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3 years ago

As air temperature approaches the dew point, what happens to the relative humidity?

2 answers:

5 0

Relative Humidity is a percentage based on how much moisture is in the air versus how much moisture the air can hold. The dewpoint is the temperature that the atmosphere must reach for a parcel of air to become fully saturated. That being said, if the temperature outside is 30 degrees (F or C doesn't matter), and the dewpoint goes from 10 to 20, then the relative humidity has increased, meaning there is more moisture present in the atmosphere.
The easiest way to look at pressure is by looking at the number of air molecules that are in a closed container. Gas molecules expand as they warm, so warming up a container of air means that these molecule will attempt to expand, thus increasing the pressure. Take for instance a soda bottle filled with air, if you heat a soda bottle, then the gases inside will attempt to expand (raising the pressure) and eventually the bottle will explode, but if you stick the same soda bottle in dry ice, the soda bottle will contract, due to decreased pressure. Source(s): Military Meteorologist
Bachelors in Natural Science

KIM [24]3 years ago

5 0

As air temperature falls toward the dew point, OR the dew point
rises toward the current air temperature, the relative humidity RISES
toward 100%, my hair dies, and my shirt becomes soaked with sweat.

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3 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

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1 year ago

A vessel contains 1 mol of O2 and 2mol of He.what is the value of'Cp/Cv ' of the mixture?

BlackZzzverrR [31]

Oxygen is diatomic, so its degree of freedom, (f1)= 5,
also its number of moles, n1= 1

Helium is monoatomic, so its degree of freedom (f2)= 3
and its number of moles given is, n2=2

Now using formula of effective degree of freedom of mixture, (f), we have:

f= (f1n1+f2n2)/(n1+n2)
= (5*1 + 3*2)/ (1+3)
=11/3
Also, from first law of thermodynamics;
U= n Cv. T = nRT(f2)
or, Cv = R. (f/2) (n & T cancel)

We know f=11/6,
substituting the value in above relation, we have:

Cv= R. 11/3*2
= R. 11/6

Also, Cp-Cv = R
or, Cp- R.(11/6)= R
or, Cp= R(11/6 )+1
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Therefore, Cp/Cv = 17/11

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3 years ago

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P and V are inversely proportional when temperature is constanst.
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