Counting Gas Particles Glenn V. Lo Department of Physical Sciences Nicholls State University

Counting gas particles “Gas Particle” – use this term as a generic way of referring to each molecule (bonded atoms) and free (nonbonded) atom in a gas sample. A free atom counts as one particle A molecule counts as one particle Compare: 1 mol CH 4 vs. 1 mol He

Moles of gas particles Measure pressure, volume, and temperature (P, V, and T) of gas Rearrange ideal gas equation, PV = nRT, to solve for moles (n) Valid as P  0; ordinary conditions (about 1% error) R = a constant = L atm mol -1 K -1 Units for P, V, and T must be consistent with units of R

Moles of gas particles at STP STP = standard temperature and pressure; K and 1 atm How many moles of gas particles are in a 22.4 L sample at STP?

Avogadro’s Law Molar volume = Volume of one mole of gas particles = V/n Since PV=nRT…. V/n = RT/P Consequence: at constant T and P: V/n = constant V is directly proportional to n.  If n , V   If n , V  The molar volume is constant: V/n = constant  V 1 /n 1 = V 2 /n 2  Equal volumes of gas at the same T, P have equal number of particles  At same T, P all gases have the same molar volume (V/n) V/n = 22.4 L/mol

Example Which of the following gas samples contains the most number of gas particles at the same temperature and pressure? Sample A L CH 4 Sample B L H 2 Sample C L He Sample D L He

Example Which of the following gas samples contains the most number of atoms at the same temperature and pressure? Sample A L CH 4 Sample B L H 2 Sample C L He Sample D L He

Partial Pressure For any gas (pure or mixture) behaving ideally: PV=nRT Therefore: P = Define: Partial Pressure = pressure that a component of a gas mixture would exert if it were alone P A = partial pressure of component “A” = n A (RT/V)

Example What are the partial pressures of O 2 and N 2 in a 25.0 L sample containing 1.00 mol O 2 and 4.00 mol N 2 at K? What is the total pressure?

Dalton’s Law of Partial Pressures For Gas Mixtures. PV = n total RT P = n total (RT/V) = (n A + n B +... ) (RT/V) = n A (RT/V) + n B (RT/V) +... = P A + P B +... Dalton’s Law: the total pressure of a gas mixture is equal to the sum of the partial pressures of the components: P = P A + P B +... P = total pressure P A = partial pressure of component “A”

Example In any enclosed vessel containing liquid water, the partial pressure of water vapor is called the vapor pressure of water; it is a constant value (at a given temperature) that we can look up. At 298K, it is 23.8 Torr. A sample of H 2 is collected over water at 298K. If the pressure of the sample is Torr, what is the partial pressure of H 2 ? (Answer: Torr)

Partial Pressure and Mole Fraction For any mixture, the MOLE FRACTION of component “A” is defined as: x A = n A / n total For a gas mixture: P A = n A (RT/V) and P = n total (RT/V) Therefore: P A /P = n A /n total = x A Or P A = x A P, P B = x B P, etc.

Example About 21% of molecules in a typical air sample are O 2 molecules. We say that the mole percent of O 2 is 21%, or that the mole fraction of O 2 is What is the partial pressure of O 2 in an air sample if the total pressure of the sample is 1.20 atm? (Answer atm)

Gas Stoichiometry Change in moles of gaseous reactant or product can be determined from P, V, T data. If only one reactant or product involved is gaseous: n=PV/RT If more than one reactant or product are gaseous, use Dalton’s Law.

Example Suppose 22.4 L of CO 2 were collected at STP from the decomposition of NaHCO 3 : 2 NaHCO 3 (s)  Na 2 O(s) + H 2 O(g) + 2 CO 2 (g) How much Na 2 O(s) was also produced?

Example A strip of Mg is reacted with HCl(aq): Mg(s) + 2 HCl(aq)  H 2 (g) + MgCl 2 (aq) and the H 2 gas was collected over water as illustrated below. As such, the gas is a mixture of H 2 and water vapor, H 2 O(g). Suppose the volume of gas collected at 298K is 2.50 L, the water level inside the tube is 10.0 cm above the water level outside, the prevailing barometric pressure is 755 Torr, how much HCl was consumed in the reaction? The vapor pressure of water at 298K is 23.8 Torr.

Determining Molar Mass of a Gas Determining molar mass of anything: need mass (m) and moles (n): M m = m/n If gas is ideal: n=PV/RT Measurements to be made for a gas: m, P, V, and T

Example Example: A sample of unknown gas, with a mass of g contains mol of particles. What is the molar mass of the unknown gas?

Example A sample of unknown gas with a mass of g occupies a volume of mL at 373K and 1.00 atm. Which of these is a possible identity of the compound? A. CH 4, B. CO, C. Cl 2

Dumas Method or Vapor Density Method From: M m = m/n, n = PV/RT Show that: M m = d(RT/P), or d (V/n), where d = m/V (the “vapor density”) Note: d= M m /V m, where V m =V/n, the molar volume

Example The density of an gas at STP is 3.17 g/L. What is the molar mass of the gas? Answer: 71.0 g mol -1

Example Ignore trace components and assume that air is only made up of N 2 and O 2. What are the mole fractions of N 2 and O 2 in air given that its density at STP is 1.29 g/L? Answer g mol -1 Hint: let x=mole fraction of O 2 In terms of x, in one mole of mixture: moles O 2 = mass O 2 = moles N 2 = mass N 2 = Equation that says total mass is 28.9 g: Solve equation for x.