Partial pressure of water vapor

[Karol]

Homework Statement

The partial pressure of water vapor in air at 20⁰ is 10[mm] mercury. according to the table of partial pressures we have to cool the air to 11.4⁰ in order to bring the air to saturation, that is100% relative humidity. this according to the book.
But when we cool the pressure and\or volume decrease according to $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, so why is P left constant? is it because V, the volume shrinks only? the atmosphere shrinks? and why doesn't P reduce, only V?

Homework Equations

From the equation of state of the ideal gases: $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

The Attempt at a Solution

Maybe, because we talk about free air, the volume isn't fixed and we can reduce V. but i am not sure we talk about atmospheric air only, it's not explicitly stated in the book.

 

[haruspex]

when we cool the pressure and\or volume decrease according to $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$,

The partial pressure discussion concerns the contribution to the total pressure from the water vapor. It is independent of the pressure from other gas species. That's what the theory of partial pressures is all about.
Say you have a container that holds only water (as gas and liquid). The pressure in the gas will depend only on the temperature. If you increase the volume but keep the temperature constant then more water will evaporate to maintain the same pressure (until there's none left). Do the same with air present and the partial pressure from the water will do exactly the same as when there was no air, while the pressure due to the air will do its thing independently of the water. The total pressure is the sum of the partial pressures.
Hope that helps.

 

[Chestermiller]

Homework Statement

The partial pressure of water vapor in air at 20⁰ is 10[mm] mercury. according to the table of partial pressures we have to cool the air to 11.4⁰ in order to bring the air to saturation, that is100% relative humidity. this according to the book.
But when we cool the pressure and\or volume decrease according to $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, so why is P left constant? is it because V, the volume shrinks only? the atmosphere shrinks? and why doesn't P reduce, only V?

Homework Equations

From the equation of state of the ideal gases: $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

The Attempt at a Solution

Maybe, because we talk about free air, the volume isn't fixed and we can reduce V. but i am not sure we talk about atmospheric air only, it's not explicitly stated in the book.

How do you know that the partial pressure of the water vapor didn't decrease when you cooled the air? Assume a constant volume of air, and use the ideal gas law to calculate the new partial pressure of the water vapor when the air is cooled to 11.4 C. This will be a little less than 10 mm Hg. How does that compare with the equilibrium vapor pressure of water vapor at 11.4 C?

Chet

 

[Karol]

How do you know that the partial pressure of the water vapor didn't decrease when you cooled the air? Assume a constant volume of air, and use the ideal gas law to calculate the new partial pressure of the water vapor when the air is cooled to 11.4 C. This will be a little less than 10 mm Hg. How does that compare with the equilibrium vapor pressure of water vapor at 11.4 C?
Chet

The table of equilibrium vapor pressure states that at 10⁰ it's 8.94 [mm] mercury and at 15⁰ it's 12.67, so at 10 [mm] it's 11.4⁰.
$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\rightarrow \frac{10\cdot V}{20}=\frac{x \cdot V}{11.4}\rightarrow x=5.7$$
And it's far from 10[mm].
But Haruspex says water vapor doesn't behave according to the ideal gas law:

Say you have a container that holds only water (as gas and liquid). The pressure in the gas will depend only on the temperature. If you increase the volume but keep the temperature constant then more water will evaporate to maintain the same pressure (until there's none left). Do the same with air present and the partial pressure from the water will do exactly the same as when there was no air, while the pressure due to the air will do its thing independently of the water. The total pressure is the sum of the partial pressures.
Hope that helps.

Does the air in the container behave differently than the water vapor? i guess it obeys the equation of state

 

[haruspex]

But Haruspex says water vapor doesn't behave according to the ideal gas law:

I described what happens when there is liquid water also present.

water will evaporate to maintain the same pressure (until there's none left)

This is what SVP is about. When the temperature rises enough, or the volume expands enough, that all the water has evaporated then it will start to behave like (roughly) an ideal gas.

 

[Karol]

What is SVP?

 

[haruspex]

What is SVP?

Saturation Vapor Pressure, the maximum partial pressure that the gas species can produce at a given temperature.

 

[Karol]

Say you have a container that holds only water (as gas and liquid). The pressure in the gas will depend only on the temperature.

