Dew Point -- will we be able to see our breath?

[il27]

Homework Statement

The dew point is the temperature at which moist air becomes saturated with water vapor. We can therefore represent how moist a particular parcel of air is by comparing its actual temperature with its dew point temperature.

We see the practical effect of reaching the dew point when on a cold day, we "see our breath" when we exhale.

Suppose it is -10 degrees Celsius outside and our breath leaves our body at 30 degrees Celsius, whereupon it mixes with the outside cold air in a one to one ratio (equal parts warm breath and cold air).

If your breath has sufficient moisture to have a dew point of 24 °C and the outside air has a dew point of -14 °C, will we be able to see our breath after it mixes with the cold outside air?

Homework Equations

rs = Ee(T)/p, e = rp/E

The Attempt at a Solution

I am not sure how to go about this. I was thinking something about using the saturation vapor pressure... and I have an idea on the answer, which is that we will be able to see our breath up until we go outside. As being outside, it will try to attain a temperature which is equal to that of air present outside. After mixing, as the dewpoint of outside air is -14oC which is lesser than the outside temperature, hence dewpoint will not be attained, which means that the air from our breath will not saturate. However i need help solving and proving this mathematically. Thank you!

 

[mfb]

You can find the vapor pressure in both your breath and the outside air, and find the vapor pressure of the mixture based on that.
Independently you can find the mixing temperature.

Then take both values and see if the dew point of the mixture is above or below the mixing temperature.

 

[Chestermiller]

What is the equilibrium vapor pressure of water at 24 C? What is the equilibrium vapor pressure of water at -14 C? What is the partial pressure of water vapor in the mixture when your breath mixes with an equal volume of cold air outside?

 

[il27]

What is the equilibrium vapor pressure of water at 24 C? What is the equilibrium vapor pressure of water at -14 C? What is the partial pressure of water vapor in the mixture when your breath mixes with an equal volume of cold air outside?

The equilibrium vapor pressure of water at 24 C = 2960 pa, and at -14 C = 210 pa.
To find the partial pressure of water vapor in the mixture... would I use the equation for saturated mixing ratio? or...
$$ r_s = \frac{\epsilon e_s(T)}{p} $$

 

[il27]

You can find the vapor pressure in both your breath and the outside air, and find the vapor pressure of the mixture based on that.
Independently you can find the mixing temperature.

Then take both values and see if the dew point of the mixture is above or below the mixing temperature.

do i find the mixing temperature of the temperature outside being -10 deg. C and one's breath being 30 deg. C, or do I take the dew point temperatures?

 

[Chestermiller]

The equilibrium vapor pressure of water at 24 C = 2960 pa, and at -14 C = 210 pa.
To find the partial pressure of water vapor in the mixture... would I use the equation for saturated mixing ratio? or...
$$ r_s = \frac{\epsilon e_s(T)}{p} $$

Those are the partial pressures of water vapor from the moist air from your mouth and the cold air outside. So, if you mix equal volumes, what is the partial pressure of water vapor in the mixture (assuming that the total volume of air doesn't change significantly)?

 

[il27]

Those are the partial pressures of water vapor from the moist air from your mouth and the cold air outside. So, if you mix equal volumes, what is the partial pressure of water vapor in the mixture (assuming that the total volume of air doesn't change significantly)?

Oh, I think i see what you mean. Do I just add the partial pressure of water vapor for the dew points of one's breath and of the outside temperature? to get the total partial pressure?

or should I use this equation of mixing ratios: $$ \frac{p_2}{p_1} = (\frac{a_1}{a_2})^.5 $$? and isolate a1/a2 to find the mixing ratio of both vapor pressrues?
Thank you for all your help by the way.

 

[Chestermiller]

Oh, I think i see what you mean. Do I just add the partial pressure of water vapor for the dew points of one's breath and of the outside temperature? to get the total partial pressure?

or should I use this equation of mixing ratios: $$ \frac{p_2}{p_1} = (\frac{a_1}{a_2})^.5 $$? and isolate a1/a2 to find the mixing ratio of both vapor pressrues?
Thank you for all your help by the way.

The final partial pressure when you mix the two equal volume parcels of air is the average of the two partial pressures.

 

[il27]

The final partial pressure when you mix the two equal volume parcels of air is the average of the two partial pressures.

Oh okay, thank you!
How would I then connect the final vapor pressure of the dew points to determine whether my hypothesis on whether you can see your breath after it mixes with the cold air outside?
I found the final vapor pressure of the original temperatures of one's breath and the temperature outside (not the dew point temperatures). and should I compare the final vapor pressure of the original temperatures to the final vapor pressure of the dew point temperatures? I got that the final vapor pressure temperature using the original temperatures is greater than the final vapor pressure using the dew point temperatures.
What would that tell me in regards to being able to see one's breath outside?

