^oC = (^oF - 32^o) ^. 5/9

^oF = ^oC^. 9/5 + 32^o

4. Density of a gas (or any liquid and solid material) is a measure of the mass of that material present in a specified volume of space:

r = density = mass / volume = M / V

Units: kg/m³ if M is in kg and V is in m³

g/cm³ if M is in g and V is in cm³

Air density is the mass of air per unit volume of air.

Number concentration is the number of gas molecules in the sample of the volume:

where N is the total number of gas molecules in the volume V.

Units: number/m³ = molecules/m³ if V is in m³

number/cm³ if V is in cm³

Mass concentration is the mass of gas molecules in the sample of the volume:

where M is the total mass of gas molecules in the volume V.

Units: kg/m³ ; g/cm³ ; mg/m³ ;

NOTE: mass concentration of a gas = gas density

• Relationship between q and n:

if each molecule has mass m, then total mass is M = N ^. m and

q = M/V = N m /V = n m

thus q = n m

NOTE: the mass of a single molecule is very small, and the number of molecules is usually enormous! For instance, the number concentration of air is n = 2.55x10¹⁹ molec./cm³ at p=1013 mbar and T = 288 K.

Mole is introduced to operate with large collections of atoms or molecules.

One mole of any substance consist of 6.022x10²³ units of that substance.

This number N_A = 6.022x10²³ is called Avogadro’s number.

Example:

1 mole of eggs is 6.022x10²³ eggs ( just as a dozen eggs is 12 eggs);

1 mole of oxygen atoms is 6.022x10²³ oxygen atoms;

The International System of Units (SI) defines the mole (abbreviation mol) as the number equal to the number of atoms in exactly 12 grams of pure ¹²C (carbon twelve).

Also, ¹²C is as the standard in the system of atomic masses. In this system ¹²C is assigned a mass of exactly 12 atomic mass units (amu), and the masses of all other atoms are given relative to the standard (more details are given in Lecture 5).

Since 6.022x10²³ atoms of ¹²C (each with a mass of 12 amu) have a mass of 12 grams, then

or

1 amu = 1.66x10^-24 g

The mass of one mole, m_g of the element is equal to its atomic mass in grams.

Example:

the mass of 1 mole of H (hydrogen) atoms is 1.008 g because hydrogen atomic mass is 1.008 amu;

the mass of 1 mole of O (oxygen) atoms is 16 g because oxygen atomic mass is 16 amu.

The mass of one mole, m_g (or molar mass, or molecular weight) of a chemical compound which is a collection of atoms is obtained by summing the masses of atoms.

NOTE: the terms molar mass and molecular weight mean exactly the same thing: the mass in grams of one mole of a compound.

Problem: What is molecular weight of methane (CH₄)?

Solution.

Mass of 1 mole of C = 12.011 g

Mass of 4 mole of H = 4 x 1.008 g

Thus, mass of 1 mol of CH₄ = 16.043 g

NOTE: that m_g relates to m as : m_g = m N_A

The number of moles can be expressed as :

m = number of moles = mass of gas / molecular weight = M / m_g

m = number of gas molecules/Avagodro’s number = N / N_A

Molar concentration is the number of moles a gas in the sample of the volume:

c = molar concentration = number of moles / volume = m /V

where m is the number of moles of a gas in the volume V.

Units: mol/m³ if V is in m³

mol/cm³ if V is in cm³

• Relationship between c and n:

since c = m/V and m = N/N_A , we obtain

c = m/V = (N/N_A)/V = (N/V)/ N_A = n/ N_A

thus c = n / N_A

• Relationship between c and q:

since n = q / m, we obtain

c = n/ N_A= (q/m) / N_A= q / (m N_A) = q / m_g

thus c = q / m_g

Problem: Although air is a mixture of gases, it is often that the term "air molecule" is used. What is an average mass of the air molecule (take molecular weight of air m_a = 28.97 g/mol)?

Solution.

One mole of air contains the Avogadro’s number of air molecules, therefore

average mass of air molecule = m_a / N_A = 28.97 (g mol^-1) / 6.022x10²³ mol^-1)

thus average mass of air molecule = 4.8096 x10^-23 g

Mixing ratio is defined as the ratio of the amount (or mass) of the substance in a given volume to the total amount (or mass) of all constituents in that volume.

Since air is a mixture of gases, the volume mixing ratio of i-th gas in the air is

r_i = N_i / N_a

where N_i is the number of molecules of i-th gas, and N_a is the number of molecules of air (total number of molecules of all individual gases in the atmosphere: N_a = S N_i ).

NOTE: r_i = N_i / N_a = n_i / n_a = m_i / m_a = c_i / c_a

NOTE: Table 3.1 presents volume mixing ratios of individual gases in the atmosphere expressed in percent.

