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Density of water vapour in the air following compression if it were an Ideal Gas

by Hendrick
Tags: compression, density, ideal, vapour, water
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Hendrick
#1
Apr17-07, 06:05 AM
P: 43
1. The problem statement, all variables and given/known data
A volume Vi = 23.2 L of air at temperature T = 22 C is compressed at constant temperature to a volume Vf = 11.0 L. The relative humidity of the air before compression is H = 59.4 %. The saturated vapour pressure of water at this temperature is 2.67 kPa, corresponding to a vapor density ρw = 19.6 g m^3.

A)What would be the density of water vapour in the air following compression if it behaved like an ideal gas instead of experiencing partial condensation?

B)What is the density of water vapour in the air following compression?

C)What mass of water condenses out because of the compression?


2. Relevant equations

PV = nRT
P1 * V1 = P2 * V2
P = rho * R/M * T
n = m/M

n = number of moles of substance
M = molar mass
m = mass
P = pressure
V = volume
rho = density
R = universal gas constant
T = absolute temperature

3. The attempt at a solution
A)P1 * V1 / V2 = P2

2670 * (23.2/11) = P2

rho = P2 * M/(R * T)

= P2 * 0.018/(8.314 * (273 + 22))

= 0.04132825

Using the molar mass of water (0.018 kg mol^-1) and P1 as the saturated vapour pressure of water at 22 C (2.67 kPa)

The actual answer is 24.6 g m^-3

----------------------------------------------------------------
B)rho = P * M/(R * T)

= 2670 * 0.018/(8.314 * (273 + 22))

= 0.019595291

Using the molar mass of water (0.018 kg mol^-1) and P as the saturated vapour pressure of water at 22 C (2.67 kPa)

Actual answer is 19.6 g m^-3

-- is this just the value of vapor density ρw = 19.6 g m^3. from the line at the beginning of the problem "corresponding to a vapor density ρw = 19.6 g m^3."

----------------------------------------------------------------
C) PV = nRT

n = PV/RT

n = m/M

P = rho * R/M * T

=> PV/RT = m/M

=> [rho * R/M * T ] *V/RT = m/M

rho * V = m

19.6 * (11/1000) = 0.2156

Using ρw = 19.6 g m^3, Vf = 11.0 L

The actual answer is 0.0545 g

-- Help would be much appreciated
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Andrew Mason
#2
Apr17-07, 11:21 AM
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Quote Quote by Hendrick View Post
1. The problem statement, all variables and given/known data
A volume Vi = 23.2 L of air at temperature T = 22 C is compressed at constant temperature to a volume Vf = 11.0 L. The relative humidity of the air before compression is H = 59.4 %. The saturated vapour pressure of water at this temperature is 2.67 kPa, corresponding to a vapor density ρw = 19.6 g m^3.

A)What would be the density of water vapour in the air following compression if it behaved like an ideal gas instead of experiencing partial condensation?

B)What is the density of water vapour in the air following compression?

C)What mass of water condenses out because of the compression?


2. Relevant equations

PV = nRT
P1 * V1 = P2 * V2
P = rho * R/M * T
n = m/M

n = number of moles of substance
M = molar mass
m = mass
P = pressure
V = volume
rho = density
R = universal gas constant
T = absolute temperature

3. The attempt at a solution
A)P1 * V1 / V2 = P2

2670 * (23.2/11) = P2
You have to specify units and be careful. You have to convert L to m^3.

The vapour pressure of fully saturated air is 2670 Pa
of which this water vapour contributes 59.4% or 1586 Pa.

Using n = PV/RT, the amount of water vapour is 1586*.0232/8.314*295 = .015 moles which has a mass of .27 g. So the density of water vapour is .27g/23.2 L = 11.6 x 10^-3 g/L = 11.6 x g/m^3 .

If you reduce the volume to 11 litres, without changing the amount of water vapour, what is the density?

AM
Hendrick
#3
Apr17-07, 09:35 PM
P: 43
Quote Quote by Andrew Mason View Post
You have to specify units and be careful. You have to convert L to m^3.

