Gases exist at 25 C and 1 atmosphere pressure (atm) under normal conditions. Gases assume the volume and shape of their
containers, are the most compressible of the states of matter, will mix eventually and completely when confined to the same
container and have much lower densities than liquids and solids.
Atmospheric Pressure is the pressure exerted by Earth’s atmosphere.
A barometer is used to measure atmospheric pressure, which is measured in atmospheres. 1 atm is equal to the pressure that
supports a column of mercury 760mm high at 0 C at sea level. 1 torr = 1mmHg; 1atm=760mmHg=760torr; 1 atm = 101,325 Pa.
(i.e. 732 mmHg x 1 atm / 760 mmHg = 0.96 atm
Boyle’s Law states that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of
the gas. The product of the pressure and volume of a gas at constant temperature and amount of gas is a constant. Volume
Decrease Pressure increase and Volume Increase Pressure decrease. Lower temperature Volume/Pressure decrease and
Higher temperature Volume/Pressure Increases. This shows that P x V is always equal to the same constant.
We use Kelvin as temperature when referring to gas. 0 K = -273.15 C. That is absolute zero, the lowest attainable temperature.
To convert Celsius to Kelvin, add 273.15.
Charles’ Law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the
absolute temperature of the gas. Another form of Charles’ Law shows that at a constant amount of gas and volume, the pressure
of a gas is proportional to temperature.
Avogadro’s Law says that at constant pressure and temperature, the volume of a gas is directly proportional to the number of
moles of the gas present.
Gas Law Summary: Boyle: V ∞ 1/P (at constant n and T); Charles’ Law: V ∞ T (at constant n and P); Avogadro’s Law: V ∞ n
(at constant P and T). We can combine these to get: V∞nT/P = RnT/P or PV / nRT where R, the proportionality constant, is the
gas constant. PV = nRT is the ideal gas equation, which describes the relationship among the four variables of P, V, T and n.
The R is a constant that does NOT change except when changing units. An ideal gas is a hypothetical gas whose pressurevolume-
temperature behavior can be completely accounted for by the ideal gas equation. My computer uses ∞ to mean
“proportional to.”
To use PV=nRT, we must evaluate for R at STP (standard temperature and pressure, which is 1 atm at 273.15 K). We know that
R, in this case, = 0.08206 L atm / mol K. To evaluate this, we had to know that there are 22.41 L of gas in 1 mol of that gas. We
can use that number in gas stoichiometry. PV=nRT is used when finding the properties of ONE GAS in ONE STATE. P1V1/n1T1
= P2V2/n2T2 is used to find the properties of two different states of a gas. You just plug in all the values given to solve for the
missing one in PV=nRT or P1V1/T1 = P2V2/T. (i.e. Calculate pressure in atm exerted by 1.82 mol of the gas in a vessel of 5.43 L at
69.5 C (or 342.65 K). PV=nRT or P=nRT/V = 1.82 mol x 0.08206 L atm / mol K x 342.65 K / 5.43 L = 51.17 / 5.43 = 9.42 atm)
(i.e. 0.55 L of air in a balloon at 1 atm is allowed to rise to an area of 0.40 atm. What is the final temperature? P 1V1 = P2T2 since
temperature remains constant. 1 atm x 0.55 L = 0.40 atm x V2. Solve for V2 algebraically and get that the balloon has 1.4 L as its
volume) (i.e. A compound has 33% Si and 67% F. At 35 C, 0.210 L exerted 1.70 atm of pressure. If the mass of 0.210 L of the
compound was 2.38g, calculate the molecular formula. First we find the empirical formula by converting grams to moles of each
substance (Si and F) and use those as subscripts to find SiF3 is the empirical formula. We find n by doing n = PV/RT = 1.70 atm
x 0.210 L / 0.08206 L atm / mol K x 208 K = 0.0141 mol. Because there are 2.38g in 0.041 mol of the compound, the mass in 1
mol, or the molar mass is given by 2.38 g / 0.0141 mol, which = 169 g/mol. Since the molar mass of the empirical SiF3 is
85.09g, the ratio is 169/85.09, which is approximately 2. So, the molecular formula is (SiF3)2, which is Si2F6.
