[Physics] Does dew point change with change of ambient temperature

Question

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.

Answer 1

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.

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