If we had in the container only water vapor, then it would behave according to the equation of state and we had to cool to a different temperature than 11.4? to which and how do i calculate

 

[haruspex]

If we had in the container only water vapor, then it would behave according to the equation of state and we had to cool to a different temperature than 11.4? to which and how do i calculate

If the partial pressure of water vapor in a sealed container at temperature T is below the SVP of water at temperature T (i.e., less than 100% relative humidity) then there will be no liquid water present. Under these circumstances you can treat the water vapor as an ideal gas. Sorry to have introduced the discussion about what happens when liquid water is present. You don't need that for this question, and it just seems to have confused you.

Back to your original question. As you wrote, this is not about a sealed container. Forget the water vapor for a moment, just consider the air. Atmospheric pressure at ground level is driven by the weight of air above, and does not depend on temperature. Thus you can take the pressure as constant here. Now consider an atmosphere consisting entirely of water vapor, no liquid water. The same would apply. So you can take the partial pressure of the water vapor to be constant as the temperatures varies, provided the temperature does not drop so low that the partial pressure exceeds the SVP.

 

[Karol]

Thank you very much Haruspex and with me don't feel at all uncomfortable that you "confused" me with another discussion, i am glad to learn and it helped me to understand, finally, the matter. it also evolved to the final post.
You said that water will evaporate to maintain the pressure if temperature rises. which pressure? the initial pressure that was in the sealed vessel or the pressure from the SVP table? from the table, when the temperature is higher the partial pressure also rises

 

[Chestermiller]

The table of equilibrium vapor pressure states that at 10⁰ it's 8.94 [mm] mercury and at 15⁰ it's 12.67, so at 10 [mm] it's 11.4⁰.
$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\rightarrow \frac{10\cdot V}{20}=\frac{x \cdot V}{11.4}\rightarrow x=5.7$$
And it's far from 10[mm].

C'mon man. Those are supposed to be absolute temperatures that you use in the ideal gas law! I get a partial pressure of 9.7 mm Hg at 11.4 C. Is that close enough to 10 mm Hg for you?

Does the air in the container behave differently than the water vapor? i guess it obeys the equation of state

In your problem, both the air and the water vapor behave very close to ideal gases. Only when you hit the dew point (11.4C) does the water vapor start to condense.

Chet

 

[Karol]

Haruspex offered another explanation and i understand that, so i ignore your explanation, Chet. also because it's a little difficult for me and i can't settle the contradiction between Haruspex's explanation and yours.
Haruspex expalined that 10[mm] remains the exact pressure even at 11.4⁰, in fact in any temp', as i understand that, while your explantion, Chet, makes what i asked about: P decreases with temperature. Haruspex explained that in the case of open atmosphere P remains constant.
And if the initial conditions weren't so close to the final, then P would have changed more.

 

[Chestermiller]

Haruspex offered another explanation and i understand that, so i ignore your explanation, Chet. also because it's a little difficult for me and i can't settle the contradiction between Haruspex's explanation and yours.
Haruspex expalined that 10[mm] remains the exact pressure even at 11.4⁰, in fact in any temp', as i understand that, while your explantion, Chet, makes what i asked about: P decreases with temperature. Haruspex explained that in the case of open atmosphere P remains constant.
And if the initial conditions weren't so close to the final, then P would have changed more.

You had originally asked what would happen if the volume were held constant, and that was what I was addressing. In that case, the partial pressure of both the water vapor and the partial pressure of the air would decrease by a factor of (273.2+11.4)/(273.2+20.0). This would lead to a water vapor partial pressure of 9.7 mm Hg at 11.4 C (compared to the original 10.0 mm Hg). Haruspex was addressing the case of open atmosphere where the pressure was held constant, and the volume was allowed to decrease (with the decrease in temperature), so that the mixture became more dense. In the latter case, the decrease in volume would be by the same factor, (273.2+11.4)/(273.2+20.0).

Hope this clarifies the distinction between the two cases and my motivation for taking the volume to be constant.

Chet

 

[Karol]

Yes, exactly right, Chet, and i thank you for addressing me since i felt uncomfortable with 2 explanations!