For instance, the volume mixing ratio of carbon dioxide (CO₂) is (see Table 3.1):

r_CO2 = 0.000365, which is very small number. Therefore, it is multiplied by 10⁶ and expressed as

r_CO2 = 365 ppmv (parts per million by volume).

In turn, mass mixing ratio of i-th gas is

r_i,m = M_i / M_a

where M_i is the mass of molecules of i-th gas, and M_a is the mass of molecules of air (total mass of molecules of all individual gases in the atmosphere: M_a = S M_i ).

NOTE: r_i,m = M_i / M_a = r_i / r_a

Mass mixing ratio and volume mixing ratio are related as:

r_i,m = r_i / r_a = m_i n_i/ m_a n_a = m_i/m_a n_i/n_a = m_i/m_a r_i

thus, mass mixing ratio of a certain component is its volume mixing ratio multiplied by its molecular weight and divided by the molecular weight of air (m_a=28.97)

Mass and volume mixing ratios are distinguished by an added v(for volume)and m(for mass), such as

ppmv is parts per million by volume ; ppmm is parts per million by mass

NOTE: often in literature v or m are omitted. In such case ppm units typically refer to volume mixing ratio unless noted in text otherwise.

one part per million 1 ppm (1x10^-6)

one part per billion 1 ppb (1x10^-9)

one part per trillion 1 ppt (1x10^-12)

Problem: A typical volume mixing ratio for ozone in urban air is r_O3 = 0.1 ppmv. What is ozone mass mixing ratio? (take molecular weight of air m_a = 28.97 g/mol)?

Solution.

Molecular weight of ozone is m_O3 = 48.0 g mol^-1. Therefore we have

r_O3,m = m_O3 ^. r_O3 / m_a = 48 (g mol^-1) ^. 0.1 ppmv / 28.97 (g mol^-1) = 0.17 ppmm

thus

mass mixing ratio for ozone is r_O3,m = 0.17 ppmm

Atmosphere is a gas mixture, therefore it obeys the gas laws. The behavior and properties of ideal gas mixtures is explained by kinetic molecular theory (part of physics).

• Atmospheric gases differ only slightly from the ideal gases, therefore ideal gas laws work in the atmosphere.

An ideal gas consists of particles (molecules or atoms) that have the following properties:

1. The particles are so small compared with the distances between them that the volume of the of the individual particles can be assumed to be negligible (zero).
2. The particles are constantly moving in a random way. They collide with each other and with the walls of the container producing the pressure exerted by the gas.
3. The particles are assumed to exert no forces on each other (they are assumed neither to attract nor to repel each other).
4. Collisions of the particles are perfectly elastic; no kinetic energy is lost during a collision. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the temperature of the gas.

 

• A particular state of a gas is described by P (pressure); T (temperature); V (volume); and n (gas concentration) or mixing ratio. These variables are called state variables and are related by gas laws.

Ideal gas law (also known as the equation of state): says that the pressure exerted by a gas is proportional to its temperature and inversely proportional to its volume:

P V = m R T

where R is the universal gas constant. If pressure P is in atmospheres (atm), volume V in liters (L) and temperature T in degrees Kelvin (K), thus R has value

R = 0.08206 L atm K^-1 mol^-1

NOTE: T must be in degree Kelvin when used in the gas laws.

Ideal gas law is a combined form of three gas laws: Boyle’s law, Charles’s law, and Avogadro’s law. These gas laws are based on experimental measurements of the properties of gases, and they are can be explained with the kinetic molecular theory of gases.

Boyle’s law: V ~ 1/P (at constant T and m)

NOTE: Robert Boyle (1627-1691), Irish chemist

Charles’s law: V ~ T (at constant P and m)

NOTE: Jacques Charles (1746-1823), French physicist

Avogadro’s law: V ~ m (at constant P and T)

NOTE: Amadeo Avogadro (1778-1850), Italian chemist

Avogadro postulated in 1811: that the equal volumes of gases at the same T and P contain the same number of molecules.

Standard temperature and pressure (abbreviated STP) conditions:

T = 0^oC = 273.2 K and P = 1 amt

Problem: What is the volume of 1 mole of ideal gas at STP?

Solution.

From the ideal gas law, the volume is given by

V = m R T / P = 1 (mol) 0.08206(L atm K^-1 mol^-1) 273.2 (K) / 1(amt) = 22.42 (L)

This volume of 22.42 liters is called the molar volume of an ideal gas at STP.

Dalton’s law of Partial Pressures states that for a fixed temperature and volume (container), the total pressure of a mixture of gases is the sum of the partial pressures that each of the individual component gases would have if placed alone in the same volume (container).

