the ideal gas law?

4. Sep 14, 2017

### john13

I do recall PV = nRT from high school - except that I'm not really sure what to plug in to n or R (we're talking about ordinary atmospheric air). T would be room temperature, or 20 degrees C (293 K).

Would you mind telling me where I go from there?

5. Sep 14, 2017

### Biker

You don't need to know n, You can know it if you know the temperature but it is not given in the question so no assumptions.

What you should do is, You have some $n_1$ moles in chamber 1 and $n_2$ of moles in chamber 2. So when they are combined you get..

Now use the ideal gas law combined with that information to get an equation that will answer any question similar to this.

6. Sep 14, 2017

### john13

Thanks for the help. Since we're actually talking about locomotive air brakes, I can safely say that we're operating at room temperature (let's say 25 degrees C).

Atmospheric air is a mixture of oxygen, carbon, nitrogen, xenon, neon, helium, hydrogen, krypton, argon and neon. I have to go to work now, so on the weekend I'll go research the atomic weights and from there try to figure out the number of moles of each.

Wow - I never figured this would get so complicated!

7. Sep 14, 2017

### jbriggs444

It is not that complicated. You do not need to know the atomic weights. You do not need to know the number of moles. You do not need to know the temperature.

Leave everything symbolic and do the algebra. See what drops out.

8. Sep 14, 2017

### john13

Hmm. Well, in that case I get:

P = (n1 + n2)RT/110

The 110 refers to litres, of course; it's 0.11 cubic metres.

But now how do I account for the overall system's pressure when I've combined the pressures of two different chambers? I imagine that the respective initial pressures, 100 kPa and 10 kPa, are critical to this calculation.

9. Sep 14, 2017

### Biker

Overall system pressure is what we need.

Why not substitute n1 and n2 using the ideal gas law? Keep the unknown variables symbolic.

10. Sep 14, 2017

### Staff: Mentor

From the ideal gas law, algebraically, in terms of T (and R), what is the initial number of moles of air in chamber 1 ($n_1)$? From the ideal gas law, algebraically, in terms of T (and R), what is the initial number of moles of air in chamber 2 ($n_2$)? From these results, what is the total initial number of moles of air in chambers 1 and 2? Does this total number of moles change between the initial state of the system and the final state of the system? What is the total volume of the two chambers? So now you know the total number of moles in the two chambers in the final state, the total volume in the final state, and the temperature. From the ideal gas law, what is the final pressure?

11. Sep 14, 2017

### john13

Okay, that makes perfect sense. I'm using cubic metres and Pascals.

So I have n1 = RT/(100,000 * 0.1) and n2 = RT/(10,000 * 0.01)

In the final system: n1 + n2 = RT/10,000 + RT/100 = 0.0001RT + 0.001RT = 0.0011RT

Therefore P = 0.0011RT(RT)/0.11 = 0.000121(RT)^2

It doesn't look all that plausible at this stage, and I'm not sure what to do with the RT.

12. Sep 14, 2017

### Staff: Mentor

These are upside down. Try again.

13. Sep 14, 2017

### john13

Ugh - how embarrassing!

So instead I have (n1 + n2) =10,100/RT = P

That looks better, no?

14. Sep 14, 2017

### Staff: Mentor

No. I get $$n_1+n_2=\frac{20200}{RT}$$where the numerator is in Joules. And please tell me why that is supposed to be equal to the final pressure, since the units don't even match.