The vapour pressure of fully saturated air is 2670 Pa
of which this water vapour contributes 59.4% or 1586 Pa.

Using n = PV/RT, the amount of water vapour is 1586*.0232/8.314*295 = .015 moles which has a mass of .27 g. So the density of water vapour is .27g/23.2 L = 11.6 x 10^-3 g/L = 11.6 x g/m^3 .

If you reduce the volume to 11 litres, without changing the amount of water vapour, what is the density?

AM
Hi,

I think I understand how to do part A) now:

n = PV/RT
m = nM
rho = m/v

n = (0.594*2670)*(23.2/1000)/8.314*(273+22)
m = n*(0.018*1000) [for molar mass in g mol^-1]
rho = m/(11/1000)
= 24.5489813
= 24.5 g m^-3 (3sf)
----------------------------------

Is part B) just the vapor density of water stated in the beginning of the equation? If so, why?

----------------------------------

I don't know where to start on C), I tried taking the change in volume and using a 'm = rho*v' equation with rho being the answer in B but it's incorrect

thanks for the help on A), I didn't even think of finding n and substituting it on find the density...

Andrew Mason
#4
Apr18-07, 03:30 AM
Sci Advisor
HW Helper
P: 6,653
Density of water vapour in the air following compression if it were an Ideal Gas

Quote Quote by Hendrick View Post
Hi,

I think I understand how to do part A) now:

n = PV/RT
m = nM
rho = m/v

n = (0.594*2670)*(23.2/1000)/8.314*(273+22)
m = n*(0.018*1000) [for molar mass in g mol^-1]
rho = m/(11/1000)
= 24.5489813
= 24.5 g m^-3 (3sf)
Correct.
Is part B) just the vapor density of water stated in the beginning of the equation? If so, why?
For part B you have to determine how much water vapour is present. You need to know the saturated vapour pressure of water at an air pressure of 213704 Pa. and T= 22C. Do you know why you must use the saturated vapour pressure?

Using n = PV/RT will give you the no. of moles from which you can work out the density.

Part C) is just the difference between A and B.

AM
Hendrick
#5
Apr18-07, 07:16 AM
P: 43
Quote Quote by Andrew Mason View Post
Correct.

For part B you have to determine how much water vapour is present. You need to know the saturated vapour pressure of water at an air pressure of 213704 Pa. and T= 22C. Do you know why you must use the saturated vapour pressure?

Using n = PV/RT will give you the no. of moles from which you can work out the density.

Part C) is just the difference between A and B.

AM
Part B):
Why is the air pressure 213704 Pa? I don't know why I must use the saturated vapour pressure.

Part C):
Do you mean the difference in masses between A and B?
I didn't think it would involve part A) as I thought it was a hypothetical question:
"What would be the density of water vapour in the air following compression if it behaved like an ideal gas instead of experiencing partial condensation?"
Anyhow, I used the rho = m/V formula to find the masses and got a difference 0.055g instead of 0.0545g
Andrew Mason
#6
Apr18-07, 10:44 AM
Sci Advisor
HW Helper
P: 6,653
Quote Quote by Hendrick View Post
Part B):
Why is the air pressure 213704 Pa? I don't know why I must use the saturated vapour pressure.
It isn't quite because I was assuming no condensation. But it is close. If n and T do not change P2V2 = P1V1. If P1 = 101325 Pa then P2 = P1V1/V2 = 101325*23.2/11.0 = 213704 Pa. In fact, however, n does change a little as some water vapour will condense.

What you have to do to get it exact is to work out the pressure of the air only at 11 L. It is 101325 - 2670 at 23.2 L.

You have to use the saturation vapour pressure because if it is 59.4% saturated at 1 atm, it will be completely saturated when compressed to less than half the volume. You know this because when you compress water vapour from 1580 Pa at 22 C in 23.2 L to 11 L at 22C, the vapour pressure increases to ________ Pa. But we know that the maximum vapour pressure is 2670 Pa. You should be able to determine from that how much precipitates out.