Gas Stoichiometry is found by taking grams of reactant, converting it to moles of the reactant, converting that to moles of the
product, then converting that to amount of product (in g or V) using 22.41 L per mol of that gas somewhere in the calculation
when converting volume to moles. (i.e. 7.40 g NH3 x 1 mol NH3 / 17.03 g NH3 x 22.41 L / 1 mol NH3 = 9.74 L) (i.e. Calculate
volume of N2 generated at 80 C (353.15K) and 823 mmHg (1.08atm) by the decomposition of 60g NaN3 (0.92 mol). First, solve
for V by doing V=nRT/P = 0.92 mol x 0.08206 L atm / mol K x 353.15 K / 1.08 atm = 24.76 L NaN3. From the balanced
equation, we see that we have to multiply this by 3 L N2 / 2 L NaN3 and get 37 L (or so)) (i.e. Pressure at 2.4 x 105
L is 7.9 x 10-3
atm at 312 K. LiOH is introduced and pressure falls to 1.2 x 10-4
atm. Find the change in pressure, which is 7.8 x 10-3 atm. This
correspond with the consumption of CO2, which lost pressure. Plug this into the equation of n =PV/RT to find 7.8 x 10-3 atm x 2.4
x 105 L / 0.08206 L atm / mol K x 312 K = 73 mol. Convert this to grams like we are asked by multiplying by 73.89 / 1mol
Li2CO3 and get 5.4 x 103 g Li2CO3)
Dalton’s Law of Partial Pressures states that the total pressure of a mixture of gases is just the sum of the pressures that each
gas would exert if it were present alone. The total pressure, therefore, is related to the partial pressures. (i.e. If we have 4.46 mol
Ne, 0.74mol Ar and 2.15mol Xe, calculate partial pressures it total pressure is 2 atm. First, find fractions of each gas of the total,
which are 0.607, 0.10, and 0.293 respectively. Multiply these by 2 atm and get 1.21 atm, 0.20 atm and 0.586 atm respectively)
(i.e. If O2 at 24C (297.15K) has 762 mmHg pressure at 128 mL, what is the mass of O2 obtained. Be sure to subtract water vapor
pressure, which is 22.4 mmHg. Find partial pressure of O2, which is 740 mmHg or 0.97 atm and plug in to the equation
PV=nRT=mRT/M, or m = PVM/RT = 0.97atm x .128L x 32g/mol / 0.08206 L atm / mol K x 297.15K = 0.164g)
Mole Fractions are dimensionless quantities that express the ratio of the number of moles of one component to the number of
moles in all components present. The mole fraction of i is found by using Xi = ni / nT where ni is the number of moles of i and nT
is the number of total moles. Now, we can express the partial pressure of A as PA = XAPT and the pressure of B as PB = XBPB,
which is just a fraction multiplied by the total pressure.
Since the joule is the unit of energy, we say 1 J = 1 N = 1 kg m2/s2. KE or kinetic energy is the energy of a moving object.
Maxwell and Boltzmann’s findings make up the kinetic molecular theory of gases, which says that:
o A gas is composed of molecules that are separated from other by great distances in proportion to their sizes.
o Gases are in constant motion in random directions and they collide elasticly—therefore, energy can be transferred in a
collision.
o Gas molecules neither attract nor exert repulsive forces on one another.
o The avg KE ∞ to the T of the gas in K. KE = ½ mu 2 , where the squiggly lines mean “average of.”
We can use the gas laws to say different things:
o Gases can be easily compressed to occupy less volume.
o The collision rate is proportional to the number density of the gas.
o Kinetic Energy is proportional to temperature and temperature increase more collisions.
o Density = n/V because pressure is proportional to density and temperature.
o If molecules don’t attract or repel, then the pressure exerted by one type of molecule is unaffected by the presence of
another gas.
· Energy is the capacity to do work (directed energy change resulting from a process). Radiant Energy comes from the sun.
Thermal Energy is associated with the random motion of atoms and molecules. Chemical Energy is stored within the structural
units of chemical substances. Potential Energy is energy available by virtue of an object’s position. Heat is the transfer of
thermal energy between two bodies at different temperatures. Thermochemistry is the study of heat change in chemical
reactions. The System is the specific part of the universe that is of interest to us. The Surroundings are the rest of the universe
outside the system. An open system can exchange mass and energy in the form of heat. A closed system allows the transfer of
energy, but not mass. An isolated system does not allow the transfer of either mass or energy. An exothermic process gives off
heat. An endothermic process needs heat to be supplied to the system by the surroundings.
· We study changes in the state of a system, which are the values of all relevant macroscopic properties like composition, energy,
temperature, pressure and volume. Energy, pressure, volume and temperature are state functions, which are determined by the
state of the system, regardless of HOW that was achieved.
· ΔV = Vf – Vi (change in volume = final volume – initial volume)
· The First Law of Thermodynamics says that energy can be converted from one form to another, but cannot be destroyed or
created. ΔE = Ef (products)- Ei (reactants). We also say that ΔEsys + ΔEsurr = 0.