 

[Chestermiller]

Oh okay, thank you!
How would I then connect the final vapor pressure of the dew points to determine whether my hypothesis on whether you can see your breath after it mixes with the cold air outside?
I found the final vapor pressure of the original temperatures of one's breath and the temperature outside (not the dew point temperatures). and should I compare the final vapor pressure of the original temperatures to the final vapor pressure of the dew point temperatures? I got that the final vapor pressure temperature using the original temperatures is greater than the final vapor pressure using the dew point temperatures.
What would that tell me in regards to being able to see one's breath outside?

I'm having trouble understanding what you are saying here.

The final temperature of the combined mixed air parcels is ?
The equilibrium vapor pressure of water at this temperature is ?
The partial pressure of water vapor in the combined mixed air parcels is ?

 

[il27]

I'm having trouble understanding what you are saying here.

The final temperature of the combined mixed air parcels is ?
The equilibrium vapor pressure of water at this temperature is ?
The partial pressure of water vapor in the combined mixed air parcels is ?

Sorry! Yes... so the

The equilibrium vapor pressure of water at the breath temperature is 42 hpa, for the temperature outside it's 2.8635 hpa.

the partial pressure of water vapor in the combined mixed air parcels is the average of the two being 22.64 hpa.

the equilibrium vapor pressure at the dew point temperature for the breath is 30.64 hpa, for outside it's 2.11 hpa (i used the clausius clapeyron equation)

the the partial pressure of water vapor in the combined mixed air parcels for the dew point temperatures is the average of the two being 16.375 hpa.

so should I find the mixing temperature? I am basically just confused on how do I connect these vapor pressures to mixing temperatures to answer the question on whether you can see your breath when going outside?
Thanks so much and sorry for the confusion.

 

[Chestermiller]

Sorry! Yes... so the

The equilibrium vapor pressure of water at the breath temperature is 42 hpa, for the temperature outside it's 2.8635 hpa.

the partial pressure of water vapor in the combined mixed air parcels is the average of the two being 22.64 hpa.

the equilibrium vapor pressure at the dew point temperature for the breath is 30.64 hpa, for outside it's 2.11 hpa (i used the clausius clapeyron equation)

the the partial pressure of water vapor in the combined mixed air parcels for the dew point temperatures is the average of the two being 16.375 hpa.

so should I find the mixing temperature? I am basically just confused on how do I connect these vapor pressures to mixing temperatures to answer the question on whether you can see your breath when going outside?
Thanks so much and sorry for the confusion.

You haven't answered my questions. Do you think that I don't know what I am doing?

 

[il27]

You haven't answered my questions. Do you think that I don't know what I am doing?

No, not at all. I'm sorry, I thought I was supposed to calculate the vapor pressures and then average them?
Is that not what you meant? I am confused on what to do :(. What would be your method to solve this problem given the original temperatures (Tbreath and Toutside) and their corresponding dew points?

Also how would I find the mixed temperatures? which temperatures should I use?

 

[Chestermiller]

The presence of the water vapor in the air has very little effect on the final temperature of the two parcels of air after they are mixed.

Equilibrium State 1:
x moles of air at 30 C and 1 atm.
x moles of air at -10 C and 1 atm.

Equilibrium State 2:
mixture of 2x moles of air at $T = T_f$ C and 1 atm.

Since we are dealing with an ideal gas, the change in internal energy is zero between states 1 and 2. So, what is the final temperature of the mixture?

 

[il27]

The presence of the water vapor in the air has very little effect on the final temperature of the two parcels of air after they are mixed.

Equilibrium State 1:
x moles of air at 30 C and 1 atm.
x moles of air at -10 C and 1 atm.

Equilibrium State 2:
mixture of 2x moles of air at $T = T_f$ C and 1 atm.

Since we are dealing with an ideal gas, the change in internal energy is zero between states 1 and 2. So, what is the final temperature of the mixture?

Oh okay. To find the number of moles though, I would need volume, but we can think of the volumes as equal right?
So would you use $$ pV = nRT $$ isolate $$ V = \frac{nRT}{V} $$ and set the two volumes together to find the ratio of moles at 30 C and -10 C?

also I don't understand what hte equation would be for the equilibrium state 2. Could you explain that?

 

[Chestermiller]

Oh okay. To find the number of moles though, I would need volume, but we can think of the volumes as equal right?
So would you use $$ pV = nRT $$ isolate $$ V = \frac{nRT}{V} $$ and set the two volumes together to find the ratio of moles at 30 C and -10 C?

also I don't understand what hte equation would be for the equilibrium state 2. Could you explain that?

My understanding is that you are mixing equal numbers of moles, not equal volumes.

Equilibrium state 2 is a thermally equilibrated mixture of x moles of dryer air at -10 C with x moles of moister air at 30 C.

