10.3 The gas law The pressure-volume relationship: Boyle's law Four variables are needed to define the physical condition or state of a gas, temperature, T, pressure, P, volume, V and amount of gas (in mole), n. The equations that express the relationships among T, P, V and n are known as the gas laws. The pressure-volume relationship: Boyle's law Boyle’s law states that the volume (V) of a fixed quantity of gas maintained at constant temperature (T) is inversely proportional to the pressure (P). V  1 P V = 1 P constant  PV = constant (k)

Relationship between pressure and volume when mass and temperature are held constant. Volume Pressure P  V 2 (V1) 2 (P1) 4 3 (V2) 1.33 (P2) 4 4 (V3) 1.0 (P3) 4 P1V1 = k = P2V2 P1V1 = P2V2

The temperature-volume relationship: Charles's law Charles’s law states that the volume (V) of a fixed amount of gas maintained at constant pressure (P) is directly proportional to its absolute temperature (T). V  T Relationship between volume and temperature when mass and pressure are held constant. V = T constant  = constant (k) V T = k V T Volume Temperature V / T 2.5 (V1) 250 (T1) 0.01 3.5 (V2) 350 (T2) 0.01 4.5 (V3) 450 (T3) 0.01 T2 T1 V1 = K = V2 T2 T1 V1 = V2

The quantity-volume-amount relationship: Avogadro’s law Avogadro’s hypothesis: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Avogadro’s law: The volume (V) of a gas maintained at constant temperature (T) and pressure (P) is directly proportional to the number of moles (n) of the gas. V  n V = constant  n V = k n

10.4 The ideal gas equation PV = nRT nT V  P nT V = R P V  1 P Boyle’s law : (constant n, T) V  T Chales’s law : (constant n, P) V  n Avogadro’s law : (constant P, T) P Combine all the three to form a single V  nT If the proportionality constant is called "R", then it becomes: V = R nT P Rearranging to a more familiar form: PV = nRT ideal gas equation

An "ideal gas" is one whose physical behavior is accurately described by the ideal-gas equation The constant R is called the gas constant The value and units of R depend on the units used in determining P, V, n and T Temperature, T, must always be expressed on an kelvin temperature scale (K) The quantity of gas, n, is normally expressed in moles The units chosen for pressure and volume are typically atmospheres (atm) and liters (l), however, other units may be chosen 273 K (0C) and 1 atm are called standard temperature and pressure (STP)

Example: If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2, what is the pressure inside the ball at 24C? 1000 144 cm3 =0.144 L V= n = 0.33 g N2  1 mol N2 28 g N2 =0.0118 mol T = 24 +273 =297 K R = 0.08206 L-atom/mol-K P = ? V P = nRT =  (297 K) ( 0.144 L ) (0.0118 mol)  ( 0.08206 L.atm / K. mol) = 2 atm

= P1V1 T1 P2V2 T2 PV = nRT PV R = nT P1V1 P2V2 = R = n1T1 n2T2 Relating the ideal-gas equation and the gas laws PV = nRT R = PV nT = R = P1V1 n1T1 P2V2 n2T2 Usually amount of gas does not change, n1 = n2 = P1V1 T1 P2V2 T2 Combined gas law

Example 1: A gas initially at 4. 0 L, 1 Example 1: A gas initially at 4.0 L, 1.2 atm, and 66C undergoes a change so that its final volume and temperature become 1.7 L and 42 C. What is its final pressure ? Assume the number of moles remains unchanged = P1V1 T1 P2V2 T2 P1= 1.2 atm V1= 4.0 L T1= 66+273 =339 K V2= 1.7 L T2= 42+273 =315 K P2= ? atm = P1V1T2 T1V2 P2 1.2 atm  4.0 L  315 K 339 K  1.7 L = = 2.62 atm

10.5 Further applications of the ideal-gas equation Gas densities and molar mass Density is defined as mass per unit volume d = m V m = mass in gram M = molar mass number of mole, n = m M Ideal gas equation, PV = nRT n = PV RT m M = PV RT d = PM RT M = dRT P V m = PM RT or

dRT M = P dRT M = P 1.17 g/L  0.08206 L.atm/K.mol  294 K = 0.974 atm Example : Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21C and 740.0 torr. M = dRT P d = 1.17 g/L P = 740.0 torr= 740.0/760 = 0.974 atm T = (21+273)= 294 K R = 0.0821 L. atm / K. mol M = dRT P 1.17 g/L  0.08206 L.atm/K.mol  294 K 0.974 atm = = 29.28 g/mol

