How to find water vapor volume at different pressures than atm?

[rhinohugger]

Homework Statement

If a known amount, let's say 300 liters of water gas (water vapor) is flowing from an atmospheric (14.7 psi) source/chamber, into a chamber of reduced pressure of 10 psi, does the volume amount and/or moles change?

Homework Equations

So I believe I would multiply the 300 liters by the pressure ratio, but not sure which pressure value is the denominator, which is the numerator

The Attempt at a Solution

I believe its either: 300 L * (10psi / 14.7 psi) which is 204.1 liters,
or it is: 300 L * (14.7 psi / 10 psi) which is 441 liters

Thanks for any help, I am leaning more towards 441 liters as my answer

 

[Chestermiller]

Hi rhinohugger. Welcome to Physics Forums.

If the new chamber were initially empty (vacuum) and you could get all the moles into the new chamber at 10 psi, and, if the temperature were held fixed, then using the ideal gas law the way you did would give you a pretty accurate answer. If you needed a more accurate answer, you would have to use a more accurate equation of state, or you could use the steam tables (which have the specific volumes at various temperatures and pressures tabulated for you).

Chet

 

[rhinohugger]

Thanks for the quick reply Chet,
1.So to clarify, the ideal gas method of 300 L * (14.7 psi / 10 psi) [NOT 10 psi /14 psi], would provide an accurate answer?
2. Also, I am looking at a steam table currently on engineering toolbox.
To use one to find a more accurate answer, I would look at the pressure on the steam chart (10 psi, in my case) and look find the specific volume which is in m^3/kg, so I would have to first calculate how many kg are in 300 liters of water vapor at 14.7 psi, and multiply it by the specific volume?

 

[Chestermiller]

Thanks for the quick reply Chet,
1.So to clarify, the ideal gas method of 300 L * (14.7 psi / 10 psi) [NOT 10 psi /14 psi], would provide an accurate answer?

Yes, as long as it is at the same temperature.

2. Also, I am looking at a steam table currently on engineering toolbox.
To use one to find a more accurate answer, I would look at the pressure on the steam chart (10 psi, in my case) and look find the specific volume which is in m^3/kg, so I would have to first calculate how many kg are in 300 liters of water vapor at 14.7 psi, and multiply it by the specific volume?

No. At 14.7, you take the specific volume at the particular temperature (≥100C) and divide it into 300 liters (0.3 m^3) to get the kg.

 

[Nickie]

.Your approach is correct. To find the volume of water vapor at a different pressure than atmospheric, you would use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. In this case, we are assuming that the temperature remains constant.

To find the volume at a different pressure, we can rearrange the equation to solve for V: V = (nRT)/P. In this case, we know the initial volume (300 liters) and pressure (14.7 psi), so we can plug those values in and solve for n. This gives us n = (300 L * 14.7 psi)/(0.0821 L*atm/mol*K * 298 K) = 12.66 moles.

Now, to find the volume at a different pressure (10 psi), we can plug in the new pressure and the number of moles we just calculated: V = (12.66 mol * 0.0821 L*atm/mol*K * 298 K)/10 psi = 309.9 liters.

So, the volume does change when the pressure changes. Your second calculation (441 liters) is incorrect because you used the wrong pressure in the denominator. It should be 10 psi, not 14.7 psi.