What is your understanding of the term "dew point."

 

[il27]

My understanding is that you are mixing equal numbers of moles, not equal volumes.

Equilibrium state 2 is a thermally equilibrated mixture of x moles of dryer air at -10 C with x moles of moister air at 30 C.

What is your understanding of the term "dew point."

Oh I see. so we would find the ratio of volumes by setting the # of moles equal to each other?
how would the second equilibrium equation be written out?

my understanding of dew point is the temperature at which the air can no longer "hold" all of the water vapor, which is mixed with it, and some of the water vapor must condense into liquid water. The dew point is always lower than (or equal to) the air temperature.

 

[Chestermiller]

Final temperature of combined air parcels = + 10 C = (-10 + 30)/2
Final partial pressure of water vapor in air mixture = (2960 + 210)/2
Final dew point = temperature at which equilibrium vapor pressure is equal to partial pressure of water vapor in mixture = ?

 

[il27]

Final temperature of combined air parcels = + 10 C = (-10 + 30)/2
Final partial pressure of water vapor in air mixture = (2960 + 210)/2
Final dew point = temperature at which equilibrium vapor pressure is equal to partial pressure of water vapor in mixture = ?

oh okay. thank you! would the final dew point be equal to:

$$ e_{final} = nRT/V $$ and then we isolate the temperature?
What would the equation for the final dew point look like?
thank you again so much.

 

[Chestermiller]

oh okay. thank you! would the final dew point be equal to:

$$ e_{final} = nRT/V $$ and then we isolate the temperature?
What would the equation for the final dew point look like?
thank you again so much.

The dew point is the temperature at which the partial pressure of the water vapor in the air is equal to the equilibrium vapor pressure.

 

[il27]

The dew point is the temperature at which the partial pressure of the water vapor in the air is equal to the equilibrium vapor pressure.

the partial pressure of water vapor in the air is the final partial pressure of 1585 pa right? however how do i determine the temperature from that?
do i set that number equal to $$ (nRT)/V $$

 

[Chestermiller]

the partial pressure of water vapor in the air is the final partial pressure of 1585 pa right? however how do i determine the temperature from that?
do i set that number equal to $$ (nRT)/V $$

No. You get a table of equilibrium vapor pressures of water vs temperature, and look up the temperature at which the equilibrium vapor pressure is equal to the present partial pressure 1585 Pa.

 

[il27]

No. You get a table of equilibrium vapor pressures of water vs temperature, and look up the temperature at which the equilibrium vapor pressure is equal to the present partial pressure 1585 Pa.

Excellent! Thank you. I looked at a table and got a temperature of about 14 degrees C. do we compare this value to dew point? and since 14 deg. C is greater than the dew point outside, does that mean we cannot see our breath when it mixes with the air outside?
Thank you again for all your help. This is really helpful :)

 

[Chestermiller]

Excellent! Thank you. I looked at a table and got a temperature of about 14 degrees C. do we compare this value to dew point? and since 14 deg. C is greater than the dew point outside, does that mean we cannot see our breath when it mixes with the air outside?
Thank you again for all your help. This is really helpful :)

14 C is the dew point. When you compare this with the actual temperature of the mixture (10 C), do you conclude that (a) we will see our breath or (b) we will not see our breath.

 

[il27]

14 C is the dew point. When you compare this with the actual temperature of the mixture (10 C), do you conclude that (a) we will see our breath or (b) we will not see our breath.

oh i see. 14 degrees C is the dew point of the mixing temperatures?
since the actual mixed temperature (10C) we conclude that we will see not see our breath. because the dew point is low, so there is not a lot of moisture in the air...?

 

[Chestermiller]

oh i see. 14 degrees C is the dew point of the mixing temperatures?
since the actual mixed temperature (10C) we conclude that we will see not see our breath. because the dew point is low, so there is not a lot of moisture in the air...?

The dew point temperature of the gas mixture is 14 C is and the actual gas mixture temperature is 10 C. When you go out on a cold morning (10C) and the dew point is 14 C, are you saying that you won't see dew on the grass?

 

[il27]

The dew point temperature of the gas mixture is 14 C is and the actual gas mixture temperature is 10 C. When you go out on a cold morning (10C) and the dew point is 14 C, are you saying that you won't see dew on the grass?

oh no. you will see dew because the dew point for hte mixed temperature is greater than the actual gas mixture temperature! right?
also one more question... i thought the dew point could never be greater than the temperature?

 

[Chestermiller]

oh no. you will see dew because the dew point for hte mixed temperature is greater than the actual gas mixture temperature! right?
also one more question... i thought the dew point could never be greater than the temperature?

Right. So. You conclude that if the temperature of the parcel got down to 10 C, it would be below the dew point and dew would form.