Thursday, October 31, 2013
Happy Halloween! To our beloved Professor Houze, I’ll give you a free pass on this one, but please don’t schedule a midterm on April 21st.
Our professor told us there are six laws that govern the atmosphere. Today, we just finished talking about Newton’s 2nd Law, which, as opposed to just being F=ma, is F = 1/(ρ)*gradient(P)*(fk X v)*av. I think. I still have to review that… today’s lecture was more confusing than why Rebecca Black ever became famous. Who knows…
Anyway, the first law we went over is the equation of state law, also known as the ideal gas law. It is given by the equation below:
P = Pressure (pascals)
V = Volume (cubic meters)
n = Amount of gas (moles)
T = Temperature of gas (kelvin)
R* = Universal Gas constant (bear with me and pretend there is a star there. I got this equation off Wikipedia and it looked nice)
This can also be written as PV=mRT, where the R has no star and m = mass. The two equations are solved in different ways. This equation is often stated as P=ρRT, where ρ = mass/volume. Sometimes, it is written as Pα=RT, where α = 1/ρ = the specific volume of the gas, i.e., the volume occupied by 1 kg of the gas at pressure P and temperature T. This is the equation we are going to focus on solving.
The gas constant R of a gas is given by R*/M where R* is the universal gas constant (8.314 Joules*Kelvin−1*moles−1) and M is the molecular weight of the gas. Therefore, the gas constant varies with each gas. If you wanted to find out the gas constant of water vapor, you would divide R* by the molecular weight of a mole of water vapor, which would be 1 oxygen + 2 hydrogen = 16 + 2 = 18 grams per mole, and this = .462 Joules*Kelvin−1*gram−1). We want this in SI units, so we multiply by 1000 to turn grams into kilograms, and we get 462, which is our gas constant for water vapor (it’s actually closer to 461.51, I just rounded the atomic masses of oxygen and hydrogen)! Water vapor plays a huge role in these gas laws; such a big role, in fact, that it has been given its own variable. The “vapor pressure” of water, or P in the equation of state above when it is solved for water, is commonly known as e. Therefore, eα_v=R_vT. I used the underscores to represent subscripts… all they denote is α and R for water vapor.
Since dry air is a mixture of gases, calculating R for it requires a two-step process. First, we add up the total mass of each gas the atmosphere and divide this sum by the total number of moles in the atmosphere. We then divide R* by molecular weight that is a conglomerate of all the components of air in the atmosphere to get R, which happens to be 287 J/kg for dry air. We do the same thing for moist air, but we add water vapor to the equation.
The reason why we can do this for a mixture of gases is because of Dalton’s law of partial pressures, which states that the total pressure exerted by a mixture of gases that do not interact chemically is equal to the sum of the partial pressures of the gases. The partial pressure of a gas is the pressure it would exert at the same temperature as the mixture if it alone occupied all of the volume that the mixture occupies.
While we’re at it, let me explain how the equation of state came to be. There were two dudes named Charles and Boyle, and they were heavily involved in the whole industry of making scientific laws that had to do with gases. Boyle discovered that if you keep your m and T constant, P is inversely proportional to V, and Charles discovered that V is directly proportional to T if you keep m and P constant AND that P is directly proportional to T if you keep m and V constant. Some smart fellow/s came around and combined Boyle’s and Charles’ findings into one beautiful law, and it remains one of the most fundamental laws in physics and chemistry today.
So we got Dalton, Boyle, and Charles. It’s like the three musketeers.
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There’s another important constant that the equation of state gives us. The ratio of the gas constant of dry air to water vapor (R_d/R_v) is equal to the ratio of the molecular weight of water vapor to dry air (M_w/M_d). This value is equal to 0.622 and is denoted by ε.
Virtual Temperature (because regular temperature wasn’t good enough)
The equation of state is useful to derive an expression for something my textbook calls “a fictitious temperature” (Wallace & Hobbes, 2006). Sounds useless to me, but turns out it is just the opposite. The virtual temperature of a parcel of air with some moisture in it is the temperature at which a theoretical dry air parcel would have an equivalent pressure and density to the moist parcel. Because moist air is less dense than dry air at the same temperature and pressure, the virtual temperature is always greater than the actual temperature. It usually doesn’t exceed it my more than a few degrees though.
Earlier, we solved for R for dry air and were able to form an equation of state suited to it. We weren’t able to do that for moist air… until now. There’s a long proof that shows that T_v = 1/.622*T, but the main thing to take away from the virtual temperature is that it allows for an equation of state that is always correct no matter how moist the air is. I’ll give the equation its own line for ease of reading.
ρ = Density (kg/m^3)
R_d = Gas constant for dry air = 287 J/kg
T_v = Virtual temperature
Whew! It’s 3:04 a.m. But you know what, this is an effective studying method because I feel like I have to not only study for myself but also put out some quality blogs for my friends… with a serious time deadline. Next!