· Also, it is important to say that ΔE = q + w, in which ΔE is the change of energy of a system, which is the sum of the heat
exchange q between the system and the surroundings and the w (work done) on (or by) the system. q is positive for an
endothermic process and negative for an exothermic process and w is positive for work done on the system by the surroundings
and negative for work done by the system on the surroundings.
o * Work done by the system on the surroundings: NEGATIVE sign
o * Work done on the system by the surroundings: POSITIVE sign
o * Heat absorbed by the system from the surroundings (ENDOthermic): POSITIVE sign
o * Heat absorbed by the surroundings from the system (EXOthermic process): NEGATIVE sign
· w = -PΔV, where ΔV is Vf – Vi. We use this work formula to find how much work is done by a gas at certain atmospheric
pressures, but we must convert what we get (L • atm) to Joules, for which we use the conversion factor of 1 L • atm = 101.325 J.
(i.e. Gas expands from 2L 6L at constant T. Find expansion against vacuum (0 atm) and against 1.2 atm constant pressure.
w=-PΔV = -(0)(6-2)L = 0.w=-PΔV = -(1.2atm)(6-2)L = -4.8 L atm. Convert to joules using 101.3 J / L atm so -4.8 L atm x 101.3
J / L atm = -4.9 x 102 J.
· The heat transferred to water is q = msΔt, where m is the mass of water in grams, s is the specific heat of water and Δt is the
temperature change. Thus, q = 100g x 4.184 J/g C x 10 C = 4184 J.
· We cannot write Δw = wf – wi or Δq = qf – qi because these are not state functions and they depend on more than just the start and
end points. ΔE is a state function and if changing the path from initial to final states increases q, it will decrease w by the same
amount and vice-versa.
· If a chemical reaction is isochoric, ΔE = q – PΔV, which is equal to qv, where v, as a subscript, reminds us this is isochoric.
· ΔE = q + w = qp – PΔV where qp = ΔE + PΔV where p, as a subscript, denotes isobaric conditions. (i.e. The work done when a
gas is compressed in a cylinder is 462 J. During this, there is heat transfer of 128 J from the gas to the surroundings. Calculate
the energy change, which is ΔE = q + w. q is -128 J because the gas is transferred from the system to the surroundings
(exothermic) and w is +462 J because it is work done by the surroundings on the system. We add these and get a total gain of
+334 J)
· Enthalpy is defined by H = E + PV. Where E is the internal energy of the system and P and V are the pressure and volume of
system, respectively. Change in enthalpy is given by: ΔH = ΔE + Δ(PV) unless pressure is held constant, in which case it is: ΔH
= ΔE + PΔV. If the reaction is isochoric, the heat change is equal to ΔE, but when the reaction is isobaric, heat change is equal to
ΔH.
· Enthalpy of reaction, ΔH, is the difference between the enthalpies of the products and the enthalpies of the reactants. ΔH =
H(products) – H(reactants). In an endothermic process (heat absorbed by the system from the surroundings), ΔH > 0, but in an
exothermic process (heat released by the system to the surroundings), ΔH < 0.
· Thermochemical equations show the enthalpy changes as well as the mass relationships. To write and interpret these equations,
specify the physical states of all reactants and products, if you multiply both sides by n, the ΔH must multiply by the same factor,
and when you reverse an equation, we change the roles of reactants and products. ΔH stays constant, but the sign changes. To
solve a thermochemical equation problem, convert grams of the substance to moles of it using simple stoichiometry and then
convert that to kJ using the kJ/mol ratio given to you in the problem. (i.e. Calculate the heat evolved when 74.6g of SO2 is
converted to SO3. ΔH is given as -99.1 kJ / mol. We convert the 74.6g to moles by dividing by 64.07 g then multiply this by
-99.1 kJ / 1 mol SO2 to get -115 kJ)
· ΔE = ΔH – PΔV. We use this calculation to show how similar ΔH and ΔE are. ΔE = ΔH – Δ(PV) = ΔH – Δ(nRT) = ΔH – RTΔn,
where Δn is defined as “number of moles of product gases – number of moles of reactant gases.” (i.e. Calculate the change in
internal energy when 2 mol CO are converted to 2 mol CO2 at 1 atm and 25 C (298.15K). We are given that ΔH = -556 kJ/mol.
Δn = n(products) – n(reactants) or nf – ni. According to the equation, the Δn = -1, so we use ΔE = ΔH – RTΔn to find that ΔE =
-566 kJ/mol – 8.314 J/mol K x 1kJ / 1000 J x 298 K x -1 = -563.5kJ/mol)