( ( Volumes of gases in chemical reactions Mole NaN3 = Gram NaN3= Example : Following the reaction if an air bag of has a volume of 36 L and is to be filled with nitrogen gas at a pressure of 1.15 atom at a temperature of 26.0 C, how many grams of NaN3 must be decomposed? 2NaN3(s) 2Na(s) + 3N2(g) V = 36.0 L P = 1.15 atm T = (26+273)= 299 K R = 0.0821 L. atm / K. mol RT n = PV =  36.0 L (1.15 atm ( 0.08206 L.atm / K. mol)  299 K = 1.69 mol N2 ( (1.69 mol N2) 2 mol NaN3 3 mol N2 Mole NaN3 = = 1.12 mol NaN3 ( 65 g NaN3 1 mol NaN3 (1.12 mol NaN3) Gram NaN3= = 72.8 g NaN3

Gas mixtures and partial pressure Partial pressure of a gas in a mixture of gases is the pressure which the gas would exert if it were the only gas present in the container. Total pressure in the container is proportional to the number of gas particles If each dot represents 1 mole of gas particles, then there are 48 moles of gas particles no particles of gas a = 12 no particles of gas b = 36 total pressure 4 atm.

12 mol  4 atm 48 mol = 1 atm 36 mol  4 atm 48 mol = 3 atm the container holds only 12 moles of gas a particles 12 mol 48 mol  4 atm pressure exerted by gas a = = 1 atm the container holds only 36 moles of gas a particles 36 mol 48 mol  4 atm pressure exerted by gas b = = 3 atm

+ Dalton’s Law of Partial Pressure Ptotal = Pgas a + Pgas b = Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. Ptotal = Pgas a + Pgas b = +

( ( ( ( ( ( P = RT V n RT V P = n RT V P = n RT V P = n RT V RT V n If total pressure of a mixture of gases is Pt and partial pressures of individual gases are P1, P2, P3… Pt=P1+P2+P3…. Let n1, n2, n3…. be the number of moles of each gas Total number of moles nt=n1+n2+n3…. P = RT V n ( ; RT V ( P = 1 n ; RT V ( P = 2 n RT V ( P = 3 n ; Pt=(n1+n2+n3…. ) RT V ( = RT V ( n t

( ( ( ( Mole H2 = Mole N2 = = 2.88 atm Pt= P + P = 2.24 + 0.64 Example : What is the total pressure exerted by a mixture of 2.0 g of H2 and 8.0 g of N2 at 273K in a 10.0 L velsel? ( (2.0 g H2) 1 mol H2 2 g H2 Mole H2 = = 1 mol H2 ( (8.0 g N2) 1 mol N2 28 g N2 Mole N2 = = 0.286 mol N2 RT V ( P = H2 n = (1 mol) ( 0.08206 L.atm / K. mol)  273 K = 2.24 atm 10.0 L RT V ( P = N2 n = (0.286 mol) ( 0.08206 L.atm / K. mol)  273 K 10.0 L = 0.64 atm Pt= P + P = 2.24 + 0.64 N2 H2 = 2.88 atm

Collecting gases over water 2KClO3 (s) 2KCl (s)+ 3O2 (g) PT = P + P O2 H2O

Example 2: Oxygen gas generated by the decomposition of potassium chlorate is collected over water. The volume of oxygen collected at 24C and atmospheric pressure of 762 mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapor at 24 C is 22.4 mmHg. PT = P + P O2 H2O P = PT - P H2O O2 = 762 mmHg – 22.4 mmHg =739.6 mmHg = 739.6 mmHg  1 atm 760 mmHg = 0.973 atm P = 0.973 atm T = 273 + 24 = 297 K V = 128 mL = 0.128 L R = 0.08206 L.atm /K.mol number of mole, n = m M M = molar mass = 32 g / mol m =mass in gram ? m M = RT PV = nRT m = PVM RT = 0.973 atm  0.128 L  32 g / mol 0.0821 L.atm / K.mol  297 K = 0.163 g