Part C):
Do you mean the difference in masses between A and B?
I didn't think it would involve part A) as I thought it was a hypothetical question:

Anyhow, I used the rho = m/V formula to find the masses and got a difference 0.055g instead of 0.0545g
In A, you determine how much water there is (all in the vapour state, since it is only 59.4% saturated). In B you determine how much is still in the vapour state. The difference is what has condensed out.

AM
Hendrick
#7
Apr19-07, 06:30 AM
P: 43
Ok, I'm pretty confused about the meaning of part A)
What would be the density of water vapour in the air following compression if it behaved like an ideal gas instead of experiencing partial condensation?
Would you be able to explain what is meant by behaving as an ideal gas instead of experiencing partial condensation?

--------------------

Quote Quote by Andrew Mason View Post
It isn't quite because I was assuming no condensation. But it is close. If n and T do not change P2V2 = P1V1. If P1 = 101325 Pa then P2 = P1V1/V2 = 101325*23.2/11.0 = 213704 Pa. In fact, however, n does change a little as some water vapour will condense.

What you have to do to get it exact is to work out the pressure of the air only at 11 L. It is 101325 - 2670 at 23.2 L.

You have to use the saturation vapour pressure because if it is 59.4% saturated at 1 atm, it will be completely saturated when compressed to less than half the volume. You know this because when you compress water vapour from 1580 Pa at 22 C in 23.2 L to 11 L at 22C, the vapour pressure increases to ________ Pa. But we know that the maximum vapour pressure is 2670 Pa. You should be able to determine from that how much precipitates out.
Does the vapour pressure increase to 302942 Pa?

I still don't quite get why you have to do all of the calculations as I thought the answer was just from
A volume Vi = 23.2 L of air at temperature T = 22 °C is compressed at constant temperature to a volume Vf = 11.0 L. The relative humidity of the air before compression is H = 59.4 %. The saturated vapour pressure of water at this temperature is 2.67 kPa, corresponding to a vapor density ρw = 19.6 g m^–3.
for part B)

--------------------
In A, you determine how much water there is (all in the vapour state, since it is only 59.4% saturated). In B you determine how much is still in the vapour state. The difference is what has condensed out.
From this I understand that by taking the difference in masses from A) and B) produces C), but I don't understand how part A determines how much water there is...

Thanks for your help
Andrew Mason
#8
Apr20-07, 09:33 AM
Sci Advisor
HW Helper
P: 6,653
Quote Quote by Hendrick View Post
Ok, I'm pretty confused about the meaning of part A)

Would you be able to explain what is meant by behaving as an ideal gas instead of experiencing partial condensation?
The molecules in an ideal gas have no attraction to each other. They do not convert any of their kinetic energy into potential energy (bonds between molecules). Water is a polar molecule. Water molecules can "stick" to each other. If there are too many at a given kinetic energy, some stick to each other and stop acting like a gas. For a given Temperature, there is a maximum pressure (related to density) of water vapour, above which the molecules stick together and stop acting like a gas (they condense out). For 22 C that pressure is 2670 Pa.

In part A, you just assume that there is no maximum vapour pressure.


--------------------

Does the vapour pressure increase to 302942 Pa?

I still don't quite get why you have to do all of the calculations as I thought the answer was just from

for part B)
Vapour pressure cannot exceed the maximum saturated vapour pressure for that temperature. For 22 C it is 2670 Pa. Work out the vapour pressure if it is 59.4% saturated (ie. 59.4% of 2670 Pa). Find the mass of water in the unsaturated air (ie all the water is in vapour state). This is what you do first in Part A.

Then find the amount of water vapour and the density of this vapour at 11 L. If none precipitated out, what would the vapour pressure be? Can it actually be that high? Why not (what is the maximum?)? So if that is the maximum, how much water is in the vapour state? How does that compare to the amount that was present in the unsaturated air (all in vapour state)? What is the difference? Where is that difference (ie. the amount not in the vapour state?)?

AM
Hendrick
#9
Apr20-07, 09:56 AM
P: 43
Thanks very much for your help Andrew.

I have everything calculation-wise but your explanation aided my comprehension of the problem...

-- Thank you


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