P = P₁ + P₂ + P₃ +.... = S P_i

where P_i is the partial pressure of i-th gas. If each gas behaves ideally, the P_i can be calculated from the ideal gas law:

P_i = m_i R T / V

Thus, the total pressure of the mixture is

P = S P_i = S (m_i R T / V) = (RT/V) S m_i = m_tot (RT/V)

where m_tot = S m_i is the total number of moles of gases in the mixture.

Mole fraction is the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture:

x_i = m_i / m_tot

From the ideal gas law

m_i = P_i (V / RT)

Therefore,

x_i = m_i / m_tot = P_i (V / RT) / {P (V / RT)} = P_i / P

rearranging this expression as

P_i = x_i P

that is, the partial pressure of a particular component of a gaseous mixture is equal to the mole fraction of that component times the total pressure.

NOTE: mole fraction is equivalent to the volume fraction.

Problem: The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N₂ in air when the atmospheric pressure is 760 torr.

Solution.

The partial pressure of N₂ can be calculated as follows:

P_N2 = x _N2 P = 0.7808 ^. 760 (torr) = 593 (torr)

derives relationships between state variables similar to that observed experimentally; and explains why gases behave in the observed fashion.

A main assumption in the kinetic molecular theory of gases is that the distribution of velocities of ideal gas molecules is described by the Maxwell-Boltzmann velocity distribution law:

f(v) = 4p {m/ 2pk_BT} ^3/2 v² exp( -mv²/2k_BT)

where

v is velocity (m/s); m is mass of a gas molecule (kg), T is temperature (K), and

k_B is Boltzmann’s constant = 1.38066x10^-23 (J/K).

NOTE: J denotes Joule which is unit of energy in SI system: J = kg m²/s²

Analysis of the expression for f(v) yields the following equation for the most probable velocity v (the velocity possessed by the greatest number of gas particles):

v = (2k_BT / m)^1/2

where n is concentration of gas molecules.

Comparing this expression to one followed from the ideal gas law, we have

n k_BT = P = m R T/ V

then, substituting n = m N_A/V, we obtain

k_B m N_A /V = m R T/ V

therefore,

k_B = R / N_A

NOTE: that gas constant and Boltzmann’s constant are related by Avogadro’s number: R= N_A k_B. In fact, it is useful to think of k_B as the gas law constant per molecule.

NOTE: R and k_B can be expressed in alternative units. For instance,

k_B = 1.38066x10^-23 (J/K) = 1.38066x10^-19 (cm³ mb K^-1) = ...

R = 8.3145 (J mol^-1 K^-1) = 8.314x10⁴ (cm³ mb mol^-1 K^-1) = ...

NOTE: the Maxwell-Boltzmann velocity distribution function gives the following equation for the mean velocity of a gas molecule:

v = (8k_BT /p m)^1/2 = (8RT /p m_g)^1/2

where m is the mass of a gas molecule, m_g is the molecular weight (or molar mass); and p is a mathematical constant p=3.141592...The mean velocity of a gas molecule if often used in various atmospheric chemistry applications.

NOTE: important to remember: the equation of state can be written in several forms:

using molar concentration of a gas, c= m/v: P = c T R

using number concentration of a gas, n = c N_A: P = n T R/N_A or P = n T k_B

using mass concentration of a gas, q = c m_g: P = q T R /m_g

NOTE: be careful to write the quantities with the proper units and to select the appropriate value for R when you use the ideal gas law.

Total atmospheric pressure can be divided into the partial pressure exerted by air (P_d) and the partial pressure by water vapor (P_n): thus

P_a = P_d + P_n

n_a = n_d + n_n

where n_d is the number concentration of dry air, n_n is the number concentration of water vapor, and n_a is the number concentration of air.

P_d = m_d R T/ V = n_d k_B T = r_d R_d T

where r_d is the mass density of dry air, and R_d is the gas constant for dry air defined as R_d = R / m_d

P_n = m_n R T/ V = n_n k_B T= r_n R_n T

where r_n is the mass density of water vapor, and R_n is the gas constant for water vapor defined as R_n = R / m_H2O

NOTE: total air = moist air = dry air + water vapor

Specific humidity is the ratio of mass of water vapor to mass of moist air.

q_s = r_n/ (r_n+ r_d) = r_n/ r_a

Relative humidity is defined as the ratio of the partial pressure of water vapor to its saturation vapor pressure at a given temperature:

RH = 100 P_n / P_n*

where the factor 100 is used because RH is usually expressed in percent, and P_n* is the saturation water vapor pressure, which is vapor pressure in the equilibrium state of the water-water vapor system at a given temperature.

NOTE: RH=90% for very humid air (summer in Florida), RH=20 % for very dry air (summer in desert)

• Condensation of water vapor releases heat, which is called the latent heat of condensation.
• Evaporation of water vapor adsorbs heat, which is called the latent heat of evaporation.
• Latent heat is one of the principal energy sources in the atmosphere.