# Ideal Gas Law Problem - Environmental Physics

1. Apr 25, 2010

### lachy

1. The problem statement, all variables and given/known data
1. Consider ideal gas in a rigid cylinder of 2 litres, at pressure of 3 Bar. The gas is connected through a thin capillary to a free expanding bag, which contains ideal gas at pressure 1Bar and volume of 1 litre. A valve in the capillary initially prevents the gases to mix. If the valve is open, calculate the volume of the bag, in thermal equilibrium with the environment. The surrounding atmosphere is at a pressure of 1Bar and temperature T. Neglect the volume of the capillary and valve, as well as the mass and elasticity of the bag. The heat capacity of the surrounding atmosphere is much larger than the heat capacity of the gas in the cylinder and bag. Note: in solving this problem, use the fact that the number of atoms of the gas in the bag and cylinder does not change after opening the valve, i.e. N1+N2 = N1’ +N2’ .Every step of the solution should be clearly given.

2. Relevant equations

PV = nRT

P1*V1/T1 = P2 * V2/T2

??

3. The attempt at a solution

I haven't really attempted as I don't know where to start as the lack of temperature given deters me.

Could someone please give me a push in the right direction?

Thanks

2. Apr 25, 2010

### Andrew Mason

You are to assume that the system is in thermal equilibrium with the environment. This means that temperature does not change. Start by finding the relative number of moles of the gas in each part (cylinder and bag) at the beginning and then at the end.

$$n1 = P1V1/RT$$

$$n2 = P2V2/RT$$

From that find the ratio n1/n2

Then work out the expression for n1+n2 after they have reached equilibrium

AM

3. May 5, 2010

### lachy

Could you explain in greater detail how to do this question, anyone?

4. May 6, 2010

### Andrew Mason

Show us what you have done. I have given you hints to start. Have you found n1/n2? n1 + n2? T is the same in both the cylinder and the bag and all processes remain at the same temperature T.

AM

5. May 6, 2010

### jimmyjack

will the bag expand so that the pressure of the system (initially at 4bar) equal that of the surrounding atmosphere (1bar)?

6. May 6, 2010

### jimmyjack

n1 = 4.35*1025

n2 = 7.25*1024

P3 (V1+V2) = (N1+N2)KBT

however, this equation assumes that both volumes are constant?

so assuming that the bag does expand until the system is in equilibrium with the atmosphere (1 Bar), for the whole system:

P1V1=P2V2

P0V0 = PtVt

Vt = (P0V0)/Pt

Vt=(4x105 Pa * 3*10-3 m3) / 1*105 Pa

= 1.2*10-2 m3

= 12*10-3 m3

=12 L

for the bag after expanding:

bag volume = system volume - cylinder volume

V = 12L - 2L
= 10L

im not sure if this is correct, acutally im fairly unconfident with it, however hopefully you can tell me if ive gone in the wrong direction?

7. May 6, 2010

anyone?

8. May 6, 2010

### lachy

Even though the temperature is constant, you do not know it ---> I do believe you cannot find the number of molecules or the number of moles as both of these are dependant upon temperature.

9. May 6, 2010

### jimmyjack

i think as temperature does not affect the system, you can place it at 1, effectively removing its value to any equation

10. May 6, 2010

### lachy

Charles’ Law describes the volume of a fixed quantity of gas maintained at constant pressure is directly proportional to its temperature. When the valve has not been released the bag has a fixed volume so I don't think you can just set the temperature to 1 (essentially cancelling it) as this would change the volume since it is not rigid...I think...

11. May 6, 2010

### Andrew Mason

Think about the problem. What is the final pressure when the two volumes of gas mix? (hint:what pressure limits the expansion of the bag?) What is the final pressure in the cylinder? What is the relationship between final P, final V and n1+n2? Can you then relate PfVf to initial values for P1V1 and P2V2? If so, you can easily find Vf.

AM

Last edited: May 6, 2010
12. May 6, 2010

### lachy

Yeah I have done this (I sent you a pm before?). I was just explaining.

13. May 6, 2010

### jimmyjack

sorry andrew, is there a chance you could not be so cryptic?

14. May 6, 2010

### Andrew Mason

What do you find cryptic in my last post? If you can figure out the final pressure, you will have solved the problem. What determines the final pressure (ie. what stops the bag from getting bigger and bigger)?

AM