2
$\begingroup$

I know dew point is the temperature at which air becomes saturated with vapor that was already in the air. So if I take some unsaturated air in a container and lower it's temperature, eventually at certain temperature the air will become saturated with vapor. To my understanding that certain temperature also known as dew point should only depend on total amount of vapor per cubic meter of air, not the initial temperature of air. But I recently have came across a physics problem where for an initial temperature dew point is given and asked to determine dew point when the temperature was lowered to some value. I thought the dew point should be the same but the given answer suggest otherwise. Can any one explain to me why dew point should change.

$\endgroup$
3
  • $\begingroup$ Did they mean lower the temperature, but keep the same relative humidity maybe? $\endgroup$
    – JMac
    Commented Aug 13, 2018 at 19:51
  • $\begingroup$ Yes, the dew point temperature should only depend on the amount of moisture in a certain volume of air. Now it's possible to change the dew point by extracting moisture from the air, so if the physics problem were describing taking the starting air below its dew point and then warming up the same air, then the final air would then have a different (and lower) dew point because moisture would have condensed out of the air when the starting air went below its dew point. $\endgroup$
    – user93237
    Commented Aug 13, 2018 at 20:19
  • 1
    $\begingroup$ Relative humidity depends on temperature and pressure. So, I'd wonder if they are taking into account the change in volume of the gas. If you start with 1 liter of gas/water and lower the temperature then the volume of the gas decrease too, thus increasing the absolute concentration of water. $\endgroup$
    – MaxW
    Commented Aug 13, 2018 at 20:22

1 Answer 1

1
$\begingroup$

The dew point temperature depends on 2 parameters: pressure and humidity.

It does not depend on temperature.

You can find it in a diagram where these two lines intersect:

  • Constant pressure line (at your level, usually 1atm).
  • Constant saturated vapor pressure line.

Because that means that you decrease temperature keeping constant pressure. At some point, lowering temperature like that will end up matching the exact point $(P,T)$ for which your actual humidity is the saturated one. Then water starts condensing.


A good way to see it is using the mass mixing ratio, which can be easily written in terms of mol mixing ratio:

$$w=\frac{m_{vap}}{m_{dryAir}}= \frac{n_{vap}}{n_{dryAir}}\cdot \frac{mm_{vap}}{mm_{dry}}\simeq \frac{n_{vap}}{n_{dryAir}}\cdot 0.622$$

So you can say that $ w= \dfrac{n_v}{n_d} \cdot 0.622 $, but the moles ratio is the same as the pressure ratio:

$$w= \dfrac{P_v}{P_d} \cdot 0.622 = \dfrac{P_v}{P-P_v} \cdot 0.622 $$

There is a particular case, when the pressure is the saturated vapor pressure:

$$w_s= \dfrac{P_s}{P-P_s} \cdot 0.622 $$

Relative humidity is strongly related to this. In fact,

$$ \frac{w}{w_s} = \frac{P_v / P_d }{P_s/P_d} = \frac{P_v}{P_s}=RH$$

So, once you are given your three parameters: $(P,T, RH)$, you can go on. Humidity must be given anyhow: RH, w, or whatever, but you need to know humidity.

You'll obviously have your dew point when $w$=$w_s$.

So you have a first equation: $w_s=constant$. What constant? The one you're given initially, initial conditions.

So you have that

$$const= \dfrac{P_s}{P-P_s} \cdot 0.622 $$

and the saturated vapor pressure depends on $T$. A good model is

$$P_s = A\cdot e^{-B/T}$$

So now you decrease temperature with constant pressure until water condenses. That's an extra equation: $P=const$. What constant? The one you're given initially.

So you finally have the equation that

$$w_0= \dfrac{P_s}{P_0-P_s} \cdot 0.622 $$

$$w_0= \dfrac{A\cdot e^{-B/T_d}}{P_0-A\cdot e^{-B/T_d}} \cdot 0.622 $$

You can get $T_d$ from here.

The constants are $A\simeq 2.53\cdot 10^9 mbar$; $B\simeq 5240 K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.